You Can Precipitate Silver Chloride From a Saturated Solution of the Compound by Adding

7.2 Precipitation and Dissolution

By the terminate of this section, y'all will be able to:

Write chemical equations and equilibrium expressions representing solubility equilibria
Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations

Solubility equilibria are established when the dissolution and atmospheric precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An agreement of the factors affecting chemical compound solubility is, therefore, essential to the constructive management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.

The Solubility Product

Call up from the affiliate on solutions that the solubility of a substance tin can vary from essentially cipher (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For case, a saturated solution of silvery chloride is i in which the equilibrium shown below has been established:

$\text{AgCl}\left(s\right)\underset{\text{precipitation}}{\overset{\text{dissolution}}{\rightleftharpoons}}{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}^-}^{\text{}}\left(aq\right)$

In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous Ag⁺ and Cl^– ions at the same charge per unit that these aqueous ions combine and precipitate to form solid AgCl (Figure 7.ii.1). Because silvery chloride is a sparingly soluble table salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.

Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker. — Effigy 7.2.one – Argent chloride is a sparingly soluble ionic solid. When information technology is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag+ and Cl– ions in equilibrium with undissolved silver chloride.

The equilibrium abiding for solubility equilibria such equally this one is called the solubility product constant, K _sp , in this case:

$\text{AgCl}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}^-}^{\text{}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{Ag}}^{\text{+}}\left(aq\right)\right]\left[{\text{Cl}^-}^{\text{}}\left(aq\right)\right]$

Recall that only gases and solutes are represented in equilibrium constant expressions, then the Thou _sp does not include a term for the undissolved AgCl. A list of solubility production constants for several sparingly soluble compounds is provided in Appendix J.

Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:

(a) AgI, silver iodide, a solid with antiseptic properties

(b) CaCO₃, calcium carbonate, the active ingredient in many over-the-counter chewable antacids

(d) Mg(NH₄)PO₄, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium

(eastward) Ca_five(PO₄)₃OH, the mineral apatite, a source of phosphate for fertilizers

Solution

$\begin{array}{ccccc}\left(\text{a}\right)\phantom{\rule{0.2em}{0ex}}\text{AgI}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{I}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & {\left[\text{Ag}}^{\text{+}}\right]\left[{\text{I}}^{\text{-}}\right]\hfill \\ \left(\text{b}\right)\phantom{\rule{0.2em}{0ex}}{\text{CaCO}}_{3}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{CO}}_{3}{}^{\text{2-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Ca}}^{\text{2+}}{\right]\left[\text{CO}}_{3}{}^{\text{2-}}\right]\hfill \\ \left(\text{c}\right)\phantom{\rule{0.2em}{0ex}}{\text{Mg}\left(\text{OH}\right)}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}\hfill \\ \left(\text{d}\right)\phantom{\rule{0.2em}{0ex}}\text{Mg}\left({\text{NH}}_{\text{4}}\right){\text{PO}}_{\text{4}}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{PO}}_{4}{}^{\text{3-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Mg}}^{\text{2+}}{\right]\left[\text{NH}}_{4}{}^{\text{+}}\right]{\left[\text{PO}}_{4}{}^{\text{3-}}\right]\hfill \\ \left(\text{e}\right)\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_5}\left({\text{PO}}_{4}\right)_3\text{OH}\left(s\right)$\rightleftharpoons${\text{5Ca}}^{\text{2+}}\left(aq\right)+{\text{3PO}}_{4}{}^{\text{3-}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & {{\left[\text{Ca}}^{\text{2+}}\right]}^{5}\left[\text{P}{\text{O}}_{4}{}^{\text{3-}}{\right]}^{3}\left[{\text{OH}}^{\text{-}}\right]\hfill \end{array}$

Check Your Learning

Write the dissolution equation and the solubility product for each of the following slightly soluble compounds:

(a) BaSO_four

(b) Ag_twoSo₄

(d) Pb(OH)Cl

Respond

$\begin{array}{ccccc}\left(\text{a}\right)\phantom{\rule{0.2em}{0ex}}{\text{BaSO}}_{4}\left(s\right)$\rightleftharpoons${\text{Ba}}^{\text{2+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Ba}}^{\text{2+}}{\right]\left[\text{SO}}_{4}{{}^{2}}^{\text{-}}\right];\hfill \\ \left(\text{b}\right)\phantom{\rule{0.2em}{0ex}}{\text{Ag}}_{2}{\text{SO}}_{4}\left(s\right)$\rightleftharpoons${\text{2Ag}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & {\left[{\text{Ag}}^{\text{+}}\right]}^{2}\left[{\text{SO}}_{4}{}^{\text{2-}}\right];\hfill \\ \left(\text{c}\right)\phantom{\rule{0.2em}{0ex}}{\text{Al}\left(\text{OH}\right)}_{3}\left(s\right)$\rightleftharpoons${\text{Al}}^{\text{3+}}\left(aq\right)+{\text{3OH}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Al}}^{\text{3+}}\right]{\left[\text{OH}}^{\text{-}}{\right]}^{\text{3}};\hfill \\ \left(\text{d}\right)\phantom{\rule{0.2em}{0ex}}\text{Pb(OH)Cl(s}\right)$\rightleftharpoons${\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)+{\text{Cl}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Pb}}^{\text{2+}}\right]{\left[\text{OH}}^{\text{-}}\right]\left[{\text{Cl}}^{\text{-}}\right]\hfill \end{array}$

One thousand _sp and Solubility

The K _sp of a slightly soluble ionic chemical compound may be merely related to its measured solubility provided the dissolution process involves just dissociation and solvation, for instance:

${\text{M}}_{p}{\text{X}}_{q}\left(s\right)$\rightleftharpoons$p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n-}}\left(aq\right)$

For cases such equally these, 1 may derive K _sp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound'due south molar solubility, measured as moles of dissolved solute per liter of saturated solution.

Fluorite, CaF₂, is a slightly soluble solid that dissolves according to the equation:

${\text{CaF}}_{2}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2F}}^{\text{-}}\left(aq\right)$

The concentration of Ca²⁺ in a saturated solution of CaF_two is 2.15 $\times$ 10^–iv M. What is the solubility product of fluorite?

Solution

According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF₂ solution is equal to twice its calcium ion molarity:

$\left[{\text{F}}^{-}\right]=\left(2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{F}}^{-}\phantom{\rule{0.2em}{0ex}}\text{/}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{Ca}}^{2+}\right)=\left(2\right)\left(2.15\phantom{\rule{0.2em}{0ex}} $\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4})\phantom{\rule{0.2em}{0ex}}M\right)=4.30\phantom{\rule{0.2em}{0ex}} $\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M$

Substituting the ion concentrations into the K _sp expression gives:

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{2}=\text{(2.15}\phantom{\rule{0.2em}{0ex}} \times\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)){\left(4.30\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)})^{2}=\text{3.98}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-11}$

Cheque Your Learning

In a saturated solution of Mg(OH)₂, the concentration of Mg²⁺ is 1.31 $\times$ ten^–4 M. What is the solubility production for Mg(OH)₂?

${\text{Mg(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)$

Respond

8.99 $\times$ 10^–12

The K _sp of copper(I) bromide, CuBr, is 6.3 $\times$ 10^–9. Calculate the tooth solubility of copper bromide.

