AP Chemistry Quiz: Introduction To Solubility Equilibria

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Introduction To Solubility EquilibriaQuestion 1 of 20

A student adds excess solid barium sulfate, $\text{BaSO}_4(s)$ , to pure water at $25^\circ\text{C}$ . After equilibrium is established, the solubility product constant is $K_{sp}(\text{BaSO}_4)=1.0\times10^{-10}$ . What is the equilibrium concentration of $\text{Ba}^{2+}(aq)$ ? (Assume volume change is negligible.)

2.0\times10^{-5}\,\text{M}

1.0\times10^{-20}\,\text{M}

1.0\times10^{-10}\,\text{M}

5.0\times10^{-11}\,\text{M}

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

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AP Chemistry Quiz: Introduction To Solubility Equilibria

Practice Introduction To Solubility Equilibria in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Introduction To Solubility Equilibria, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

$2.0\times10^{-5}\,\text{M}$
$1.0\times10^{-20}\,\text{M}$
$1.0\times10^{-10}\,\text{M}$
$5.0\times10^{-11}\,\text{M}$
$1.0\times10^{-5}\,\text{M}$ (correct answer)

Explanation: This question tests your understanding of solubility equilibria (quantitative). The dissolution equation is BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). To construct the ICE table, we set initial concentrations of Ba²⁺ and SO₄²⁻ to 0 M, the change as +x for both ions due to 1:1 stoichiometry, and equilibrium concentrations as x for both. The Ksp expression is [Ba²⁺][SO₄²⁻] = x · x = x² = 1.0×10^{-10}, so we solve for the single variable x by taking the square root to get x = 1.0×10^{-5} M. We use only one variable because the equal stoichiometric coefficients link the ion concentrations directly. A tempting distractor is 1.0×10^{-20} M (choice E), which results from squaring the Ksp instead of taking the square root. For simple salts, Ksp problems reduce to one variable on AP MCQs, so always set up the expression based on stoichiometry and solve accordingly.

Question 2

Excess solid copper(I) chloride, $\text{CuCl}(s)$ , is placed in pure water at $25^\circ\text{C}$ until equilibrium is established. The $K_{sp}$ of $\text{CuCl}$ is $1.0\times10^{-6}$ .

Dissolution: $\text{CuCl}(s)\rightleftharpoons \text{Cu}^+(aq)+\text{Cl}^-(aq)$

Starting with zero ion concentrations, what is the equilibrium concentration of $\text{Cl}^-(aq)$ ?

ICE table (in M):

	$\text{Cu}^+$	$\text{Cl}^-$
I	0	0
C	+ $s$	+ $s$
E	$s$	$s$

$1.0\times10^{-6}\,\text{M}$
$1.0\times10^{-3}\,\text{M}$ (correct answer)
$5.0\times10^{-4}\,\text{M}$
$2.0\times10^{-3}\,\text{M}$
$1.0\times10^{3}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation CuCl(s) ⇌ Cu⁺(aq) + Cl⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Cu⁺ and Cl⁻. The Ksp expression is Ksp = [Cu⁺][Cl⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻⁶, so s = √(1.0×10⁻⁶) = 1.0×10⁻³ M, which equals [Cl⁻]. Choice A (1.0×10⁻⁶) incorrectly uses Ksp as the concentration. For any 1:1 salt, both ion concentrations equal the molar solubility at equilibrium.

Question 3

Excess solid barium sulfate, $\text{BaSO}_4(s)$ , is stirred in pure water at $25^\circ\text{C}$ until equilibrium is reached. The $K_{sp}$ of $\text{BaSO}_4$ is $1.0\times10^{-10}$ .

Dissolution: $\text{BaSO}_4(s)\rightleftharpoons \text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq)$

Starting from zero ion concentrations, what is the equilibrium concentration of $\text{Ba}^{2+}(aq)$ ?

