Write the K_sp expression for Ag_2CrO_4(s) 2Ag^+ + CrO_4^2-. [2 points]

Two solutions are mixed and the ion product Q is found to be greater than K_sp. State what happens. [1 point]

College Board AP 2023 (style)-style practice: Section II (long FRQ, part). The solubility product of PbCl_2 is K_sp = 1.6 × 10^-5. (a) Write the dissolution equation and the K_sp expression. (b) Let the molar solubility be s; write [Pb^2+] and [Cl^-] in terms of s. (c) Calculate the molar solubility s. (d) Justify why Cl^- is squared in the K_sp expression.

A 4-point quantitative FRQ on Ksp. (a) Equation and expression (1 point): PbCl_2(s) Pb^2+(aq) + 2Cl^-(aq); K_sp = [Pb^2+][Cl^-]^2 (the solid is omitted). (b) In terms of s (1 point): dissolving s mol/L gives [Pb^2+] = s and [Cl^-] = 2s. (c) Molar solubility (1 point): K_sp = (s)(2s)^2 = 4s^3 = 1.6 × 10^-5, so s^3 = 4.0 × 10^-6 and s = 1.6 × 10^-2 M. (d) Justify (1 point): each formula unit releases two Cl^- ions, so the coefficient of Cl^- is 2, and equilibrium expressions raise each concentration to its coefficient, giving [Cl^-]^2. Markers reward the equation and K_sp expression, the ion concentrations in terms of s, the molar solubility, and the squared-chloride reasoning.

College Board AP 2021 (style)-style practice: Section I (multiple choice). For AgCl(s) Ag^+(aq) + Cl^-(aq), the K_sp expression is (A) [Ag^+][Cl^-][AgCl] (B) [Ag^+][Cl^-] (C) [Ag^+] + [Cl^-] (D) 1[Ag^+][Cl^-]. Justify your choice.

A 1-point conceptual MCQ. The answer is (B). The pure solid AgCl is omitted from the equilibrium expression (its activity is constant), so K_sp = [Ag^+][Cl^-], the product of the ion concentrations. The trap is (A): including the solid in the denominator is incorrect.

United StatesChemistrySyllabus dot point

How does the solubility product constant describe the equilibrium of a slightly soluble salt, and how is it used?

Topic 7.11 Introduction to Solubility Equilibria: write the solubility product expression Ksp for a slightly soluble salt and relate Ksp to molar solubility and ion concentrations.

A focused answer to AP Chemistry Topic 7.11, covering the solubility product constant Ksp, writing the Ksp expression, relating Ksp to molar solubility, and using Q versus Ksp to predict precipitation, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A slightly soluble salt in water reaches an equilibrium between the solid and its dissolved ions. The solubility product $K_\text{sp}$ is the equilibrium constant for this dissolution: the product of the ion concentrations, each raised to its coefficient, with the pure solid omitted. For $\text{M}_a\text{X}_b(s) \rightleftharpoons a\text{M}^+ + b\text{X}^-$ , $K_\text{sp} = [\text{M}^+]^a[\text{X}^-]^b$ . The molar solubility $s$ is the moles of salt that dissolve per liter; you express each ion concentration in terms of $s$ using the stoichiometry, substitute into $K_\text{sp}$ , and solve. Comparing the ion product Q with $K_\text{sp}$ predicts precipitation: if $Q > K_\text{sp}$ a precipitate forms, if $Q < K_\text{sp}$ the solution is unsaturated, and if $Q = K_\text{sp}$ the solution is saturated.

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What this topic is asking
The solubility product expression
Molar solubility
Predicting precipitation with Q
Try this

What this topic is asking

The College Board (Topic 7.11) wants you to write the solubility product expression $K_\text{sp}$ for a slightly soluble salt and relate $K_\text{sp}$ to molar solubility and ion concentrations. This applies the general equilibrium machinery to the dissolving of ionic solids, a frequent quantitative theme on the exam.

The solubility product expression

So for $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ , $K_\text{sp} = [\text{Ag}^+][\text{Cl}^-]$ , and for $\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^-$ , $K_\text{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2$ . The coefficients become exponents, just as in any equilibrium expression. A smaller $K_\text{sp}$ means a less soluble salt.