Solution

The dissolution equation and solubility product expression are:

$\text{CuBr}\left(s\right)$\rightleftharpoons${\text{Cu}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{-}}\left(aq\right)$

${K}_{\text{sp}}=\left[{\text{Cu}}^{\text{+}}\right]\left[{\text{Br}}^{\text{-}}\right]$

Following the ICE arroyo to this calculation yields the tabular array:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of,

Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields:

${K}_{\text{sp}}={\left[\text{Cu}}^{\text{+}}\right]\left[{\text{Br}}^{\text{-}}\right]$

$6.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}=\left(x\right)\left(x\right)={x}^{2}$

$x=\sqrt{\left(6.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\right)}=\text{7.9}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M$

Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the tooth solubility of CuBr is 7.9 $\times$ 10^–5 M.

Check Your Learning

The Thou _sp of AgI is 1.five $\times$ 10^–16. Calculate the molar solubility of silvery iodide.

Answer

ane.2 $\times$ x^–eight M

The Thousand _sp of calcium hydroxide, Ca(OH)₂, is ane.3 $\times$ 10^–vi. Calculate the tooth solubility of calcium hydroxide.

Solution

The dissolution equation and solubility product expression are:

${\text{Ca(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)$

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

The Water ice table for this system is:

Substituting terms for the equilibrium concentrations into the solubility product expression and solving for x gives:

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

$1.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6\text{}}=\left(x\right)\left(2x{\right)}^{2}=\left(x\right)\left(4{x}^{2}\right)=4{x}^{3}$

$x=\sqrt[3]{\phantom{\rule{0.2em}{0ex}}\frac{1.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}}{4}\phantom{\rule{0.2em}{0ex}}}=\text{7.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$

Every bit divers in the Ice table, ten is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:ane relation between moles of calcium ion in solution and moles of compound dissolved, and and then, the molar solubility of Ca(OH)_ii is vi.nine $\times$ x^–3 M.

Check Your Learning

The One thousand _sp of PbI₂ is one.4 $\times$ 10^–eight. Calculate the tooth solubility of pb(II) iodide.

Answer

1.v $\times$ x^–3 M

Many of the pigments used past artists in oil-based paints (Figure 7.ii.two) are sparingly soluble in h2o. For example, the solubility of the artist's pigment chrome yellow, PbCrO₄, is iv.6 $\times$ x^–six g/Fifty. Determine the solubility production for PbCrO₄.

A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface. — Figure 7.2.2 – Oil paints contain pigments that are very slightly soluble in water. In add-on to chrome yellow (PbCrO₄), examples include Prussian blue (Fe₇(CN)₁₈), the ruby-red-orange color vermilion (HgS), and greenish color veridian (Cr₂O₃). (credit: Sonny Abesamis)

Solution

Earlier calculating the solubility product, the provided solubility must exist converted to molarity:

$\left[{\text{PbCrO}}_{4}]=\frac{4.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}{\text{g PbCrO}}_{4}}{1\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.4em}{0ex}}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.4em}{0ex}}=\frac{1{\phantom{\rule{0.2em}{0ex}}\text{mol PbCrO}}_{4}}{323.2{\phantom{\rule{0.2em}{0ex}}\text{g PbCrO}}_{4}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}\frac{1.4\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-8}{\phantom{\rule{0.2em}{0ex}}\text{mol PbCrO}}_{4}}{1\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}1.4\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}M$

The dissolution equation for this chemical compound is:

${\text{PbCrO}}_{4}\left(s\right)$\rightleftharpoons${\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{CrO}}_{4}{}^{\text{2-}}\left(aq\right)$

The dissolution stoichiometry shows a one:1 relation between the molar amounts of compound and its two ions, and and so both [Lead^two+] and $\left[{\text{CrO}}_{4}{}^{\text{2-}}\right]$ are equal to the molar solubility of PbCrO₄:

$\left[{\text{Pb}}^{\text{2+}}\right]=\left[{\text{CrO}}_{4}{}^{\text{2-}}\right]=\text{1.4}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}M$

K _sp = [Pb²⁺] $\left[{\text{CrO}}_{4}{^{2-}}^{\text{}}\right]$ = (ane.4 $\times$ 10^–8)(one.iv $\times$ x^–eight) = 2.0 $\times$ 10^–xvi

Check Your Learning

The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is existence isolated from ores, is 3.12 grams per liter at 20 °C. What is its solubility product?

Answer

two.08 $\times$ x^–4

Calomel, Hg_twoCl_two, is a compound equanimous of the diatomic ion of mercury(I), ${\text{Hg}}_{2}{}^{\text{2+}},$ and chloride ions, Cl^–. Although near mercury compounds are now known to exist poisonous, eighteenth-century physicians used calomel equally a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested past its very small M _sp:

${\text{Hg}}_{2}{\text{Cl}}_{2}\left(s\right)$\rightleftharpoons${\text{Hg}}_{2}{}^{\text{2+}}\left(aq\right)+{\text{2Cl}}^{\text{-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.1}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}$

Calculate the tooth solubility of Hg₂Cl₂.

Solution

The dissolution stoichiometry shows a i:1 relation between the corporeality of compound dissolved and the amount of mercury(I) ions, and and then the molar solubility of Hg₂Cl_ii is equal to the concentration of ${\text{Hg}}_{2}{}^{\text{2+}}$ ions.

Post-obit the Ice arroyo results in:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of,

Substituting the equilibrium concentration terms into the solubility product expression and solving for x gives:

${K}_{\text{sp}}=\left[{\text{Hg}}_{\text{2}}{}^{\text{2+}}\right]{\left[{\text{Cl}}^{\text{-}}\right]}^{2}$

$1.1\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}=\left(x\right)\left(2x{\right)}^{2}$

$4{x}^{3}=\text{1.1}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}$

$x=\sqrt[3]{\left(\frac{1.1\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}}{4}\right)}=\text{6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M$

$\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]=\text{6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M=\text{6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M$

$\left[{\text{Cl}}^{\text{-}}\right]=2x=\text{2(6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\right))=1.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M$

The dissolution stoichiometry shows the molar solubility of Hg₂Cl₂ is equal to $\left[{\text{Hg}}_{\text{2}}{}^{\text{2+}}\right],$ or 6.5 $\times$ x^–7 Thou.

Bank check Your Learning

Make up one's mind the tooth solubility of MgF_two from its solubility product: K _sp = six.iv $\times$ ten^–9.

Respond

1.2 $\times$ 10^–3 K

Using Barium Sulfate for Medical Imaging

Various types of medical imaging techniques are used to help diagnoses of illnesses in a noninvasive manner. Ane such technique utilizes the ingestion of a barium compound earlier taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested past the patient. Since the K _sp of barium sulfate is two.3 $\times$ ten^–eight, very little of information technology dissolves as it coats the lining of the patient'due south abdominal tract. Barium-coated areas of the digestive tract and then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 7.ii.3).

This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white. — Figure seven.2.3 – A interruption of barium sulfate coats the abdominal tract, permitting greater visual item than a traditional X-ray. (credit modification of work by "glitzy queen00"/Wikimedia Commons)

Medical imaging using barium sulfate tin can be used to diagnose acid reflux illness, Crohn's disease, and ulcers in addition to other conditions.

Visit this website for more information on how barium is used in medical diagnoses and which weather condition information technology is used to diagnose.