ICE table (in M):

	$\text{Ba}^{2+}$	$\text{SO}_4^{2-}$
I	0	0
C	+ $s$	+ $s$
E	$s$	$s$

$1.0\times10^{-10}\,\text{M}$
$1.0\times10^{-5}\,\text{M}$ (correct answer)
$5.0\times10^{-6}\,\text{M}$
$2.0\times10^{-5}\,\text{M}$
$1.0\times10^{5}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Ba²⁺ and SO₄²⁻. The Ksp expression is Ksp = [Ba²⁺][SO₄²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻¹⁰, so s = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. Choice A (1.0×10⁻¹⁰) incorrectly uses Ksp as the concentration without solving. For 1:1 salts, always take the square root of Ksp to find the molar solubility.

Question 4

Excess solid barium sulfate, $\text{BaSO}_4(s)$ , is mixed with pure water at $25^\circ\text{C}$ until equilibrium is established. For $\text{BaSO}_4(s)\rightleftharpoons \text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq)$ , $K_{sp}=1.0\times10^{-10}$ . What is the equilibrium concentration of $\text{Ba}^{2+}$ in the saturated solution?

$1.0\times10^{-5}\,\text{M}$ (correct answer)
$1.0\times10^{-10}\,\text{M}$
$5.0\times10^{-6}\,\text{M}$
$1.0\times10^{-20}\,\text{M}$
$2.0\times10^{-5}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). For BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq), both ions form in a 1:1 ratio, so [Ba²⁺] = [SO₄²⁻] = x at equilibrium. Substituting into Ksp = [Ba²⁺][SO₄²⁻] = x² = 1.0×10⁻¹⁰. Solving gives x = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. A common mistake is reporting the Ksp value (1.0×10⁻¹⁰ M) as the concentration. For simple 1:1 salts, molar solubility equals the square root of Ksp.

Question 5

Excess solid barium sulfate, $\text{BaSO}_4(s)$ , is added to pure water at $25^\circ\text{C}$ and stirred to equilibrium. The solubility-product constant is $K_{sp}(\text{BaSO}_4)=1.0\times 10^{-10}$ . What is the equilibrium concentration of $\text{Ba}^{2+}(aq)$ ?\n\nUse the following ICE table setup:\n\n| | $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$ | $[\text{Ba}^{2+}]$ (M) | $[\text{SO}_4^{2-}]$ (M) |\n|------|-------------------------------------------------------------------------|------------------------|--------------------------|\n| I | | 0 | 0 |\n| C | | $+s$ | $+s$ |\n| E | | $s$ | $s$ |

$1.0\times 10^{-5}\,\text{M}$ (correct answer)
$1.0\times 10^{-10}\,\text{M}$
$3.2\times 10^{-5}\,\text{M}$
$5.0\times 10^{-6}\,\text{M}$
$2.0\times 10^{-5}\,\text{M}$

Explanation: This problem tests your understanding of solubility equilibria (quantitative). When $\text{BaSO}_4$ dissolves, it produces equal moles of $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ , so if $s$ = molar solubility, then $[\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = s$ at equilibrium. The Ksp expression is $K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2$ . Substituting: $1.0 \times 10^{-10} = s^2$ , so $s = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \, \text{M}$ . A common error is confusing this 1:1 salt with more complex stoichiometries or using Ksp directly as the concentration. For simple 1:1 salts, the calculation reduces to taking the square root of Ksp.

Question 6

Excess solid zinc sulfide, $\text{ZnS}(s)$ , is added to pure water at $25^\circ\text{C}$ . The equilibrium is $\text{ZnS}(s) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{S}^{2-}(aq)$ with $K_{sp}=1.0\times10^{-24}$ . What is the molar solubility of $\text{ZnS}$ (in $\text{mol·L}^{-1}$ )?

