Solubility Product

$K_{sp}$ is the product of the concentrations of ions (in mol/dm³) raised to their stoichiometric coefficients in the ionization equation.

How to write Ksp expressions

1. Write the correct and balanced ionization equation.
2. Stoichiometric coefficients become the powers of respective ions.

For a general sparingly soluble salt $A_xB_y$:

$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$

$K_{sp} = [A^{y+}]^n [B^{x-}]^m$

Examples

1. $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-$

   $K_{sp} = [Al^{3+}] [OH^-]^3$

2. $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$

   $K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$

3. $Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}$

   $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$

Significance of Ksp

1. $K_{sp}$ helps predict whether a precipitate will form when ions are mixed in a solution.
2. If the product of ion concentrations exceeds $K_{sp}$, precipitation occurs.

Examples: determining Ksp from solubility

Example 1: AgI

Given:

Solubility of AgI = $1.22 \times 10^{-8}$ mol/L

Equation: $AgI \rightleftharpoons Ag^+ + I^-$

$K_{sp} = [Ag^+] [I^-] = (1.22 \times 10^{-8})^2 = 1.4884 \times 10^{-16} \text{ mol}^2 L^{-2}$

Example 2: PbCl₂

Given:

$[Pb^{2+}] = 1.62 \times 10^{-2}$ mol/L

Equation: $PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-$

$K_{sp} = [Pb^{2+}] [Cl^-]^2 = (1.62 \times 10^{-2})(2 \times 1.62 \times 10^{-2})^2 = 1.7005 \times 10^{-5} \text{ mol}^3 L^{-3}$

Example 3: PbCrO₄

Given:

Solubility of $PbCrO_4$ = $4.3 \times 10^{-5}$ g/L

Molar mass of $PbCrO_4$ = 323 g/mol

$\text{Moles of } PbCrO_4 = \frac{4.3 \times 10^{-5}}{323} = 1.33 \times 10^{-7} \text{ mol/L}$

$K_{sp} = [Pb^{2+}][CrO_4^{2-}] = (1.33 \times 10^{-7})^2 = 1.76 \times 10^{-14} \text{ mol}^2 L^{-2}$

Determining molar solubility from Ksp

Example 1: Ag₂CrO₄

Given:

$K_{sp} = 2.4 \times 10^{-12}$ M³

Equation: $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$

$K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2S)^2 (S) = 4S^3$

$2.4 \times 10^{-12} = 4S^3$

$S = 8.434 \times 10^{-5} \text{ mol/L}$

Example 2: CaF₂

Given:

$K_{sp} = 1.7 \times 10^{-10}$ M³

Equation: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$

$K_{sp} = [Ca^{2+}][F^-]^2 = S(2S)^2 = 4S^3$

$1.7 \times 10^{-10} = 4S^3$

$S = 3.489 \times 10^{-4} \text{ mol/L}$

These examples demonstrate how to calculate $K_{sp}$ and molar solubility for sparingly soluble salts.

The solubility of sparingly soluble salts is lowered by the presence of a second solute that furnishes (produces) common ions. Since the concentration of the common ion is higher than the equilibrium concentration, some ions will combine to restore the equilibrium (Le Chatelier's Principle).

The solubility equilibrium of CaF₂ is affected by adding either Ca²⁺ or F⁻ ions. This shifts the equilibrium to the left, reducing its solubility.

Find the molar solubility of CaF₂ ($K_{sp}$ = $3.9 \times 10^{-11}$ M³) in a solution containing 0.01M Ca(NO₃)₂.

Solution

Since Ca(NO₃)₂ dissociates completely:

$Ca(NO_3)_2(s) \rightarrow Ca^{2+}(aq) + 2NO_3^-(aq)$

Initial: 0.01 → 0.01 + (2 × 0.01)

For CaF₂, let the solubility be S:

$CaF_2(s) \rightarrow Ca^{2+}(aq) + 2F^-(aq)$

Equilibrium: S → S + 2S

From Ca(NO₃)₂, $[Ca^{2+}] = 0.01 + S$. Since $K_{sp}$ is small, we approximate S:

$K_{sp} = [Ca^{2+}][F^-]^2$

$3.9 \times 10^{-11} = 0.01 \times (2S)^2$

$3.9 \times 10^{-11} = 0.04 S^2$

$S^2 = 9.75 \times 10^{-10}$

$S = 3.122 \times 10^{-5} \text{ mol/L}$

Conclusion: The solubility of CaF₂ is reduced from $2.13 \times 10^{-4}$ M to $3.122 \times 10^{-5}$ M due to the common ion effect.

Calculate the mass of PbBr₂ that dissolves in 1 liter of 0.1M HBr at 25°C ($K_{sp}$ = $3.9 \times 10^{-8}$ M³, Pb = 207, Br = 80).

Solution

$PbBr_2(s) \rightarrow Pb^{2+}(aq) + 2Br^-(aq)$

Initial: 0, 0

Equilibrium: S, 2S + 0.1 (from HBr)

$K_{sp} = [Pb^{2+}][Br^-]^2$

$3.9 \times 10^{-8} = S \times (0.1)^2$

$S = 3.9 \times 10^{-6} \text{ mol/L}$

Mass of $PbBr_2$:

$M = S \times \text{Molar Mass}$

$M = (3.9 \times 10^{-6}) \times (207 + 2 \times 80)$

$M = 1.43 \times 10^{-3} \text{ g/L}$

Conclusion: The mass of PbBr₂ dissolved is $1.43 \times 10^{-3}$ g/L.

Precipitation occurs when the reaction quotient $Q_{sp}$ exceeds $K_{sp}$. This can be predicted as:

If $Q_{sp} < K_{sp}$: No precipitation (unsaturated solution).

If $Q_{sp} = K_{sp}$: Equilibrium (saturated solution).

If $Q_{sp} > K_{sp}$: Precipitation occurs.

Example 3: Will nickel carbonate precipitate?

The concentration of Ni²⁺ ions is $1.5 \times 10^{-6}$ M, and CO₃²⁻ is $6.04 \times 10^{-4}$ M. Will NiCO₃ precipitate ($K_{sp}$ = $6.6 \times 10^{-9}$)?

Solution

$Q_{sp} = [Ni^{2+}][CO_3^{2-}]$

$Q_{sp} = (1.5 \times 10^{-6})(6.04 \times 10^{-4})$

$Q_{sp} = 9.06 \times 10^{-10}$

Since $Q_{sp} < K_{sp}$, no precipitation occurs.