Predicting Precipitation

The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:

$\begin{array}{ccc}{\text{CaCO}}_{3}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{CO}}_{3}^{\text{2-}}^{}\left(aq\right)& & \hfill {K}_{sp}=\left[{\text{Ca}}^{2+}\right]\left[{\text{CO}}_{3}{}^{2-}\right]=8.7\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\end{array}$

Information technology is important to realize that this equilibrium is established in any aqueous solution containing Ca²⁺ and CO_three ^ii– ions, non just in a solution formed by saturating water with calcium carbonate. Consider, for case, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction caliber, Q_sp , that exceeds the solubility product, Thousand _sp, and then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, so precipitation will occur, lowering those concentrations until equilibrium is established (Q_sp = K _sp). The comparing of Q_sp to K _sp to predict atmospheric precipitation is an example of the general approach to predicting the direction of a reaction showtime introduced in the chapter on equilibrium. For the specific case of solubility equilibria:

Q_sp < 1000 _sp: the reaction gain in the forrard management (solution is not saturated; no precipitation observed)

Q_sp > Thou _sp: the reaction proceeds in the reverse direction (solution is supersaturated; atmospheric precipitation volition occur)

This predictive strategy and related calculations are demonstrated in the next few example exercises.

The starting time step in the preparation of magnesium metal is the atmospheric precipitation of Mg(OH)₂ from sea h2o by the addition of lime, Ca(OH)_two, a readily available cheap source of OH^– ion:

${\text{Mg(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{8.9}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-12}$

The concentration of Mg²⁺(aq) in sea h2o is 0.0537 G. Volition Mg(OH)_two precipitate when enough Ca(OH)_two is added to give a [OH^–] of 0.0010 M?

Solution

Adding of the reaction quotient nether these atmospheric condition is shown here:

$Q=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}=\text{(0.0537)(}{\text{0.0010)}}^{2}=\text{5.4}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}$

Because Q is greater than K _sp (Q = 5.iv $\times$ 10^–8 is larger than Thou _sp = 8.nine $\times$ 10^–12), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations accept been sufficiently lowered, so that Q_sp = K _sp.

Check Your Learning

Predict whether CaHPO₄ volition precipitate from a solution with [Caⁱⁱ⁺] = 0.0001 M and $\left[{\text{HPO}}_{4}{}^{\text{2-}}\right]$ = 0.001 Yard.

Answer

No precipitation of CaHPO_four; Q = ane $\times$ x^–7, which is less than G _sp (7 $\times$ 10^–7)

Does silver chloride precipitate when equal volumes of a 2.0 $\times$ x^–4 Chiliad solution of AgNO₃ and a 2.0 $\times$ 10^–4 M solution of NaCl are mixed?

Solution

The equation for the equilibrium between solid argent chloride, silver ion, and chloride ion is:

$\text{AgCl}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{-}}\left(aq\right)$

The solubility product is 1.6 $\times$ 10^–10 (encounter Appendix J).

AgCl will precipitate if the reaction caliber calculated from the concentrations in the mixture of AgNO₃ and NaCl is greater than K _sp. Considering the volume doubles when equal volumes of AgNO₃ and NaCl solutions are mixed, each concentration is reduced to half its initial value:

$\frac{1}{2}\left(2.0\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}M=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M)$

The reaction caliber, Q, is greater than G _sp for AgCl, and so a supersaturated solution is formed:

$Q=\left[{\text{Ag}}^{\text{+}}\right]{\left[\text{Cl}}^{\text{-}}\right]=\text{(1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right))\left(1.0\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4})\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}>{K}_{\text{sp}}$

AgCl will precipitate from the mixture until the dissolution equilibrium is established, with Q equal to K _sp.

Check Your Learning

Will KClO_iv precipitate when 20 mL of a 0.050-1000 solution of Grand⁺ is added to 80 mL of a 0.50 M solution of ${\text{ClO}}_{4}{}^{\text{-}}?$ (Hint: Use the dilution equation to summate the concentrations of potassium and perchlorate ions in the mixture.)

Solution

No, Q = 4.0 $\times$ 10^–3, which is less than Thou _sp = 1.05 $\times$ 10^–2

Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, ${\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}},$ for this purpose (Figure 7.two.four). At sufficiently high concentrations, the calcium and oxalate ions class solid, CaC₂O_four·H₂O (calcium oxalate monohydrate). The concentration of Ca²⁺ in a sample of blood serum is 2.2 $\times$ 10^–3 Grand. What concentration of ${\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}$ ion must be established earlier CaC_twoO₄·H₂O begins to precipitate?

A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial. — Figure vii.two.four – Anticoagulants can exist added to claret that volition combine with the Ca²⁺ ions in blood serum and prevent the blood from clotting. (credit: modification of piece of work by Neeta Lind)

Solution

The equilibrium expression is:

${\text{CaC}}_{2}{\text{O}}_{4}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\left(aq\right)$

For this reaction:

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=1.96\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-8}$

(encounter Appendix J)

Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:

$Q={K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]\phantom{\rule{0.2em}{0ex}}=\text{1.96}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}$

$\left(2.2\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-3}\right))\left[{\text{C}}_{\text{2}}{\text{O}}_{4}{}^{\text{2-}}\right]=\text{1.96}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}$

$\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{1.96\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}}{2.2\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-3}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\text{8.9}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M$

A concentration of $\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]$ = 8.9 $\times$ x^–6 K is necessary to initiate the atmospheric precipitation of CaC₂O₄ nether these conditions.

Check Your Learning

If a solution contains 0.0020 mol of ${\text{CrO}}_{4}{}^{\text{2-}}$ per liter, what concentration of Ag⁺ ion must be reached by adding solid AgNO_three before Ag₂CrO_iv begins to precipitate? Neglect any increase in book upon adding the solid silvery nitrate.

Answer

half-dozen.7 $\times$ 10^–5 M

Wearable done in water that has a manganese [Mn^two+(aq)] concentration exceeding 0.1 mg/L (one.8 $\times$ 10^–6 1000) may be stained by the manganese upon oxidation, just the corporeality of Mn²⁺ in the water can be decreased by adding a base to precipitate Mn(OH)_two. What pH is required to keep [Mn²⁺] equal to ane.8 $\times$ ten^–six M?

Solution

The dissolution of Mn(OH)₂ is described past the equation:

${\text{Mn(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Mn}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{2}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-13}$

At equilibrium:

${K}_{\text{sp}}=\left[{\text{Mn}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

$\left(1.8\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right))\left[{\text{OH}{}^{\text{-}}\right]^{2}=\text{2}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-13}$

and then

$\left[{\text{OH}}^{\text{-}}\right]=\text{3.3}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M$

Calculate the pH from the pOH:

$\begin{array}{c}\text{pOH}=\text{−log}\left[{\text{OH}}^{\text{-}}\right]=\text{−log}\left(3.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}10^{-4}\right))=\text{3.48}\\ \text{pH}=\text{14.00}-\text{pOH}=\text{14.00}-\text{3.80}=\text{10.5}\end{array}$

(final outcome rounded to i meaning digit, limited by the certainty of the K _sp)

Check Your Learning

The kickoff stride in the training of magnesium metal is the precipitation of Mg(OH)₂ from sea water by the addition of Ca(OH)₂. The concentration of Mg²⁺(aq) in sea water is v.37 $\times$ 10^–2 M. Calculate the pH at which [Mgⁱⁱ⁺] is decreased to 1.0 $\times$ ten^–five M

Reply

10.97

In solutions containing 2 or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective atmospheric precipitation may be used to remove private ions from solution. By increasing the counter ion concentration in a controlled style, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the to the lowest degree soluble compound will precipitate starting time (at the everyman concentration of counter ion), with the other ions afterwards precipitating as their compound's solubilities are reached. As an analogy of this technique, the next example exercise describes separation of a two halide ions via precipitation of ane as a silver salt.

The Office of Precipitation in Wastewater Treatment

Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Effigy 7.2.5). Specifically, selective atmospheric precipitation is used to remove contaminants from wastewater earlier it is released back into natural bodies of h2o. For example, phosphate ions ${\text{(PO}}_{4}{}^{\text{3-}}\right))$ are oftentimes nowadays in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to abound, which impacts the amount of oxygen available for marine life also as making h2o unsuitable for human consumption.

