Introduction to Solubility Equilibria - AP Chemistry

1. 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?
   Answer: $1.0\times10^{-5}\,\text{M}$ 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.
2. Excess solid barium sulfate, $\text{BaSO}_4(s)$, is added to pure water at $25^\circ\text{C}$. The equilibrium is $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$ with $K_{sp}=1.0\times10^{-10}$. What is the equilibrium concentration of $\text{Ba}^{2+}$ in the saturated solution (in $\text{mol}\cdot\text{L}^{-1}$)?
   Answer: $1.0\times10^{-5}\ \text{mol}\cdot\text{L}^{-1}$ This problem tests solubility equilibria (quantitative). For BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq), the dissolution produces equal concentrations of both ions, so if 's' mol/L dissolves, then [Ba²⁺] = [SO₄²⁻] = s at equilibrium. The Ksp expression is Ksp = [Ba²⁺][SO₄²⁻] = s × s = s², so 1.0×10⁻¹⁰ = s², giving s = 1.0×10⁻⁵ mol/L. Since the question asks for [Ba²⁺], which equals s, the answer is 1.0×10⁻⁵ mol/L. A common mistake is confusing Ksp with the ion concentration directly. For 1:1 salts, the equilibrium concentration of each ion equals the square root of Ksp.
3. Solid barium sulfate, $\text{BaSO}_4(s)$, is added to pure water at $25^\circ\text{C}$ until equilibrium is established. $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$ For this compound, $K_{sp}=1.0\times10^{-10}$. What is the equilibrium concentration of $\text{Ba}^{2+}(aq)$ in the saturated solution?
   Answer: $1.0\times10^{-5}\ \text{M}$ This problem tests solubility equilibria (quantitative). For BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq), this is a 1:1 salt where both ions have equal concentrations at equilibrium. If x mol/L dissolves, then [Ba²⁺] = [SO₄²⁻] = x, and Ksp = [Ba²⁺][SO₄²⁻] = x² = 1.0×10⁻¹⁰. Taking the square root gives x = 1.0×10⁻⁵ M for [Ba²⁺]. A common error is using the Ksp value directly as the concentration (choice A). For 1:1 salts, the calculation reduces to a simple square root of Ksp, making these the most straightforward solubility problems.
4. 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 $\text{Cl}^-(aq)$? | | $\text{Cu}^+$ | $\text{Cl}^-$ | | - | ------------- | ------------- | | I | 0 | 0 | | C | +$s$ | +$s$ | | E | $s$ | $s$ | ICE table (in M):
   Answer: $1.0\times10^{-3}\,\text{M}$ 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.
5. 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 solubility-product constant is $K_{sp}(\text{AgCl}) = 1.6\times 10^{-10}$. Assuming the initial ion concentrations are zero, what is the equilibrium concentration of $\text{Ag}^+(aq)$? Use the following ICE table setup:

   	$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$	$[\text{Ag}^+]$ (M)	$[\text{Cl}^-]$ (M)
   I		0	0
   C		$+s$	$+s$
   E		$s$	$s$

