FAD1018 Solubility Product ($K_{sp}$) — Step-by-Step

Seven question types. Each one has the same first step: write the dissolution equation, then follow the recipe.

Step 0: Write the Dissociation Equation + $K_{sp}$ Expression

Salt Dissociation $K_{sp}$
AgCl $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ $K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s^2$
CaF$_2$ $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-$ $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = 4s^3$
Ag$_2$CrO$_4$ $\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$ $K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = 4s^3$
Fe(OH)$_3$ $\text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+} + 3\text{OH}^-$ $K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 = 27s^4$
Ca$_3$(PO$_4$)$_2$ $\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-}$ $K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 = 108s^5$

For a general salt A$_a$B$b$: $K{sp} = (as)^a(bs)^b$

Type 1: Given molar solubility $s$, find $K_{sp}$

Step Action
1 Write dissociation equation
2 Express each ion's concentration in terms of $s$
3 Substitute into $K_{sp} = [\text{cation}]^a[\text{anion}]^b$
4 Compute

Worked — PbS

Solubility of PbS = $8.36 \times 10^{-14}$ mol L$^{-1}$. Find $K_{sp}$.

$$\text{PbS}(s) \rightleftharpoons \text{Pb}^{2+} + \text{S}^{2-}$$ $$[\text{Pb}^{2+}] = s, \quad [\text{S}^{2-}] = s$$ $$K_{sp} = (s)(s) = (8.36 \times 10^{-14})^2 = 6.99 \times 10^{-27}$$

Worked — Pb(IO$_3$)$_2$

Molar solubility of Pb(IO$_3$)$2$ = $4.0 \times 10^{-5}$ mol L$^{-1}$. Find $K{sp}$.

$$\text{Pb(IO}_3)_2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{IO}_3^-$$ $$[\text{Pb}^{2+}] = s, \quad [\text{IO}3^-] = 2s$$ $$K{sp} = (s)(2s)^2 = 4s^3 = 4(4.0 \times 10^{-5})^3 = 2.56 \times 10^{-13}$$

Type 2: Given $K_{sp}$, find molar solubility $s$

Step Action
1 Write dissociation equation
2 Let $s$ = molar solubility; express each ion as $as$ or $bs$
3 $K_{sp} = (as)^a(bs)^b$, solve for $s$

Quick formulas

Ratio $K_{sp} =$ $s =$
1:1 (AgCl) $s^2$ $\sqrt{K_{sp}}$
1:2 (CaF$_2$) $4s^3$ $\sqrt[3]{K_{sp}/4}$
1:3 (Fe(OH)$_3$) $27s^4$ $\sqrt[4]{K_{sp}/27}$
2:3 (Ca$_3$(PO$_4$)$_2$) $108s^5$ $\sqrt[5]{K_{sp}/108}$

Worked — Ag$_2$CrO$_4$

$K_{sp} = 1.9 \times 10^{-12}$. Find $[\text{Ag}^+]$ and $[\text{CrO}_4^{2-}]$ in saturated solution.

$$\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+ + \text{CrO}4^{2-}$$ $$K{sp} = (2s)^2(s) = 4s^3 = 1.9 \times 10^{-12}$$ $$s = \sqrt[3]{\frac{1.9 \times 10^{-12}}{4}} = 7.8 \times 10^{-5} \text{ M}$$ $$[\text{Ag}^+] = 2s = 1.6 \times 10^{-4} \text{ M}, \quad [\text{CrO}_4^{2-}] = s = 7.8 \times 10^{-5} \text{ M}$$

Type 3: Solubility in g L$^{-1}$ from $K_{sp}$

Step Action
1 Find molar solubility $s$ (mol L$^{-1}$) using Type 2
2 Multiply by molar mass: $\text{solubility (g L}^{-1}) = s \times M$

Worked — AgCl

$K_{sp} = 1.56 \times 10^{-10}$, molar mass = 143.5 g mol$^{-1}$. Find solubility in g L$^{-1}$.