A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground. — Figure 7.ii.five – Wastewater handling facilities, such as this one, remove contaminants from wastewater before the h2o is released dorsum into the natural environment. (credit: "eutrophication&hypoxia"/Wikimedia Commons)

One common way to remove phosphates from h2o is past the addition of calcium hydroxide, or lime, Ca(OH)₂. As the water is fabricated more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca₅(PO4)₃OH, which then precipitates out of the solution:

$5{\text{Ca}}^{\text{2+}}+{\text{3PO}}_{4}{}^{\text{3-}}+{\text{OH}}^{\text{-}}$\rightleftharpoons${\text{Ca}}_{5}{{\text{(PO}}_{4}\right))}_{\text{3}}\text{}\text{OH}\left(s\right)$

Because the corporeality of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the improver of CO₂ in a recarbonation process. Other chemicals tin also be used for the removal of phosphates past precipitation, including iron(Three) chloride and aluminum sulfate.

View this site for more information on how phosphorus is removed from wastewater.

A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO₃ is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?

Solution

The two equilibria involved are:

$\begin{array}{cccccc}\hfill \text{AgCl}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{-}}\left(aq\right)\hfill & & & \hfill {K}_{\text{sp}}& =\hfill & \text{1.6}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-10}\hfill \\ \hfill \text{AgBr}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{-}}\left(aq\right)\hfill & & & \hfill {K}_{\text{sp}}& =\hfill & \text{5.0}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-13}\hfill \end{array}$

If the solution contained about equal concentrations of Cl^– and Br^–, then the silver salt with the smaller K _sp (AgBr) would precipitate first. The concentrations are not equal, withal, so the [Ag⁺] at which AgCl begins to precipitate and the [Ag⁺] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag⁺] precipitates first.

AgBr precipitates when Q equals One thousand _sp for AgBr:

$Q=\left[{\text{Ag}}^{\text{+}}\right]{\left[\text{Br}}^{\text{-}}\right]=\left[{\text{Ag}}^{+}\right]\text{(0.00010)}=\text{5.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-13}$

$\left[{\text{Ag}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{\text{5.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-13}}{0.00010}\phantom{\rule{0.2em}{0ex}}=\text{5.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M$

AgI begins to precipitate when [Ag⁺] is 5.0 $\times$ 10^–nine Chiliad.

For AgCl: AgCl precipitates when Q equals K _sp for AgCl (ane.six $\times$ 10^–10). When [Cl^–] = 0.10 M:

${Q}_{\text{sp}}=\left[{\text{Ag}}^{\text{+}}\right]{\left[\text{Cl}}^{\text{-}}\right]=\left[{\text{Ag}}^{\text{+}}\right]\text{(0.10)}=\text{1.6}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

$\left[{\text{Ag}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{1.6\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-10}}{0.10}=\text{1.6}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M$

AgCl begins to precipitate when [Ag⁺] is i.6 $\times$ x^–nine M.

AgCl begins to precipitate at a lower [Ag⁺] than AgBr, so AgCl begins to precipitate first. Annotation the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, then silver chloride precipitated first despite having a K _sp greater than that of silver bromide.

Bank check Your Learning

If silver nitrate solution is added to a solution which is 0.050 M in both Cl^– and Br^– ions, at what [Ag⁺] would precipitation begin, and what would be the formula of the precipitate?

Answer

[Ag⁺] = 1.0 $\times$ 10^–xi Thou; AgBr precipitates first

Common Ion Effect

Compared with pure water, the solubility of an ionic chemical compound is less in aqueous solutions containing a common ion (ane too produced by dissolution of the ionic chemical compound). This is an example of a phenomenon known as the mutual ion event, which is a outcome of the law of mass action that may be explained using Le ChÂtelier's principle. Consider the dissolution of silver iodide:

$\text{AgI}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{I}}^{\text{-}}\left(aq\right)$

This solubility equilibrium may be shifted left by the add-on of either silver(I) or iodide ions, resulting in the atmospheric precipitation of AgI and lowered concentrations of dissolved Ag⁺ and I^–. In solutions that already contain either of these ions, less AgI may exist dissolved than in solutions without these ions.

This effect may also be explained in terms of mass action as represented in the solubility product expression:

${K}_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{I}}^{-}\right]$

The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in 1 ion's concentration must be balanced by a proportional decrease in the other.

View this simulation to explore diverse aspects of the common ion consequence.

View this simulation to meet how the common ion effect works with dissimilar concentrations of salts.

What is the effect on the amount of solid Mg(OH)_two and the concentrations of Mg²⁺ and OH^– when each of the post-obit are added to a saturated solution of Mg(OH)₂?

(a) MgCl_two

(b) KOH

(d) Mg(OH)_ii

Solution

The solubility equilibrium is:

$\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+2{\text{OH}}^{\text{-}}\left(aq\right)$

(a) The reaction shifts to the left to relieve the stress produced past the additional Mg²⁺ ion, in accord with Le Châtelier'south principle. In quantitative terms, the added Mg²⁺ causes the reaction caliber to exist larger than the solubility production (Q > K _sp), and Mg(OH)_two forms until the reaction quotient once more equals Grand _sp. At the new equilibrium, [OH^–] is less and [Mg²⁺] is greater than in the solution of Mg(OH)₂ in pure water. More solid Mg(OH)₂ is present.

(b) The reaction shifts to the left to relieve the stress of the additional OH^– ion. Mg(OH)_two forms until the reaction quotient over again equals K _sp. At the new equilibrium, [OH^–] is greater and [Mgⁱⁱ⁺] is less than in the solution of Mg(OH)₂ in pure water. More solid Mg(OH)_ii is nowadays.

(c) The concentration of OH^– is reduced every bit the OH^– reacts with the acid. The reaction shifts to the right direction.

(a) Adding a mutual ion, Mg²⁺, will increment the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the corporeality of undissolved magnesium hydroxide.

(b) Adding a common ion, OH^–, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.

(c) The added compound does non contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.

(d) Adding more solid magnesium hydroxide volition increment the corporeality of undissolved chemical compound in the mixture. The solution is already saturated, though, then the concentrations of dissolved magnesium and hydroxide ions will remain the same:

$Q=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium abiding.

Check Your Learning

What is the effect on the corporeality of solid NiCO_iii and the concentrations of Ni²⁺ and ${\text{CO}}_{3}{}^{\text{2-}}$ when each of the post-obit are added to a saturated solution of NiCO₃?

(a) Ni(NO₃)_ii

(b) KClO₄

(d) K_iiCO₃

Answer

(a) mass of NiCO_three(s) increases, [Niⁱⁱ⁺] increases, $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ decreases

(b) no appreciable upshot

(d) mass of NiCO₃(due south) increases, [Ni²⁺] decreases, $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ increases

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr₂). The Chiliad _sp of CdS is one.0 $\times$ ten^–28.

Solution

This adding can exist performed using the ICE approach:

$\text{CdS}\left(s\right)$\rightleftharpoons${\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2-}}\left(aq\right)$

${K}_{\text{sp}}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{\text{2-}}\right]=1.0\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-28}$

$\left(0.010+x\right)\left(x\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-28}$

Because K _sp is very small, assume x << 0.010 and solve the simplified equation for x:

$\left(0.010\right)\left(x\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-28}$

$x=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-26}\phantom{\rule{0.2em}{0ex}}M$

The molar solubility of CdS in this solution is 1.0 $\times$ 10^–26 M.

