Solubility Equilibria & Ksp practice questions

Question 1

A student measures the solubility of silver chloride, $\text{AgCl}(s)$, in pure water. At $50^\circ\text{C}$, the solubility product constant, $K_{sp}$, for $\text{AgCl}(s)$ is $1.6 \times 10^{-9}$. What is the molar solubility of $\text{AgCl}(s)$ in pure water at $50^\circ\text{C}$?

Accepted Answer

$4.0 \times 10^{-5}\text{ M}$

Answer

$8.0 \times 10^{-5}\text{ M}$

Answer

$4.0 \times 10^{-4}\text{ M}$

Answer

$8.0 \times 10^{-4}\text{ M}$

Question 2

A student tests whether chloride ions in rinse water will form a precipitate when treated with silver ions. The student mixes equal volumes of $6.0 \times 10^{-5}\text{ M AgNO}_3$ and $6.0 \times 10^{-6}\text{ M NaCl}$ at $25^\circ\text{C}$. The volumes are assumed to be additive.

$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)}+\text{Cl}^-\text{(aq)}\qquad K_{sp}=1.8\times10^{-10}$

Which of the following correctly predicts whether a precipitate forms immediately after the solutions are mixed?

Accepted Answer

No precipitate forms because mixing equal volumes halves each ion concentration, giving \(Q_{sp}=9.0\times10^{-11}

Answer

A precipitate forms because \(Q_{sp}=9.0\times10^{-11}

Answer

A precipitate forms because $Q_{sp}=(6.0\times10^{-5})(6.0\times10^{-6})=3.6\times10^{-10}>K_{sp}$.

Answer

No precipitate forms because mixing equal volumes makes each ion concentration one-fourth its initial value, giving \(Q_{sp}=2.25\times10^{-11}

Question 3

A student evaluating mineral deposits in a heated water-treatment line studies the hypothetical sparingly soluble salt $\text{MX(s)}$. At each temperature, excess solid is allowed to reach equilibrium with pure water according to the following equation.

$\text{MX(s)} \rightleftharpoons \text{M}^{2+}\text{(aq)}+\text{X}^{2-}\text{(aq)}$

| Quantity | Value |
|---|---|
| $\Delta H^\circ_{\text{soln}}$ | $+60.0\text{ kJ mol}^{-1}$ |
| $\Delta S^\circ_{\text{soln}}$ | $+80.0\text{ J mol}^{-1}\text{ K}^{-1}$ |

Assume that $\Delta H^\circ_{\text{soln}}$ and $\Delta S^\circ_{\text{soln}}$ remain constant from $298\text{ K}$ to $318\text{ K}$. Which statement correctly compares the equilibrium molar solubility at $318\text{ K}$ with that at $298\text{ K}$ and justifies the comparison?

Accepted Answer

The solubility increases, because $-T\Delta S^\circ_{\text{soln}}$ becomes more negative while $\Delta H^\circ_{\text{soln}}$ remains approximately constant, so $\Delta G^\circ_{\text{soln}}$ decreases.

Answer

The solubility decreases, because the positive $\Delta H^\circ_{\text{soln}}$ means that heating makes the dissolution less thermodynamically favorable.

Answer

The solubility decreases, because increasing $T$ makes $T\Delta S^\circ_{\text{soln}}$ larger and therefore makes $\Delta G^\circ_{\text{soln}}$ larger.

Answer

The solubility increases, because heating changes $\Delta H^\circ_{\text{soln}}$ from positive to negative.

Question 4

A student has two separate saturated aqueous solutions at equilibrium: Beaker 1 contains solid calcium carbonate, $\text{CaCO}_3\text{(s)}$, in contact with its ions, and Beaker 2 contains solid silver chloride, $\text{AgCl(s)}$, in contact with its ions. A few drops of concentrated nitric acid, $\text{HNO}_3\text{(aq)}$, are added to each beaker, significantly decreasing the pH of both solutions without noticeably changing the total volume. Which of the following predictions best describes the effect of decreasing the pH on the solubility of each solid?

Accepted Answer

The solubility of $\text{CaCO}_3\text{(s)}$ increases because $\text{CO}_3^{2-}\text{(aq)}$ reacts with $\text{H}^+\text{(aq)}$, whereas the solubility of $\text{AgCl(s)}$ remains unchanged because $\text{Cl}^-\text{(aq)}$ is the conjugate base of a strong acid.

Answer

The solubility of both $\text{CaCO}_3\text{(s)}$ and $\text{AgCl(s)}$ increases because $\text{H}^+\text{(aq)}$ reacts with the anions in both solutions to form weak acids.

Answer

The solubility of $\text{CaCO}_3\text{(s)}$ remains unchanged, whereas the solubility of $\text{AgCl(s)}$ increases because $\text{Ag}^+\text{(aq)}$ forms a soluble complex with nitrate ions.