$1.0\times10^{-24}\ \text{mol·L}^{-1}$
$1.0\times10^{-12}\ \text{mol·L}^{-1}$ (correct answer)
$5.0\times10^{-13}\ \text{mol·L}^{-1}$
$1.0\times10^{-6}\ \text{mol·L}^{-1}$
$2.0\times10^{-12}\ \text{mol·L}^{-1}$

Explanation: This problem tests solubility equilibria (quantitative). For ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq), both ions have 1:1 stoichiometry, so if 's' mol/L dissolves, then [Zn²⁺] = [S²⁻] = s at equilibrium. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = s × s = s², so 1.0×10⁻²⁴ = s², giving s = 1.0×10⁻¹² mol/L. This equals the molar solubility since one formula unit produces one of each ion. A tempting error is to take the fourth root instead of square root (if confused with more complex salts). For 1:1 salts, the calculation is straightforward: molar solubility = √Ksp.

Question 7

Excess solid zinc sulfide, $\text{ZnS}(s)$ , is added to pure water at $25^\circ\text{C}$ and allowed to reach equilibrium. The $K_{sp}$ of $\text{ZnS}$ is $1.0\times10^{-24}$ .

Dissolution: $\text{ZnS}(s)\rightleftharpoons \text{Zn}^{2+}(aq)+\text{S}^{2-}(aq)$

Assuming initial ion concentrations are zero, what is the molar solubility of $\text{ZnS}$ ?

ICE table (in M):

	$\text{Zn}^{2+}$	$\text{S}^{2-}$
I	0	0
C	+ $s$	+ $s$
E	$s$	$s$

$1.0\times10^{-12}\,\text{M}$ (correct answer)
$1.0\times10^{-24}\,\text{M}$
$5.0\times10^{-13}\,\text{M}$
$1.0\times10^{-8}\,\text{M}$
$2.0\times10^{-12}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Zn²⁺ and S²⁻. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻²⁴, so s = √(1.0×10⁻²⁴) = 1.0×10⁻¹² M. Choice B (1.0×10⁻²⁴) incorrectly uses the Ksp value as the molar solubility without taking the square root. For extremely insoluble salts like ZnS, the molar solubility is still found by taking the square root of Ksp.

Question 8

A student adds excess solid silver chloride, $\text{AgCl}(s)$ , to pure water at $25^\circ\text{C}$ . The equilibrium is $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ with $K_{sp}=1.6\times 10^{-10}$ . What is the equilibrium concentration of $\text{Ag}^+$ in the saturated solution (in $\text{mol·L}^{-1}$ )?

$1.6\times10^{-10}\ \text{mol·L}^{-1}$
$4.0\times10^{-5}\ \text{mol·L}^{-1}$
$1.3\times10^{-5}\ \text{mol·L}^{-1}$ (correct answer)
$8.0\times10^{-10}\ \text{mol·L}^{-1}$
$2.5\times10^{-5}\ \text{mol·L}^{-1}$

Explanation: This problem tests solubility equilibria (quantitative). For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), we set up an ICE table where initially both ion concentrations are 0, and at equilibrium both equal 's' (the molar solubility) due to the 1:1 stoichiometry. The Ksp expression becomes Ksp = [Ag⁺][Cl⁻] = s × s = s², so 1.6×10⁻¹⁰ = s², giving s = 1.3×10⁻⁵ mol/L. A common error is using Ksp directly as the concentration (choice A), forgetting that Ksp equals the product of ion concentrations, not the individual concentrations. For 1:1 salts like AgCl, remember that molar solubility equals the square root of Ksp.

Question 9

Excess solid calcium carbonate, $\text{CaCO}_3(s)$ , is added to pure water at $25^\circ\text{C}$ . The equilibrium is $\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq)$ with $K_{sp}=9.0\times10^{-9}$ . What is the equilibrium concentration of $\text{CO}_3^{2-}$ in the saturated solution (in $\text{mol·L}^{-1}$ )?