Bank check Your Learning

Calculate the molar solubility of aluminum hydroxide, Al(OH)₃, in a 0.015-K solution of aluminum nitrate, Al(NO₃)₃. The K _sp of Al(OH)₃ is 2 $\times$ 10^–32.

Reply

4 $\times$ x^–11 G

Key Concepts and Summary

The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Chiliad _sp, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid K_pTen_q and its ions M^m+ and X^n–:

${\text{M}}_{p}{\text{X}}_{q}\left(s\right)$\rightleftharpoons$p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n-}}\left(aq\right)$

the solubility product expression is:

${K}_{\text{sp}}={{\text{[M}}^{\text{m+}}]\right]}^{p}{{\text{[X}}^{\text{n-}}]\right]}^{q}$

The solubility production of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its K _sp, provided the simply significant reaction that occurs when the solid dissolves is the germination of its ions. A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.

Key Equations

${\text{M}}_{p}{\text{X}}_{q}\left(s\right)$\rightleftharpoons$p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}={\left[\text{M}}^{\text{m+}}{\right]}^{p}{{\left[\text{X}}^{\text{n-}}\right]}^{q}$

Some useful links related to waster water treatment and its process and application in chemic engineering:

https://www.chemengonline.com/water-treatment-technologies/
https://www-sciencedirect-com.ezproxy.library.ubc.ca/science/article/pii/S004565351631061X?via%3Dihub
http://Chemic Water and Wastewater Handling Six
https://world wide web.intechopen.com/books/wastewater-handling-engineering/biological-and-chemical-wastewater-treatment-processes
https://softeningwater.com/4-best-methods-of-water-softening/
https://iaspub.epa.gov/tdb/pages/treatment/treatmentOverview.do?treatmentProcessId=-2062922688
https://www.waterfiltermag.com/
https://extensionpublications.unl.edu/assets/pdf/g1491.pdf
https://www2.cambridgema.gov/CityOfCambridge_Content/documents/Drinking%20WaterMy%20edition.pdf (Case Report)

News Manufactures:

https://globalnews.ca/news/6134683/inside-the-investigation-tainted-h2o-canada/
http://www.watertech.ca/solutions/water-handling.html
http://cwn-rce.ca/
https://world wide web.ccme.ca/en/resource/water/from_source_to_tap_the_multi_barrier_approach.html

Water purification methods to get tap water:

https://bestpurification.com/purify-tap-water-home-learn-various-purification-methods/
https://www.bryceviewlodge.com/make-h2o-rubber-drinking-uncomplicated-water-purification-techniques/
https://world wide web.safewater.org/fact-sheets-1/2017/i/23/oil-spills
https://www.epa.gov/basis-water-and-drinking-h2o/emergency-disinfection-drinking-water
https://world wide web.healthlinkbc.ca/health-topics/tf6354

(1) Complete the changes in concentrations for each of the post-obit reactions:

(1a) $\begin{array}{ccc}\text{AgI}\left(s\right)\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}& {\text{Ag}}^{\text{+}}\left(aq\right)& +\phantom{\rule{0.2em}{0ex}}{\text{I}}^{\text{-}}\left(aq\right)\\ \\ \underset{_}{}\end{array}$

(1b) $\begin{array}{ccc}{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Ca}}^{\text{2+}}\left(aq\right)+& {\text{CO}}_{3}{^{2-}}^{\text{}}\left(aq\right)\\ \\ & & \underset{_}{}\end{array}$

(1c) $\begin{array}{ccc}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Mg}}^{\text{2+}}\left(aq\right)+& 2{\text{OH}^-}^{\text{}}\left(aq\right)\\ \\ & & \underset{_}{}\end{array}$

(1d) $\begin{array}{ccc}{\text{Mg}}_{3}\left({\text{PO}}_{4}{\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 3{\text{Mg}}^{\text{2+}}\left(aq\right)+& 2{\text{PO}}_{4}{^{3-}}^{\text{}}\left(aq\right)\\ \\ & & \underset{_}{}\end{array}$

(1e) $\begin{array}{cccc}{\text{Ca}}_{5}\left({\text{PO}}_{4}{\right)}_{3}\text{OH}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 5{\text{Ca}}^{\text{2+}}\left(aq\right)+& 3{\text{PO}}_{4}{^{3-}}^{\text{}}\left(aq\right)+& {\text{OH}^-}^{\text{}}\left(aq\right)\\ \\ & & & \underset{_}{}\end{array}$

Solutions

(2) Complete the changes in concentrations for each of the following reactions:

(2a) $\begin{array}{ccc}{\text{BaSO}}_{4}\left(s\right)\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}& {\text{Ba}}^{\text{2+}}\left(aq\right)+& {\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\\ \\ & x& \underset{_}{x}\end{array}$

(2b) $\begin{array}{ccc}{\text{Ag}}_{2}{\text{SO}}_{4}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 2{\text{Ag}}^{\text{+}}\left(aq\right)+& {\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\\ \\ & 2x & \underset{_}{x}\end{array}$

(2c) $\begin{array}{ccc}\text{Al}{\left(\text{OH}\right)}_{3}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Al}}^{\text{3+}}\left(aq\right)+& 3{\text{OH}}^{\text{-}}\left(aq\right)\\ \\ & x& \underset{_}{3x}\end{array}$

(2d) $\begin{array}{cccc}\text{Pb}\left(\text{OH}\right)\text{Cl}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Pb}}^{\text{2+}}\left(aq\right)+& {\text{OH}}^{\text{-}}\left(aq\right)+& {\text{Cl}}^{\text{-}}\left(aq\right)\\ \\ & x\phantom{\rule{0.2em}{0ex}}& x& \underset{_}{x}\end{array}$

(2e) $\begin{array}{ccc}{\text{Ca}}_{3}\left({\text{AsO}}_{4}{\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 3{\text{Ca}}^{\text{2+}}\left(aq\right)+& 2{\text{AsO}}_{4}{}^{\text{3-}}\left(aq\right)\\ \\ & 3x&\underset{_}{2x}\end{array}$

(iii) How do the concentrations of Ag⁺ and ${\text{CrO}}_{4}{}^{\text{2-}}$ in a saturated solution higher up 1.0 g of solid Ag₂CrO₄ alter when 100 1000 of solid Ag₂CrO_four is added to the arrangement? Explain.

Solution

In that location is no change. A solid has an activity of 1 whether in that location is a little or a lot.

(iv) How do the concentrations of Pbⁱⁱ⁺ and Southward^two– change when K₂S is added to a saturated solution of PbS?

(five) What additional data do we need to reply the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?

Solution

The solubility of silverish bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.

(vi) Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility production because of hydrolysis of the anion present: CoSO₃, CuI, PbCO₃, PbCl₂, Tl_iiS, KClO₄?

(7) Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion nowadays: AgCl, BaSO₄, CaF_ii, Hg_iiI₂, MnCO₃, ZnS, PbS?