Answer

The solubility of both $\text{CaCO}_3\text{(s)}$ and $\text{AgCl(s)}$ remains unchanged because $\text{H}^+\text{(aq)}$ does not react with metal cations in aqueous solution.

Question 5

A student investigates the solubility of three generic sparingly soluble salts in pure water at $25^\circ\text{C}$. The chemical formula and solubility product constant ($K_{sp}$) for each salt are provided in the table below.

| Salt | Formula | $K_{sp}$ at $25^\circ\text{C}$ |
|---|---|---|
| Salt 1 | $\text{MX}$ | $1.0 \times 10^{-10}$ |
| Salt 2 | $\text{QX}_2$ | $4.0 \times 10^{-12}$ |
| Salt 3 | $\text{ZX}$ | $1.0 \times 10^{-14}$ |

Which of the following correctly ranks the three salts in order of greatest to least molar solubility in pure water at $25^\circ\text{C}$?

Accepted Answer

$\text{Salt 2} > \text{Salt 1} > \text{Salt 3}$

Answer

$\text{Salt 1} > \text{Salt 2} > \text{Salt 3}$

Answer

$\text{Salt 1} > \text{Salt 3} > \text{Salt 2}$

Answer

$\text{Salt 3} > \text{Salt 1} > \text{Salt 2}$

Question 6

A student prepares a saturated solution of calcium fluoride, $\text{CaF}_2(s)$, at $25^\circ\text{C}$ and determines that the molar solubility of $\text{CaF}_2$ is $2.0 \times 10^{-4} \text{ M}$. Based on this measurement, what is the value of the solubility product constant, $K_{sp}$, for $\text{CaF}_2$ at $25^\circ\text{C}$?

Accepted Answer

$3.2 \times 10^{-11}$

Answer

$6.4 \times 10^{-11}$

Answer

$8.0 \times 10^{-8}$

Answer

$1.6 \times 10^{-7}$

Question 7

A student models the recovery of lead(II) iodide from a pigment-waste stream using the dissolution equilibrium

$\text{PbI}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)}$

At the temperature of the experiment, $K_{sp}=4.0\times10^{-9}$. Excess $\text{PbI}_2\text{(s)}$ is added to a $0.020\text{ M}$ solution of $\text{NaI(aq)}$, which dissociates completely. Assume that the iodide ions produced by dissolution of $\text{PbI}_2\text{(s)}$ make a negligible contribution to the total iodide concentration. What is the approximate molar solubility of $\text{PbI}_2\text{(s)}$ in the solution?

Accepted Answer

$1.0\times10^{-5}\text{ M}$

Answer

$5.0\times10^{-6}\text{ M}$

Answer

$4.5\times10^{-4}\text{ M}$

Answer

$1.0\times10^{-3}\text{ M}$

Question 8

A student compares the molar solubility of excess solid $\text{Mg(OH)}_2$ in pure water with its molar solubility in a pH-controlled acidic solution at the same temperature. The acidic solution does not initially contain $\text{Mg}^{2+}$. The relevant equilibria are

$\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^{-}\text{(aq)}$

$\text{H}_3\text{O}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)} \rightleftharpoons 2\text{H}_2\text{O(l)}$, which is strongly product-favored.

Which of the following correctly predicts and explains how the molar solubility of $\text{Mg(OH)}_2$ in the acidic solution compares with its molar solubility in pure water?

Accepted Answer

It is higher because $\text{H}_3\text{O}^{+}$ consumes $\text{OH}^{-}$, causing the dissolution equilibrium to shift toward the dissolved ions.

Answer

It is lower because $\text{H}_3\text{O}^{+}$ is a common ion in the dissolution equilibrium and shifts the equilibrium toward the solid.

Answer

It is lower because neutralization removes $\text{OH}^{-}$, causing the dissolution equilibrium to shift toward the solid.

Answer

It is higher because decreasing the pH increases the value of $K_{sp}$ for $\text{Mg(OH)}_2$.

Question 9

A student models a halide-exchange step that can occur when silver halide solids contact an aqueous sample. At $25^\circ\text{C}$, the equilibrium constants for the dissolution reactions are given below.

| Equilibrium | Equilibrium constant |
|---|---|
| $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$ | $K_1=2.0\times10^{-10}$ |
| $\text{AgBr(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)}$ | $K_2=5.0\times10^{-13}$ |

The student combines the equilibria to represent the net reaction $\text{AgCl(s)}+\text{Br}^-\text{(aq)}\rightleftharpoons\text{AgBr(s)}+\text{Cl}^-\text{(aq)}$. What is the value of $K$ for this net reaction at $25^\circ\text{C}$?

Accepted Answer

$4.0\times10^2$

Answer

$1.0\times10^{-22}$

Answer

$2.0\times10^{-10}$

Answer

$2.5\times10^{-3}$

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