$$s = \sqrt{K_{sp}} = \sqrt{1.56 \times 10^{-10}} = 1.25 \times 10^{-5} \text{ M}$$ $$\text{solubility} = 1.25 \times 10^{-5} \times 143.5 = 1.79 \times 10^{-3} \text{ g L}^{-1}$$

Type 4: Given solubility in g L$^{-1}$, find $K_{sp}$

Step Action
1 Convert to molar solubility: $s = \dfrac{\text{g L}^{-1}}{\text{molar mass}}$
2 Then use Type 1 to find $K_{sp}$

Type 5: Will a precipitate form? ($Q$ vs $K_{sp}$)

Step Action
1 Write the dissolution equation
2 Calculate initial ion concentrations after mixing (account for dilution!)
3 Compute $Q = [\text{cation}]_0^a[\text{anion}]_0^b$
4 Compare: $Q < K_{sp}$ = no precipitate; $Q > K_{sp}$ = precipitate forms

Worked — Mixed Solutions

200 mL of 0.004 M BaCl$_2$ + 600 mL of 0.008 M K$_2$SO$_4$. Will BaSO$4$ precipitate? ($K{sp} = 1.1 \times 10^{-10}$)

Total volume = 0.8 L. After mixing: $$[\text{Ba}^{2+}] = \frac{0.004 \times 0.200}{0.800} = 0.001 \text{ M}$$ $$[\text{SO}4^{2-}] = \frac{0.008 \times 0.600}{0.800} = 0.006 \text{ M}$$ $$Q = (0.001)(0.006) = 6 \times 10^{-6}$$ $$Q > K{sp} \quad \rightarrow \quad \text{Precipitate forms}$$

Worked — [OH$^-$] given

$[\text{Mg}^{2+}] = 1.5 \times 10^{-6}$ M, $[\text{OH}^-] = 1.0 \times 10^{-4}$ M. Will Mg(OH)$2$ precipitate? ($K{sp} = 5.6 \times 10^{-12}$)

$$Q = (1.5 \times 10^{-6})(1.0 \times 10^{-4})^2 = 1.5 \times 10^{-14}$$ $$Q < K_{sp} \quad \rightarrow \quad \text{No precipitate (unsaturated)}$$

Type 6: Common-Ion Effect — find $s$ in presence of common ion

Step Action
1 Write dissociation equation
2 Let $s$ = molar solubility
3 Set up ICE table — the common ion has an initial concentration $\ne 0$
4 Assume $s \ll C_{\text{common}}$, so $C_{\text{common}} + s \approx C_{\text{common}}$
5 Substitute into $K_{sp}$ and solve for $s$
6 Compare to $s$ in pure water (should be much smaller)

Worked — Mg(OH)$_2$ in 0.1 M NaOH

$K_{sp} = 1.2 \times 10^{-11}$. Find molar solubility of Mg(OH)$_2$ in 0.1 M NaOH.

Mg(OH)$_2(s)$ $\rightleftharpoons$ Mg$^{2+}$ 2OH$^-$
Initial — 0 0.1 (from NaOH)
Change $-s$ $+s$ $+2s$
Equilibrium — $s$ $0.1 + 2s \approx 0.1$

$$K_{sp} = (s)(0.1)^2 = 1.2 \times 10^{-11}$$ $$s = \frac{1.2 \times 10^{-11}}{0.01} = 1.2 \times 10^{-9} \text{ M}$$

In pure water: $s = \sqrt[3]{1.2 \times 10^{-11}/4} = 1.44 \times 10^{-4}$ M. $10^5$ times more soluble without common ion.

Worked — AgI in 0.274 M NaI

$K_{sp} = 8.52 \times 10^{-17}$. Find molar solubility of AgI in 0.274 M NaI.

$$\text{AgI}(s) \rightleftharpoons \text{Ag}^+ + \text{I}^-$$ $$[\text{Ag}^+] = s, \quad [\text{I}^-] = 0.274 + s \approx 0.274$$ $$K_{sp} = (s)(0.274) = 8.52 \times 10^{-17}$$ $$s = \frac{8.52 \times 10^{-17}}{0.274} = 3.11 \times 10^{-16} \text{ M}$$

In pure water: $s = \sqrt{8.52 \times 10^{-17}} = 9.23 \times 10^{-9}$ M.