Solution

CaF_two, MnCO₃, and ZnS

(8) Write the ionic equation for dissolution and the solubility production (K _sp) expression for each of the following slightly soluble ionic compounds:

(8a) PbCl₂

(8b) Ag_twoS

(8c) Sr₃(PO₄)_ii

(8d) SrSO₄

(9) Write the ionic equation for the dissolution and the K _sp expression for each of the following slightly soluble ionic compounds:

(9a) LaF₃

(9b) CaCO_three

(9c) Ag₂So₄

(9d) Pb(OH)_two

Solution

(10) The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Considering these compounds are simply slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

(10a) BaSiF_{half dozen}, 0.026 g/100 mL (contains ${\text{SiF}}_{6}{}^{\text{2-}}$ ions)

(10b) Ce(IO_iii)_four, ane.5 $\times$ ten^–2 1000/100 mL

(10c) Gd₂(And then₄)₃, 3.98 k/100 mL

(10d) (NH_four)₂PtBr_half-dozen, 0.59 g/100 mL (contains ${\text{PtBr}}_{6}{}^{\text{2-}}$ ions)

(11) The Handbook of Chemistry and Physics gives solubilities of the post-obit compounds in grams per 100 mL of water. Considering these compounds are just slightly soluble, presume that the volume does non change on dissolution and calculate the solubility product for each.

(11a) BaSeO₄, 0.0118 k/100 mL

(11b) Ba(BrO₃)_two·H_twoO, 0.30 one thousand/100 mL

(11c) NH_fourMgAsO₄·6H₂O, 0.038 1000/100 mL

(11d) La_two(MoO₄)₃, 0.00179 one thousand/100 mL

Solution

(a) i.77 $\times$ ten^–7; (b) one.6 $\times$ x^–6; (c) 2.2 $\times$ 10^–ix; (d) seven.91 $\times$ ten^–22

(12) Use solubility products and predict which of the following salts is the nearly soluble, in terms of moles per liter, in pure h2o: CaF₂, Hg₂Cl₂, PbI₂, or Sn(OH)₂.

(13) Assuming that no equilibria other than dissolution are involved, summate the molar solubility of each of the following from its solubility product:

(13a) KHC₄H_ivO₆

(13b) PbI₂

(13c) Ag_iv[Atomic number 26(CN)_half-dozen], a common salt containing the ${\text{Fe(CN)}}_{4}{}^{\text{-}}$ ion

(13d) Hg₂I_two

Solution

(a) 2 $\times$ x^–2 M; (b) 1.v $\times$ 10^–3 M; (c) 2.27 $\times$ x^–9 K; (d) ii.2 $\times$ 10^–ten M

(14) Bold that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

(14a) Ag_twoSO_four

(14b) PbBr_two

(14c) AgI

(14d) CaC_iiO₄·H_iiO

(15) Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a mutual ion. Show that changes in the initial concentrations of the common ions can be neglected.

(15a) AgCl(southward) in 0.025 M NaCl

(15b) CaF₂(s) in 0.00133 M KF

(15c) Ag₂Then_iv(s) in 0.500 50 of a solution containing 19.50 grand of Grand₂SO_iv

(15d) Zn(OH)₂(due south) in a solution buffered at a pH of 11.45

Solution

(xvi) Assuming that no equilibria other than dissolution are involved, summate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a mutual ion. Show that changes in the initial concentrations of the common ions can exist neglected.

(16a) TlCl(s) in one.250 Thousand HCl

(16b) PbI₂(due south) in 0.0355 K CaI₂

(16c) Ag₂CrO_iv(s) in 0.225 L of a solution containing 0.856 g of K₂CrO₄

(16d) Cd(OH)₂(s) in a solution buffered at a pH of 10.995

(17) Bold that no equilibria other than dissolution are involved, summate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that information technology is not appropriate to fail the changes in the initial concentrations of the common ions.

(17a) TlCl(s) in 0.025 M TlNO₃

(17b) BaF₂(south) in 0.0313 Yard KF

(17c) MgC₂O₄ in ii.250 Fifty of a solution containing 8.156 m of Mg(NO₃)_two

(17d) Ca(OH)₂(s) in an unbuffered solution initially with a pH of 12.700

Solution

(a) [Cl^–] = 7.6 $\times$ x⁻³ M

Check: $\frac{7.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-3}}{0.025}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=30%$

This value is as well large to drop ten. Therefore solve by using the quadratic equation:

[Ti⁺] = iii.1 $\times$ 10^–two Thou
[Cl^–] = 6.i $\times$ 10^–3

(b) [Ba²⁺] = 7.7 $\times$ x^–4 M

Check: $\frac{7.7\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-4}}{0.0313}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=2.4%$

Therefore, the condition is satisfied.

[Baⁱⁱ⁺] = vii.7 $\times$ x^–4 Chiliad
[F^–] = 0.0321 G;

$\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=2.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}$

Cheque: $\frac{2.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}}{0.02444}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=0.12%$

The status is satisfied; the above value is less than 5%.

$\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=2.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}\phantom{\rule{0.2em}{0ex}}M$

[Mg²⁺] = 0.0244 Thou

(d) [OH^–] = 0.0501 M

[Ca^two+] = iii.15 $\times$ 10^–3

Check: $\frac{3.15\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}}{0.050}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=6.28%$

This value is greater than 5%, so a more exact method, such equally successive approximations, must be used.

[Ca²⁺] = 2.eight $\times$ 10^–3 M
[OH^–] = 0.053 $\times$ 10^–two M

(18) Explain why the changes in concentrations of the common ions in (Figure) can be neglected.

(xix) Explain why the changes in concentrations of the common ions in (Figure) cannot exist neglected.

Solution

The changes in concentration are greater than v% and thus exceed the maximum value for disregarding the change.

(20) Calculate the solubility of aluminum hydroxide, Al(OH)₃, in a solution buffered at pH 11.00.

(21) Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is virtually soluble in moles per liter and which is virtually soluble in grams per liter.

Solution

CaSO_four∙2H_twoO is the virtually soluble Ca salt in mol/Fifty, and it is also the nearly soluble Ca salt in g/L.

(22) Virtually barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure vii.2.3). This use of BaSO_iv is possible because of its low solubility. Calculate the molar solubility of BaSO₄ and the mass of barium nowadays in i.00 50 of water saturated with BaSO₄.

(23) Public Health Service standards for drinking water set a maximum of 250 mg/L (ii.lx $\times$ 10^–3 K) of ${\text{SO}}_{4}{}^{\text{2-}}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO₄ ("gyp" water) as a outcome or passing through soil containing gypsum, CaSO₄·2H₂O, meet these standards? What is the concentration of ${\text{SO}}_{4}{}^{\text{2-}}$ in such h2o?

Solution

4.viii $\times$ 10^–3 M = $\left[{\text{SO}}_{4}{}^{\text{2-}}\right]$ = [Ca²⁺]; Since this concentration is higher than 2.60 $\times$ x^–3 M, "gyp" water does not meet the standards.

(24) Perform the following calculations:

(24a) Summate [Ag⁺] in a saturated aqueous solution of AgBr.

(24b) What will [Ag⁺] exist when enough KBr has been added to make [Br^–] = 0.050 M?

(24c) What volition [Br^–] exist when enough AgNO_three has been added to brand [Ag⁺] = 0.020 M?

(25) The solubility product of CaSO_four·2H₂O is ii.four $\times$ ten^–5. What mass of this salt will dissolve in i.0 L of 0.010 1000 ${\text{SO}4}_{}{}^{\text{2-}}?$

Solution

Mass (CaSO_four·2H₂O) = 0.72 g/50

(26) Bold that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the post-obit (see Appendix J for solubility products).