$6.0\times10^{-5}\ \text{mol·L}^{-1}$
$9.5\times10^{-5}\ \text{mol·L}^{-1}$
$9.0\times10^{-9}\ \text{mol·L}^{-1}$
$3.0\times10^{-4}\ \text{mol·L}^{-1}$
$3.0\times10^{-5}\ \text{mol·L}^{-1}$ (correct answer)

Explanation: This problem tests solubility equilibria (quantitative). For CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq), the 1:1 stoichiometry means if 's' mol/L dissolves, then [Ca²⁺] = [CO₃²⁻] = s at equilibrium. The Ksp expression is Ksp = [Ca²⁺][CO₃²⁻] = s × s = s², so 9.0×10⁻⁹ = s², giving s = 3.0×10⁻⁵ mol/L. Since the question asks for [CO₃²⁻], which equals s, the answer is 3.0×10⁻⁵ mol/L. A common mistake is taking the wrong root or confusing the carbonate concentration with total dissolved amount. For 1:1 salts, the concentration of each ion at equilibrium equals the molar solubility.

Question 10

A student adds excess solid silver chloride, $\text{AgCl}(s)$ , to $1.00\,\text{L}$ of pure water at $25^\circ\text{C}$ and stirs until equilibrium is established. The $K_{sp}$ of $\text{AgCl}$ at $25^\circ\text{C}$ is $1.6\times10^{-10}$ .

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

Assuming the initial ion concentrations are zero, what is the equilibrium concentration of $\text{Ag}^+(aq)$ ?

ICE table (in M):

	$\text{Ag}^+$	$\text{Cl}^-$
I	0	0
C	+ $s$	+ $s$
E	$s$	$s$

$4.0\times10^{-5}\,\text{M}$
$1.3\times10^{-5}\,\text{M}$ (correct answer)
$8.0\times10^{-11}\,\text{M}$
$1.6\times10^{-10}\,\text{M}$
$2.5\times10^{5}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) shows a 1:1 stoichiometry, so if s moles of AgCl dissolve, we get s moles of Ag⁺ and s moles of Cl⁻. The Ksp expression is Ksp = [Ag⁺][Cl⁻] = (s)(s) = s². Solving: s² = 1.6×10⁻¹⁰, so s = √(1.6×10⁻¹⁰) = 1.26×10⁻⁵ M. Choice A (8.0×10⁻¹¹) incorrectly divides Ksp by 2 instead of taking the square root. For simple 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

Question 11

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

Assuming the initial ion concentrations are zero, what is the equilibrium concentration of $\text{Ag}^+(aq)$ ?

ICE table (in M):

	$\text{Ag}^+$	$\text{Cl}^-$
I	0	0
C	+ $s$	+ $s$
E	$s$	$s$

$2.5\times10^{5}\,\text{M}$
$4.0\times10^{-5}\,\text{M}$
$8.0\times10^{-11}\,\text{M}$
$1.3\times10^{-5}\,\text{M}$ (correct answer)
$1.6\times10^{-10}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation $\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)$ shows a 1:1 stoichiometry, so if s moles of AgCl dissolve, we get s moles of Ag⁺ and s moles of Cl⁻. The Ksp expression is $K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] = (s)(s) = s^2$ . Solving: $s^2 = 1.6\times10^{-10}$ , so $s = \sqrt{1.6\times10^{-10}} = 1.26\times10^{-5} \, \text{M}$ . Choice A ( $8.0\times10^{-11}$ ) incorrectly divides Ksp by 2 instead of taking the square root. For simple 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

Question 12

Excess solid iron(II) hydroxide, $\text{Fe(OH)}_2(s)$ , is added to pure water at $25^\circ\text{C}$ until equilibrium is established. The $K_{sp}$ of $\text{Fe(OH)}_2$ is $4.0\times10^{-14}$ .

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

If the initial ion concentrations are zero, what is the equilibrium concentration of $\text{OH}^-(aq)$ ?