   Answer: $1.3\times 10^{-5}\,\text{M}$ This problem tests your understanding of solubility equilibria (quantitative). When AgCl dissolves, it produces equal moles of Ag⁺ and Cl⁻, so if we let s = molar solubility, then [Ag⁺] = [Cl⁻] = s at equilibrium. The Ksp expression is Ksp = [Ag⁺][Cl⁻] = s × s = s². Substituting the given Ksp value: 1.6 × 10⁻¹⁰ = s², so s = √(1.6 × 10⁻¹⁰) = 1.26 × 10⁻⁵ M ≈ 1.3 × 10⁻⁵ M. A common error is using Ksp directly as the concentration (choice A), forgetting to take the square root. For 1:1 salts like AgCl, always remember that Ksp = s² where s is the molar solubility.
6. 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 $\text{mol}\cdot\text{L}^{-1}$)?
   Answer: $3.0\times10^{-5}\ \text{mol}\cdot\text{L}^{-1}$ 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.
7. Solid silver chloride, $\text{AgCl}(s)$, is added to pure water at $25^\circ\text{C}$ until equilibrium is established. The dissolution is $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ with $K_{sp}=1.6\times10^{-10}$. What is the equilibrium concentration of $\text{Ag}^+(aq)$ in the saturated solution?
   Answer: $1.3\times10^{-5}\ \text{M}$ 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 x since the dissolution produces one Ag⁺ and one Cl⁻ per formula unit. The Ksp expression is Ksp = [Ag⁺][Cl⁻] = x·x = x², so 1.6×10⁻¹⁰ = x², giving x = 1.3×10⁻⁵ M. A common error is using Ksp directly as the concentration (choice A), forgetting to take the square root. For 1:1 salts like AgCl, the molar solubility equals the concentration of either ion, making these problems straightforward single-variable calculations.
8. 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:

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

   Answer: $2.0\times 10^{-4}\,\text{M}$ 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.
9. Excess solid copper(I) chloride, $\text{CuCl}(s)$, is added to pure water at $25^\circ\text{C}$. The equilibrium is $\text{CuCl}(s) \rightleftharpoons \text{Cu}^+(aq) + \text{Cl}^-(aq)$ with $K_{sp}=1.0\times10^{-6}$. What is the molar solubility of $\text{CuCl}$ (in $\text{mol}\cdot\text{L}^{-1}$)?
   Answer: $1.0\times10^{-3}\ \text{mol}\cdot\text{L}^{-1}$ This problem tests solubility equilibria (quantitative). For CuCl(s) ⇌ Cu⁺(aq) + Cl⁻(aq), the 1:1 stoichiometry means if 's' mol/L dissolves, then [Cu⁺] = [Cl⁻] = s at equilibrium. The Ksp expression is Ksp = [Cu⁺][Cl⁻] = s × s = s², so 1.0×10⁻⁶ = s², giving s = 1.0×10⁻³ mol/L. This represents both the molar solubility and the concentration of each ion in solution. A common mistake is confusing Ksp with solubility or taking the wrong root. For 1:1 salts, molar solubility is simply the square root of the Ksp value.
10. A student adds excess solid strontium sulfate, $\text{SrSO}_4(s)$, to pure water at $25^\circ\text{C}$. The solubility product constant is $K_{sp}(\text{SrSO}_4)=1.0\times10^{-6}$. What is the molar solubility of $\text{SrSO}_4$ in water?
    Answer: $1.0\times10^{-3}\ \text{mol\,L}^{-1}$ This problem tests solubility equilibria (quantitative). For SrSO₄(s) ⇌ Sr²⁺(aq) + SO₄²⁻(aq), the molar solubility s equals the concentration of each ion since they form in a 1:1 ratio. The Ksp expression is Ksp = [Sr²⁺][SO₄²⁻] = s² = 1.0×10⁻⁶. Solving gives s = √(1.0×10⁻⁶) = 1.0×10⁻³ M. Choice E (1.0×10⁻¹² M) is incorrect because it might result from squaring the Ksp instead of taking its square root. For 1:1 salts, molar solubility is always the square root of Ksp.
11. 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.)
    Answer: $1.3\times10^{-5}\,\text{M}$ 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.
12. 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 $\text{mol}\cdot\text{L}^{-1}$)?
    Answer: $1.3\times10^{-5}\ \text{mol}\cdot\text{L}^{-1}$ 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.
13. 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.)
    Answer: $1.3\times10^{-5}\,\text{M}$ 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.
14. 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 | $s$ | $s$ |​
    Answer: $1.3\times10^{-5}\,\text{M}$ 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.
15. Excess solid silver bromide, $\text{AgBr}(s)$, is shaken with pure water at $25^\circ\text{C}$ until equilibrium is reached. The solubility product constant is $K_{sp}(\text{AgBr})=2.5\times10^{-13}$. What is the equilibrium concentration of $\text{Br}^-(aq)$? (Assume volume change is negligible.)
    Answer: $5.0\times10^{-7}\,\text{M}$ This question tests your understanding of solubility equilibria (quantitative). The dissolution equation is AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq). To construct the ICE table, we set initial concentrations of Ag⁺ and Br⁻ 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 [Ag⁺][Br⁻] = x · x = x² = 2.5×$10^{-13}$, so we solve for the single variable x by taking the square root to get x = 5.0×$10^{-7}$ M. We use only one variable because the equal stoichiometric coefficients link the ion concentrations directly. A tempting distractor is 2.5×$10^{-13}$ M (choice A), which is just the Ksp value 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.
16. Excess solid silver bromide, $\text{AgBr}(s)$, is added to pure water at $25^\circ\text{C}$. The value of $K_{sp}(\text{AgBr})$ is $2.5\times10^{-13}$. What is the equilibrium concentration of $\text{Br}^-(aq)$?
    Answer: $5.0\times10^{-7}\ \text{M}$ This problem tests solubility equilibria (quantitative). For AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq), both ions form in a 1:1 ratio, so [Ag⁺] = [Br⁻] = x at equilibrium. The Ksp expression is Ksp = [Ag⁺][Br⁻] = x² = 2.5×10⁻¹³. Solving gives x = √(2.5×10⁻¹³) = 5.0×10⁻⁷ M. Choice A (2.5×10⁻¹³ M) is incorrect because it's the Ksp value itself, not the square root. For 1:1 salts like AgBr, the ion concentrations equal the square root of Ksp.
17. 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 $\text{mol}\cdot\text{L}^{-1}$)?
    Answer: $5.0\times10^{-7}\ \text{mol}\cdot\text{L}^{-1}$ 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.
18. Solid silver chloride, $\text{AgCl}(s)$, is added to pure water at $25^\circ\text{C}$ until equilibrium is established. The solubility-product constant is $K_{sp}=1.6\times10^{-10}$ for the reaction $\text{AgCl}(s)\rightleftharpoons \text{Ag}^+(aq)+\text{Cl}^-(aq)$. What is the equilibrium concentration of $\text{Ag}^+(aq)$ in the saturated solution?
    Answer: $1.3\times10^{-5}\ \text{M}$ 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 AgCl dissociates in a 1:1 ratio, [Ag⁺] = [Cl⁻] = x at equilibrium. The Ksp expression becomes Ksp = [Ag⁺][Cl⁻] = x·x = x² = 1.6×10⁻¹⁰. Solving for x gives x = √(1.6×10⁻¹⁰) = 1.26×10⁻⁵ M ≈ 1.3×10⁻⁵ M. A common error (choice E) would be to take 2x thinking both ions contribute separately. For 1:1 salts like AgCl, the molar solubility equals the ion concentration, making these problems straightforward single-variable calculations.
19. Solid silver bromide, $\text{AgBr}(s)$, is added to pure water at $25^\circ\text{C}$ until equilibrium is established. $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$ For this compound, $K_{sp}=2.5\times10^{-13}$. What is the molar solubility of $\text{AgBr}$ in water?
    Answer: $5.0\times10^{-7}\ \text{M}$ This problem tests solubility equilibria (quantitative). For AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq), this is a 1:1 salt where molar solubility equals the concentration of either ion. Setting up the ICE table with x as the molar solubility gives [Ag⁺] = [Br⁻] = x at equilibrium. The Ksp expression is Ksp = [Ag⁺][Br⁻] = x² = 2.5×10⁻¹³, so x = √(2.5×10⁻¹³) = 5.0×10⁻⁷ M. A common mistake is using Ksp directly (choices A or D) without taking the square root. For simple 1:1 salts on AP exams, always remember that molar solubility = √Ksp.
20. 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?
    Answer: $6.3\times10^{-7}\,\text{M}$ 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.

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