(26a) TlCl

(26b) BaF₂

(26c) Ag₂CrO₄

(26d) CaC₂O_four·H₂O

(26e) the mineral anglesite, PbSO_four

(27) Assuming that no equilibria other than dissolution are involved, summate the concentrations of ions in a saturated solution of each of the following (come across Appendix J for solubility products):

(27a) AgI

(27b) Ag₂SO₄

(27c) Mn(OH)₂

(27d) Sr(OH)_ii·8H_iiO

(27e) the mineral brucite, Mg(OH)₂

Solution

(a) [Ag⁺] = [I^–] = 1.3 $\times$ ten^–5 1000

(b) [Ag⁺] = 2.88 $\times$ 10^–2 Thou, $\left[{\text{SO}}_{4}{}^{\text{2-}}\right]$ = 1.44 $\times$ ten^–2 M

(d) [Srⁱⁱ⁺] = 4.3 $\times$ ten^–ii M, [OH^–] = 8.6 $\times$ 10^–2 M

(east) [Mgⁱⁱ⁺] = one.3 $\times$ 10^–4 M, [OH^–] = 2.half-dozen $\times$ 10^–4 Thou

(28) The post-obit concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K _sp for each of the slightly soluble solids indicated:

(28a) AgBr: [Ag⁺] = v.7 $\times$ 10^–7 M, [Br^–] = 5.7 $\times$ x^–seven One thousand

(28b) CaCO₃: [Ca^two+] = 5.iii $\times$ 10^–3 M, $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ = 9.0 $\times$ 10^–vii M

(28c) PbF₂: [Atomic number 82²⁺] = two.one $\times$ ten^–three M, [F^–] = four.2 $\times$ 10^–iii Grand

(28d) Ag_iiCrO₄: [Ag⁺] = 5.3 $\times$ 10^–5 M, three.2 $\times$ 10^–3 M

(28e) InF₃: [In^three+] = two.3 $\times$ ten^–3 Grand, [F^–] = 7.0 $\times$ 10^–3 M

(29) The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Yard _sp for each of the slightly soluble solids indicated:

(29a) TlCl: [Tl⁺] = 1.21 $\times$ 10^–2 Yard, [Cl^–] = 1.2 $\times$ x^–2 G

(29b) Ce(IO₃)₄: [Ce⁴⁺] = i.8 $\times$ x^–4 Thou, $\left[{\text{IO}}_{3}{}^{\text{-}}\right]$ = two.6 $\times$ x^–13 M

(29c) Gd_ii(And so₄)₃: [Gd^three+] = 0.132 Yard, $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$ = 0.198 M

(29d) Ag₂SO_iv: [Ag⁺] = 2.40 $\times$ 10^–2 Thousand, $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$ = two.05 $\times$ x^–two M

(29e) BaSO₄: [Ba²⁺] = 0.500 G, $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$ = 4.six $\times$ 10⁻⁸ One thousand

Solution

(a) ane.7 $\times$ ten^–four

(b) 8.2 $\times$ 10^–55

(d) 1.eighteen $\times$ ten^–five

(e) 1.08 $\times$ 10^–10

(30) Which of the following compounds precipitates from a solution that has the concentrations indicated? (Run into Appendix J for K _sp values.)

(30a) KClO₄: [K⁺] = 0.01 M, $\left[{\text{ClO}}_{4}{}^{\text{-}}\right]$ = 0.01 Thousand

(30b) K_twoPtCl_vi: [Chiliad⁺] = 0.01 G, $\left[{\text{PtCl}}_{\text{6}}{}^{\text{2-}}\right]$ = 0.01 M

(30c) PbI_two: [Lead²⁺] = 0.003 M, [I^–] = 1.3 $\times$ x^–3 Thou

(30d) Ag_twoS: [Ag⁺] = 1 $\times$ 10^–10 Thousand, [S^2–] = ane $\times$ 10^–13 M

(31) Which of the post-obit compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for M _sp values.)

(31a) CaCO₃: [Ca²⁺] = 0.003 M, $\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$ = 0.003 One thousand

(31b) Co(OH)₂: [Co²⁺] = 0.01 M, [OH^–] = 1 $\times$ ten^–7 One thousand

(31c) CaHPO₄: [Ca²⁺] = 0.01 M, $\left[{\text{HPO}}_{\text{4}}{}^{\text{2-}}\right]$ = ii $\times$ ten^–6 M

(31d) Pb₃(PO_four)₂: [Pb²⁺] = 0.01 M, $\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\right]$ = 1 $\times$ ten^–13 Grand

Solution

(a) CaCO₃ does precipitate. (b) The compound does non precipitate. (c) The compound does not precipitate. (d) The chemical compound precipitates.

(32) Calculate the concentration of Tl⁺ when TlCl only begins to precipitate from a solution that is 0.0250 M in Cl^–.

(33) Calculate the concentration of sulfate ion when BaSO₄ just begins to precipitate from a solution that is 0.0758 G in Ba²⁺.

Solution

3.03 $\times$ 10⁻⁷ M.

(34) Calculate the concentration of Sr²⁺ when SrF₂ starts to precipitate from a solution that is 0.0025 M in F^–.

(35) Calculate the concentration of ${\text{PO}}_{4}{}^{\text{3-}}$ when Ag_threePO_iv starts to precipitate from a solution that is 0.0125 G in Ag⁺.

Solution

9.2 $\times$ 10^−xiii M

(36) Calculate the concentration of F^– required to begin precipitation of CaF_two in a solution that is 0.010 M in Caⁱⁱ⁺.

(37) Summate the concentration of Ag⁺ required to begin atmospheric precipitation of Ag_iiCO₃ in a solution that is 2.50 $\times$ ten^–6 M in ${\text{CO}}_{3}{}^{\text{2-}}.$

Solution

[Ag⁺] = 1.8 $\times$ ten^–3 1000

(38) What [Ag⁺] is required to reduce $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ to 8.2 $\times$ 10^–four M by atmospheric precipitation of Ag₂CO₃?

(39) What [F^–] is required to reduce [Caⁱⁱ⁺] to 1.0 $\times$ 10^–4 K past atmospheric precipitation of CaF₂?

Solution

6.iii $\times$ 10^–iv

(40) A volume of 0.800 L of a 2 $\times$ x^–4–M Ba(NO_iii)₂ solution is added to 0.200 L of 5 $\times$ x^–4 Grand Li_iiAnd so_iv. Does BaSO_iv precipitate? Explain your answer.

Perform these calculations for nickel(II) carbonate.

(41a) With what book of h2o must a precipitate containing NiCO_iii be washed to dissolve 0.100 g of this compound? Presume that the launder water becomes saturated with NiCO₃ (K _sp = 1.36 $\times$ 10^–7).

(41b) If the NiCO_three were a contaminant in a sample of CoCO₃ (K _sp = i.0 $\times$ 10^–12), what mass of CoCO_three would have been lost? Keep in mind that both NiCO₃ and CoCO_three dissolve in the same solution.

Solution

(a) ii.25 L; (b) seven.2 $\times$ 10^–seven thousand

(42) Iron concentrations greater than 5.4 $\times$ 10^–6 M in water used for laundry purposes tin can crusade staining. What [OH^–] is required to reduce [Fe²⁺] to this level past precipitation of Fe(OH)₂?

(43) A solution is 0.010 One thousand in both Cu²⁺ and Cd²⁺. What percentage of Cd²⁺ remains in the solution when 99.9% of the Cu²⁺ has been precipitated as CuS by adding sulfide?

Solution

100% of it is dissolved

(44) A solution is 0.15 M in both Lead²⁺ and Ag⁺. If Cl^– is added to this solution, what is [Ag⁺] when PbCl₂ begins to precipitate?

(45) What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases, it may be necessary to control the pH. (Hint: Consider the G _sp values given in Appendix J.)