ICE table (in M):

	$\text{Fe}^{2+}$	$\text{OH}^-$
I	0	0
C	+ $s$	+ $2s$
E	$s$	$2s$

$2.0\times10^{-5}\,\text{M}$ (correct answer)
$1.0\times10^{-10}\,\text{M}$
$1.0\times10^{-5}\,\text{M}$
$5.0\times10^{-5}\,\text{M}$
$4.0\times10^{-14}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation $\text{Fe(OH)}_2(s) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{OH}^{-}(aq)$ shows that dissolving s moles produces s moles of $\text{Fe}^{2+}$ and 2s moles of $\text{OH}^{-}$ . The $K_{sp}$ expression is $K_{sp} = [\text{Fe}^{2+}][\text{OH}^{-}]^2 = s \cdot (2s)^2 = 4s^3$ . Solving: $4s^3 = 4.0 \times 10^{-14}$ , so $s^3 = 1.0 \times 10^{-14}$ , and $s = 1.0 \times 10^{-5} \, \text{M}$ , making $[\text{OH}^{-}] = 2s = 2.0 \times 10^{-5} \, \text{M}$ . Choice C ( $4.0 \times 10^{-14}$ ) incorrectly uses the Ksp value as the concentration. For hydroxide salts, remember to multiply the molar solubility by 2 to get the OH⁻ concentration.

Question 13

Excess solid zinc sulfide, $\text{ZnS}(s)$ , is added to pure water at $25^\circ\text{C}$ and allowed to reach equilibrium. The $K_{sp}$ of $\text{ZnS}$ is $1.0\times10^{-24}$ .

Dissolution: $\text{ZnS}(s)\rightleftharpoons \text{Zn}^{2+}(aq)+\text{S}^{2-}(aq)$

Assuming initial ion concentrations are zero, what is the molar solubility of $\text{ZnS}$ ?

ICE table (in M):

	$\text{Zn}^{2+}$	$\text{S}^{2-}$
I	0	0
C	+ $s$	+ $s$
E	$s$	$s$

$1.0\times10^{-12}\,\text{M}$ (correct answer)
$1.0\times10^{-24}\,\text{M}$
$5.0\times10^{-13}\,\text{M}$
$1.0\times10^{-8}\,\text{M}$
$2.0\times10^{-12}\,\text{M}$

Question 14

Excess solid iron(II) hydroxide, $\text{Fe(OH)}_2(s)$ , is added to pure water at $25^\circ\text{C}$ until equilibrium is established. The $K_{sp}$ of $\text{Fe(OH)}_2$ is $4.0\times10^{-14}$ .

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

If the initial ion concentrations are zero, what is the equilibrium concentration of $\text{OH}^-(aq)$ ?

ICE table (in M):

	$\text{Fe}^{2+}$	$\text{OH}^-$
I	0	0
C	+ $s$	+ $2s$
E	$s$	$2s$

$2.0\times10^{-5}\,\text{M}$ (correct answer)
$4.0\times10^{-14}\,\text{M}$
$1.0\times10^{-5}\,\text{M}$
$5.0\times10^{-5}\,\text{M}$
$1.0\times10^{-10}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation Fe(OH)₂(s) ⇌ Fe²⁺(aq) + 2OH⁻(aq) shows that dissolving s moles produces s moles of Fe²⁺ and 2s moles of OH⁻. The Ksp expression is Ksp = [Fe²⁺][OH⁻]² = (s)(2s)² = 4s³. Solving: 4s³ = 4.0×10⁻¹⁴, so s³ = 1.0×10⁻¹⁴, and s = 1.0×10⁻⁵ M, making [OH⁻] = 2s = 2.0×10⁻⁵ M. Choice C (4.0×10⁻¹⁴) incorrectly uses the Ksp value as the concentration. For hydroxide salts, remember to multiply the molar solubility by 2 to get the OH⁻ concentration.

Question 15

A student adds excess solid silver chloride, $\text{AgCl}(s)$ , to $1.00\,\text{L}$ of pure water at $25^\circ\text{C}$ and allows the system to reach equilibrium. The solubility product constant is $K_{sp}(\text{AgCl}) = 1.6\times10^{-10}$ . What is the equilibrium concentration of $\text{Ag}^+(aq)$ ? (Assume volume change is negligible.)