Molar solubility

For $\text{PbCl}_2$ , dissolving $s$ mol/L gives $[\text{Pb}^{2+}] = s$ and $[\text{Cl}^-] = 2s$ (two chlorides per formula unit). Substituting, $K_\text{sp} = (s)(2s)^2 = 4s^3$ , which you solve for $s$ . The relationship between $K_\text{sp}$ and molar solubility depends on the salt's formula, so you must derive it each time from the stoichiometry rather than memorizing a single formula.

Predicting precipitation with Q

To decide whether a precipitate forms when solutions are mixed, compute the ion product $Q$ (the same expression as $K_\text{sp}$ but with the actual ion concentrations) and compare with $K_\text{sp}$ . If $Q > K_\text{sp}$ , the solution is supersaturated and a precipitate forms until $Q = K_\text{sp}$ . If $Q < K_\text{sp}$ , the solution is unsaturated and no precipitate forms. If $Q = K_\text{sp}$ , the solution is exactly saturated. This is the Q-versus-K rule applied to solubility.

Finding molar solubility from Ksp

Calculate the molar solubility of $\text{CaF}_2$ , given $K_\text{sp} = 3.9 \times 10^{-11}$ .

step 1 Write the dissolution and Ksp

$\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$ ; $K_\text{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$ .

step 2 Express ions in terms of s

$[\text{Ca}^{2+}] = s$ , $[\text{F}^-] = 2s$ .

step 3 Substitute

$K_\text{sp} = (s)(2s)^2 = 4s^3 = 3.9 \times 10^{-11}$ .

step 4 Solve for s

$s^3 = \dfrac{3.9 \times 10^{-11}}{4} = 9.75 \times 10^{-12}$ , so $s = (9.75 \times 10^{-12})^{1/3} = 2.1 \times 10^{-4}$ M.

Try this

Q1. Write the $K_\text{sp}$ expression for $\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$ . [2 points]

Cue. $K_\text{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$ .

Q2. Two solutions are mixed and the ion product $Q$ is found to be greater than $K_\text{sp}$ . State what happens. [1 point]

Cue. A precipitate forms until $Q$ falls to $K_\text{sp}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). The solubility product of

\text{PbCl}_2

K_\text{sp} = 1.6 \times 10^{-5}

. (a) Write the dissolution equation and the

K_\text{sp}

expression. (b) Let the molar solubility be

s

; write

[\text{Pb}^{2+}]

and

[\text{Cl}^-]

in terms of

s

. (c) Calculate the molar solubility

s

. (d) Justify why

\text{Cl}^-

is squared in the

K_\text{sp}

expression.

Show worked answer →

A 4-point quantitative FRQ on Ksp.

(a) Equation and expression (1 point): $\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)$ ; $K_\text{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2$ (the solid is omitted).
(b) In terms of s (1 point): dissolving $s$ mol/L gives $[\text{Pb}^{2+}] = s$ and $[\text{Cl}^-] = 2s$ .
(c) Molar solubility (1 point): $K_\text{sp} = (s)(2s)^2 = 4s^3 = 1.6 \times 10^{-5}$ , so $s^3 = 4.0 \times 10^{-6}$ and $s = 1.6 \times 10^{-2}$ M.
(d) Justify (1 point): each formula unit releases two $\text{Cl}^-$ ions, so the coefficient of $\text{Cl}^-$ is 2, and equilibrium expressions raise each concentration to its coefficient, giving $[\text{Cl}^-]^2$ .

Markers reward the equation and $K_\text{sp}$ expression, the ion concentrations in terms of $s$ , the molar solubility, and the squared-chloride reasoning.

AP 2021 (style)1 marksSection I (multiple choice). For

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

, the

K_\text{sp}

expression is (A)

\dfrac{[\text{Ag}^+][\text{Cl}^-]}{[\text{AgCl}]}

(B)

[\text{Ag}^+][\text{Cl}^-]

(C)

[\text{Ag}^+] + [\text{Cl}^-]

(D)

\dfrac{1}{[\text{Ag}^+][\text{Cl}^-]}

. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The pure solid AgCl is omitted from the equilibrium expression (its activity is constant), so $K_\text{sp} = [\text{Ag}^+][\text{Cl}^-]$ , the product of the ion concentrations. The trap is (A): including the solid in the denominator is incorrect.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)