Type 7: Selective Precipitation — which salt precipitates first?

Step Action
1 For each salt, write its $K_{sp}$ expression
2 Calculate $[\text{reagent}]$ needed to begin precipitation: solve $K_{sp} = [\text{cation}][\text{reagent}]^b$
3 The one needing smaller $[\text{reagent}]$ precipitates first
4 Separation window: $[\text{reagent}]$ between the two values

Worked — Ag$^+$ + Pb$^{2+}$ with Cl$^-$

Solution: $1.0 \times 10^{-2}$ M Ag$^+$ and $2.0 \times 10^{-2}$ M Pb$^{2+}$. Add Cl$^-$. AgCl ($K_{sp}=1.8\times10^{-10}$), PbCl$2$ ($K{sp}=1.7\times10^{-5}$).

AgCl begins to precipitate when: $$[\text{Cl}^-] = \frac{K_{sp}}{[\text{Ag}^+]} = \frac{1.8 \times 10^{-10}}{1.0 \times 10^{-2}} = 1.8 \times 10^{-8} \text{ M}$$

PbCl$2$ begins to precipitate when: $$[\text{Cl}^-] = \sqrt{\frac{K{sp}}{[\text{Pb}^{2+}]}} = \sqrt{\frac{1.7 \times 10^{-5}}{2.0 \times 10^{-2}}} = 2.9 \times 10^{-2} \text{ M}$$

AgCl precipitates first (needs $10^6$ times less Cl$^-$).

Separation window: keep $[\text{Cl}^-]$ between $1.8 \times 10^{-8}$ M and $2.9 \times 10^{-2}$ M.

[!tip] Same stoichiometry? Smaller $K_{sp}$ = precipitates first. Different stoichiometry? You must calculate — don't guess.

Quick Pick — Which Type?

graph TD
    Q["What is the question asking?"]
    Q -->|"Find Ksp"| T1["Type 1: s → Ksp<br/>Substitute s into Ksp expression"]
    Q -->|"Find molar solubility"| T2{"Given Ksp?"}
    T2 -->|Yes| T2a["Type 2: Ksp → s<br/>Solve (as)ᵃ(bs)ᵇ = Ksp"]
    T2 -->|No, g/L given| T2b["Type 4: g/L → s → Ksp"]
    Q -->|"Solubility in g/L"| T3["Type 3: s × molar mass"]
    Q -->|"Will ppt form?"| T5["Type 5: Compute Q, compare to Ksp"]
    Q -->|"Solubility with common ion"| T6["Type 6: ICE table with initial [common ion]"]
    Q -->|"Which precipitates first?"| T7["Type 7: Calculate [reagent] needed for each"]

Common Exam Traps

Trap Fix
Forgetting to account for dilution when mixing solutions Calculate $C_{\text{new}} = C_1V_1/V_{\text{total}}$
Using wrong exponent in $K_{sp}$ expression Exponent = stoichiometric coefficient: CaF$_2$ → $[\text{F}^-]^2$, not $[\text{F}^-]$
Using $K_{sp}$ formula for wrong ratio AgCl: $s^2$; CaF$_2$: $4s^3$; Fe(OH)$_3$: $27s^4$
Thinking smaller $K_{sp}$ always means less soluble Only true for same stoichiometry. Compare AgCl ($s^2$) to Ag$_2$CrO$_4$ ($4s^3$) by calculating $s$
Common-ion effect: forgetting the common ion is already present ICE table: common ion starts at its given concentration, not zero
Selective precipitation: comparing $K_{sp}$ values for different stoichiometries Must calculate $[\text{reagent}]$ — PbCl$2$ has larger $K{sp}$ but needs higher [Cl$^-$] to begin precipitating

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