$1.6\times10^{-10}\,\text{M}$
$1.3\times10^{-5}\,\text{M}$ (correct answer)
$4.0\times10^{-20}\,\text{M}$
$8.0\times10^{-11}\,\text{M}$
$2.5\times10^{-5}\,\text{M}$

Explanation: This question tests your understanding of solubility equilibria (quantitative). The dissolution equation is AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). To construct the ICE table, we set initial concentrations of Ag⁺ and Cl⁻ to 0 M, the change as +x for both ions due to 1:1 stoichiometry, and equilibrium concentrations as x for Ag⁺ and x for Cl⁻. The Ksp expression is [Ag⁺][Cl⁻] = x · x = x² = 1.6×10^{-10}, so we solve for the single variable x by taking the square root to get x ≈ 1.3×10^{-5} M. We use only one variable because the equal stoichiometric coefficients link the ion concentrations directly. A tempting distractor is 1.6×10^{-10} M (choice A), which results from mistakenly using the Ksp value directly without solving the equation. For simple salts, Ksp problems reduce to one variable on AP MCQs, so always set up the expression based on stoichiometry and solve accordingly.

Question 16

Excess solid copper(I) bromide, $\text{CuBr}(s)$ , is added to pure water at $25^\circ\text{C}$ . After equilibrium is established, the solubility-product constant is $K_{sp}(\text{CuBr})=4.0\times 10^{-8}$ . What is the equilibrium concentration of $\text{Br}^-(aq)$ ?

Use the following ICE table setup:

	$[\text{Cu}^+]$ (M)	$[\text{Br}^-]$ (M)
I	0	0
C	$+s$	$+s$
E	$s$	$s$

$4.0\times 10^{-8}\,\text{M}$
$4.0\times 10^{-4}\,\text{M}$
$1.0\times 10^{-4}\,\text{M}$
$6.3\times 10^{-4}\,\text{M}$
$2.0\times 10^{-4}\,\text{M}$ (correct answer)

Explanation: This problem tests your understanding of solubility equilibria (quantitative). When CuBr dissolves, it produces equal moles of Cu⁺ and Br⁻, so if s = molar solubility, then [Cu⁺] = [Br⁻] = s at equilibrium. The Ksp expression is Ksp = [Cu⁺][Br⁻] = s × s = s². Substituting: 4.0 × 10⁻⁸ = s², so s = √(4.0 × 10⁻⁸) = 2.0 × 10⁻⁴ M. A common error is using the Ksp value directly or miscalculating the square root. For 1:1 salts on AP exams, the molar solubility equals the concentration of each ion.

Question 17

A student adds excess solid silver chloride, $\text{AgCl}(s)$ , to $1.00\,\text{L}$ of pure water at $25^\circ\text{C}$ . The equilibrium expression is $K_{sp}=[\text{Ag}^+][\text{Cl}^-]$ , and $K_{sp}(\text{AgCl})=1.6\times10^{-10}$ . What is the equilibrium concentration of $\text{Ag}^+$ in the solution? (Assume volume change is negligible.)

$4.0\times10^{-20}\,\text{M}$
$1.6\times10^{-10}\,\text{M}$
$4.0\times10^{-5}\,\text{M}$
$8.0\times10^{-10}\,\text{M}$
$1.3\times10^{-5}\,\text{M}$ (correct answer)

Explanation: This problem tests solubility equilibria (quantitative). For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), we set up an ICE table where both ions start at 0 and increase by x at equilibrium. Since Ksp = [Ag⁺][Cl⁻] = 1.6×10⁻¹⁰ and both concentrations equal x, we get x² = 1.6×10⁻¹⁰. Solving gives x = √(1.6×10⁻¹⁰) = 1.26×10⁻⁵ M, which rounds to 1.3×10⁻⁵ M. A common error is forgetting to take the square root and reporting 1.6×10⁻¹⁰ M. For 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

Question 18

Excess solid iron(II) sulfide, $\text{FeS}(s)$ , is placed in pure water at $25^\circ\text{C}$ until equilibrium is reached. For $\text{FeS}(s)\rightleftharpoons \text{Fe}^{2+}(aq)+\text{S}^{2-}(aq)$ , $K_{sp}=6.4\times10^{-19}$ . What is the equilibrium concentration of $\text{Fe}^{2+}$ in the saturated solution?