(45a) ${\text{Hg}}_{2}{}^{\text{2+}}$ and Cu^two+

(45b) ${\text{SO}}_{4}{}^{\text{2-}}$ and Cl^–

(45c) Hg²⁺ and Coⁱⁱ⁺

(45d) Zn²⁺ and Sr²⁺

(45e) Ba²⁺ and Mg²⁺

(45f) ${\text{CO}}_{3}{}^{\text{2-}}$ and OH^–

Solution

(a) ${\text{Hg}}_{2}{}^{\text{2+}}$ and Cu²⁺: Add together ${\text{SO}}_{4}{}^{\text{2-}}.$

(b) ${\text{SO}}_{4}{}^{\text{2-}}$ and Cl^–: Add Ba²⁺.

(d) Znⁱⁱ⁺ and Sr²⁺: Add OH^– until [OH^–] = 0.050 K.

(e) Ba²⁺ and Mg²⁺: Add ${\text{SO}}_{4}{}^{\text{2-}}.$

(f) ${\text{CO}}_{3}{}^{\text{2-}}$ and OH^–: Add together Ba²⁺.

(46) A solution contains 1.0 $\times$ 10^–5 mol of KBr and 0.ten mol of KCl per liter. AgNO₃ is gradually added to this solution. Which forms outset, solid AgBr or solid AgCl?

(47) A solution contains i.0 $\times$ x^–2 mol of KI and 0.10 mol of KCl per liter. AgNO₃ is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

Solution

AgI will precipitate first.

(48) The calcium ions in human blood serum are necessary for coagulation (Figure 7.2.four). Potassium oxalate, Grand₂C₂O₄, is used as an anticoagulant when a blood sample is fatigued for laboratory tests because information technology removes the calcium as a precipitate of CaC₂O₄·H₂O. It is necessary to remove all simply i.0% of the Ca²⁺ in serum in order to forbid coagulation. If normal blood serum with a buffered pH of seven.40 contains 9.5 mg of Ca²⁺ per 100 mL of serum, what mass of K₂C₂O₄ is required to prevent the coagulation of a ten mL blood sample that is 55% serum past volume? (All volumes are authentic to two significant figures. Note that the volume of serum in a x-mL blood sample is v.v mL. Assume that the K_sp value for CaC₂O₄ in serum is the aforementioned as in h2o.)

(49) About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca₃(PO₄)₂. The normal mid range calcium content excreted in the urine is 0.10 grand of Ca²⁺ per twenty-four hours. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine tin contain before a calculus begins to form?

Solution

1.5 $\times$ x⁻¹² Grand

(51) Magnesium metal (a component of alloys used in aircraft and a reducing amanuensis used in the product of uranium, titanium, and other active metals) is isolated from sea water past the following sequence of reactions:

${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{Ca(OH)}}_{2}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{Mg(OH)}}_{2}\left(s\right)+{\text{Ca}}^{\text{2+}}\left(aq\right)$

${\text{Mg(OH)}}_{2}\left(s\right)+\text{2HCl(}aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{MgCl}}_{2}\left(s\right)+{\text{2H}}_{2}\text{O}\left(l\right)$

${\text{MgCl}}_{2}\left(l\right)\stackrel{\phantom{\rule{0.7em}{0ex}}\text{electrolysis}\phantom{\rule{0.7em}{0ex}}}{\to }\text{Mg}\left(s\right)+{\text{Cl}}_{2}\left(g\right)$

Sea water has a density of i.026 g/cm³ and contains 1272 parts per million of magnesium as Mg²⁺(aq) past mass. What mass, in kilograms, of Ca(OH)₂ is required to precipitate 99.9% of the magnesium in 1.00 $\times$ ten³ L of sea water?

Solution

3.99 kg

(52) Hydrogen sulfide is bubbled into a solution that is 0.ten M in both Leadⁱⁱ⁺ and Fe^two+ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H₂S ([H_iiS] = 0.10 M). What concentrations of Pb²⁺ and Fe²⁺ remain in the solution? For a saturated solution of H₂South we can use the equilibrium:

${\text{H}}_{2}\text{S}\left(aq\right)+{\text{2H}}_{2}\text{O}\left(l\right)$\rightleftharpoons${\text{2H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{\text{2-}}\left(aq\right)\phantom{\rule{4em}{0ex}}K=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-26}$

(Hint: The $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ changes as metallic sulfides precipitate.)

Perform the post-obit calculations involving concentrations of iodate ions:

(52a) The iodate ion concentration of a saturated solution of La(IO₃)_iii was found to be 3.1 $\times$ 10^–3 mol/L. Find the Thousand _sp.

(52b) Discover the concentration of iodate ions in a saturated solution of Cu(IO₃)₂ (K _sp = 7.4 $\times$ x^–8).

Solution

(a) 3.1 $\times$ 10^–eleven; (b) [Cuⁱⁱ⁺] = 2.6 $\times$ 10^–3; $\left[{\text{IO}}_{3}{}^{\text{-}}\right]$ = 5.iii $\times$ 10^–three

(53) Calculate the molar solubility of AgBr in 0.035 M NaBr (K _sp = five $\times$ 10^–xiii).

(54) How many grams of Pb(OH)_two will deliquesce in 500 mL of a 0.050-M PbCl₂ solution (K _sp = 1.2 $\times$ 10^–15)?

Solution

1.eight $\times$ 10^–5 thousand Pb(OH)₂

(55) Apply the simulation from the earlier Link to Learning to complete the following exercise. Using 0.01 g CaF₂, give the K_sp values constitute in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.

(56) How many grams of Milk of Magnesia, Mg(OH)₂ (s) (58.iii yard/mol), would be soluble in 200 mL of water. Yard _sp = 7.ane $\times$ 10^–12. Include the ionic reaction and the expression for Thousand _sp in your answer. (K _{due west} = 1 $\times$ x^–fourteen = [H₃O⁺][OH^–])

Solution

${\text{Mg(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}+{\text{2OH}}^{\text{-}}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$
ane.23 $\times$ 10^−three m Mg(OH)₂

(57) Two hypothetical salts, LM₂ and LQ, accept the same molar solubility in H_twoO. If M _sp for LM₂ is 3.20 $\times$ 10^–five, what is the K _sp value for LQ?

(58) The carbonate ion concentration is gradually increased in a solution containing divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will form start? Which of the following will class last? Explicate.

(58a) ${\text{MgCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=3.5\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-8}$

(58b) ${\text{CaCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=4.2\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-7}$

(58c) ${\text{SrCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=3.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-9}$

(58d) ${\text{BaCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=4.4\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-5}$

(58e) ${\text{MnCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=5.1\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-9}$

Solution

MnCO₃ will form commencement, since it has the smallest One thousand _sp value it is the least soluble. MnCO_three will be the final to precipitate, it has the largest Thousand _sp value.

(59) How many grams of Zn(CN)_ii(due south) (117.44 g/mol) would be soluble in 100 mL of H₂O? Include the balanced reaction and the expression for K _sp in your answer. The Thousand _sp value for Zn(CN)₂(southward) is 3.0 $\times$ 10^–sixteen.

Glossary

: effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a subtract in the ionization of a weak acid or base

: solubility of a compound expressed in units of moles per liter (mol/L)

: process in which ions are separated using differences in their solubility with a given precipitating reagent

: equilibrium abiding for the dissolution of an ionic compound

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Source: https://pressbooks.bccampus.ca/inorganicchemistrychem250/chapter/precipitation-and-dissolution/

You Can Precipitate Silver Chloride From a Saturated Solution of the Compound by Adding

7.2 Precipitation and Dissolution

The Solubility Product

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One thousand sp and Solubility

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Using Barium Sulfate for Medical Imaging

Predicting Precipitation

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The Office of Precipitation in Wastewater Treatment

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Common Ion Effect

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Key Concepts and Summary

Key Equations

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One thousand _sp and Solubility