Solubility Product ($K_{sp}$)

The equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. $K_{sp}$ indicates how soluble the solid is in water.

Definition

For a general dissolution:

[A].[B]

$$\text{A}_a\text{B}_b(s) \rightleftharpoons a\text{A}^{m+}(aq) + b\text{B}^{n-}(aq)$$

Molar Solubility
$s$

$$K_{sp} = [\text{A}^{m+}]^a[\text{B}^{n-}]^b = (as)^a(bs)^b$$

$$\text{unit } K_{sp} = (\text{concentration})^{a+b}$$

  • Pure solid is omitted (activity = 1)
  • $K_{sp}$ is temperature-dependent
  • The smaller the $K_{sp}$ value, the less soluble the compound in water
  • $K_{sp} \downarrow$, solubility $\downarrow$; $T \uparrow$, solubility $\uparrow$, $K_{sp} \uparrow$

Solubility vs Molar Solubility

Solubility Molar Solubility
Definition Maximum amount of solute (g) that dissolves in a known volume of solvent (L) to form a saturated solution Number of moles of solute in 1 L (or 1 dm³) of saturated solution
Unit g L⁻¹ or g dm⁻³ mol L⁻¹ or mol dm⁻³
Conversion $\div$ Molar mass $\times$ Molar mass

[!important] $K_{sp}$ is not the same as solubility $K_{sp}$ is an equilibrium constant; solubility is a concentration.

Molar Solubility Calculations

1:1 Ratio (e.g., AgCl)

[Ag+].[Cl-]

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$$ $$K_{sp} = s^2 \quad \Rightarrow \quad s = \sqrt{K_{sp}}$$

1:2 or 2:1 Ratio (e.g., CaF₂)

[Ca+2].[F-].[F-]

$$\text{CaF}2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$$ $$K{sp} = 4s^3 \quad \Rightarrow \quad s = \sqrt[3]{\frac{K_{sp}}{4}}$$

1:3 or 3:1 Ratio (e.g., LaF₃)

[La+3].[F-].[F-].[F-]

$$\text{LaF}3(s) \rightleftharpoons \text{La}^{3+}(aq) + 3\text{F}^-(aq)$$ $$s = \sqrt[4]{\frac{K{sp}}{27}}$$

Common Ion Effect

If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium shifts to the left (Le Chatelier's principle) and the solubility of the salt decreases.

In general, the solubility of a slightly soluble salt decreases in the presence of a second solute that provides a common ion.

[!important] $K_{sp}$ remains constant At a given temperature, only the solubility is altered by the common-ion effect. The solubility product ($K_{sp}$), being an equilibrium constant, remains the same whether or not other substances are present.

Example — AgI in NaI:

[Ag+].[I-]
  • In pure water: $s = 9.23 \times 10^{-9}$ M
  • In 0.274 M NaI: $s = 3.11 \times 10^{-16}$ M

The molar solubility of AgI is much smaller in NaI solution due to the common ion $\text{I}^-$.

Example — CaF₂:

  • In pure water: $s = 2.1 \times 10^{-4}$ M
  • In 0.010 M $\text{Ca}^{2+}$: $s = 3.1 \times 10^{-5}$ M
  • In 0.010 M $\text{F}^-$: $s = 3.9 \times 10^{-7}$ M

The effect of $\text{F}^-$ is more pronounced than $\text{Ca}^{2+}$ because $[\text{F}^-]$ appears to the second power in the $K_{sp}$ expression for $\text{CaF}_2$, whereas $\text{Ca}^{2+}$ appears to the first power.

Solubility Quotient ($Q$)

The ion product $Q$ (also called the solubility quotient) uses initial concentrations (not equilibrium concentrations). Its expression has the same form as $K_{sp}$:

$$Q = [\text{A}^+]_0[\text{B}^-]_0$$

The subscript 0 reminds us that these are initial concentrations.

Precipitation Prediction

Condition Meaning
$Q < K_{sp}$ Unsaturated solution; no precipitate forms
$Q = K_{sp}$ Saturated solution; at equilibrium (point of precipitation)
$Q > K_{sp}$ Supersaturated solution; precipitate will form

A precipitate will form only when $Q > K_{sp}$.

Procedure for mixed solutions:

  1. Calculate moles of each ion before mixing
  2. Find total volume after mixing
  3. Calculate new concentrations
  4. Compute $Q$ and compare to $K_{sp}$

Selective Precipitation

Separating ions by controlled precipitation based on different $K_{sp}$ values. Separation by using a reagent that forms a precipitate with one or more (but not all) ions is called selective precipitation.

Example — separating Ag⁺ and Pb²⁺ by adding Cl⁻:

Solution contains $1.0 \times 10^{-2}$ M $\text{Ag}^+$ and $2.0 \times 10^{-2}$ M $\text{Pb}^{2+}$.

  • $\text{AgCl}$ ($K_{sp} = 1.8 \times 10^{-10}$)
  • $\text{PbCl}2$ ($K{sp} = 1.7 \times 10^{-5}$)
    [Pb+2].[Cl-].[Cl-]

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

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

Separation window: Keep $[\text{Cl}^-]$ between $1.8 \times 10^{-8}$ M and $2.9 \times 10^{-2}$ M to selectively precipitate $\text{Ag}^+$ while $\text{Pb}^{2+}$ remains in solution.

[!tip] Which salt precipitates first? The salt with the smaller $K_{sp}$ precipitates first only if the stoichiometry is the same. For different stoichiometries, calculate the required reagent concentration for each.

Common $K_{sp}$ Values (at 25°C)

Compound $K_{sp}$
$\text{Al(OH)}_3$ $1.8 \times 10^{-33}$
$\text{BaCO}_3$ $8.1 \times 10^{-9}$
$\text{BaSO}_4$ $1.1 \times 10^{-10}$
$\text{CaCO}_3$ $8.7 \times 10^{-9}$
$\text{CaF}_2$ $4.0 \times 10^{-11}$
$\text{Ca(OH)}_2$ $8.0 \times 10^{-6}$
$\text{Ca}_3(\text{PO}_4)_2$ $1.2 \times 10^{-26}$
$\text{Fe(OH)}_3$ $1.1 \times 10^{-36}$
$\text{Mg(OH)}_2$ $1.2 \times 10^{-11}$
AgCl $1.6 \times 10^{-10}$
AgBr $7.7 \times 10^{-13}$
AgI $8.3 \times 10^{-17}$
$\text{Ag}_2\text{CrO}_4$ $1.4 \times 10^{-5}$
$\text{Ag}_2\text{S}$ $6.0 \times 10^{-51}$
$\text{PbCl}_2$ $2.4 \times 10^{-4}$
$\text{PbCrO}_4$ $2.0 \times 10^{-14}$
$\text{PbS}$ $3.4 \times 10^{-28}$
$\text{Zn(OH)}_2$ $1.8 \times 10^{-14}$

Related Topics

  • Ionic Equilibria — Acid-base equilibria
  • Chemical Equilibrium — General equilibrium concepts
  • Le Chatelier's Principle — Equilibrium shifts
  • Solubility Rules — Predicting precipitation

Sources