$8.0\times10^{-10}\,\text{M}$ (correct answer)
$3.2\times10^{-19}\,\text{M}$
$6.4\times10^{-19}\,\text{M}$
$4.0\times10^{-10}\,\text{M}$
$8.0\times10^{-19}\,\text{M}$

Explanation: This problem tests solubility equilibria (quantitative). For FeS(s) ⇌ Fe²⁺(aq) + S²⁻(aq), both ions have equal concentrations at equilibrium: [Fe²⁺] = [S²⁻] = x. Substituting into Ksp = [Fe²⁺][S²⁻] = x² = 6.4×10⁻¹⁹. Solving gives x = √(6.4×10⁻¹⁹) = 8.0×10⁻¹⁰ M. A common mistake is confusing the ion concentration with the Ksp value itself (6.4×10⁻¹⁹ M). For 1:1 salts, the molar solubility equals the square root of the solubility product constant.

Question 19

A student places excess solid silver bromide, $\text{AgBr}(s)$ , into pure water at $25^\circ\text{C}$ and allows the system to reach equilibrium. For $\text{AgBr}(s)\rightleftharpoons \text{Ag}^+(aq)+\text{Br}^-(aq)$ , $K_{sp}=4.0\times10^{-13}$ . What is the equilibrium concentration of $\text{Br}^-$ in the saturated solution?

$2.0\times10^{-6}\,\text{M}$
$4.0\times10^{-13}\,\text{M}$
$4.0\times10^{-26}\,\text{M}$
$2.0\times10^{-13}\,\text{M}$
$6.3\times10^{-7}\,\text{M}$ (correct answer)

Explanation: This problem tests solubility equilibria (quantitative). For AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq), both ions have equal concentrations at equilibrium, so [Ag⁺] = [Br⁻] = x. The Ksp expression gives x² = 4.0×10⁻¹³. Taking the square root: x = 6.32×10⁻⁷ M, which rounds to 6.3×10⁻⁷ M. A common error is confusing this with the Ksp value itself (4.0×10⁻¹³ M). For 1:1 salts, the equilibrium concentration of each ion equals the square root of Ksp.

Question 20

A student adds excess solid silver bromide, $\text{AgBr}(s)$ , to pure water at $25^\circ\text{C}$ . The equilibrium is $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$ with $K_{sp}=2.5\times10^{-13}$ . What is the molar solubility of $\text{AgBr}$ (in $\text{mol·L}^{-1}$ )?

$5.0\times10^{-7}\ \text{mol·L}^{-1}$ (correct answer)
$2.5\times10^{-13}\ \text{mol·L}^{-1}$
$1.0\times10^{-6}\ \text{mol·L}^{-1}$
$1.6\times10^{-7}\ \text{mol·L}^{-1}$
$2.0\times10^{-7}\ \text{mol·L}^{-1}$

Explanation: This problem tests solubility equilibria (quantitative). For AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq), the 1:1 stoichiometry means that if 's' mol/L dissolves, then [Ag⁺] = [Br⁻] = s at equilibrium. The Ksp expression is Ksp = [Ag⁺][Br⁻] = s × s = s², so 2.5×10⁻¹³ = s², giving s = 5.0×10⁻⁷ mol/L. This value represents both the molar solubility and the concentration of each ion. A common error is using the Ksp value directly as the answer (choice B). For simple 1:1 salts on AP exams, molar solubility always equals √Ksp.