Solubility product (Ksp)

Even 'insoluble' ionic compounds may dissolve to a very small extent in water.

An equilibrium is established between the undissolved solid and its ions in a saturated solution:

M_aX_b(s) ⇌ aMⁿ⁺_(aq) + bX^m-_(aq)

The equilibrium constant expression for this process is:

${\text{K}}_{\text{c}}=\frac{{[{\text{M}}_{\text{(aq)}}^{\text{n}+}]}^{\text{a}}{[{\text{X}}_{\text{(aq)}}^{\text{m}-}]}^{\text{b}}}{[{\text{M}}_{\text{a}}{\text{X}}_{\text{b(s)}}]}$

As the concentration of a solid is constant, it can be combined with K_c to give the solubility product (K_sp):

K_sp = [Mⁿ⁺_(aq)]^a [X^m-_(aq)]^b

Where:

• a = number of Mⁿ⁺ cations in one formula unit.
• b = number of X^m- anions in one formula unit.

For example, the K_sp expression for Fe₂S₃ is:

K_sp = [Fe³⁺_(aq)]² [S^2-_(aq)]³

The units of K_sp depend on the number of each type of ion in the formula, for example, the units of K_sp for Fe₂S₃ are mol⁵ dm^-15.

The lower the K_sp value, the lower the solubility of the compound. K_sp is only applicable to sparingly soluble salts.

Worked example 1 - Calculating Ksp from solubility

Calculate the K_sp of silver bromide (AgBr) if its solubility is $5.71\times {10}^{-7}{\text{ mol dm}}^{-3}$ at 298 K. Give your answer to 3 significant figures.

Step 1: Write the equilibrium equation

AgBr_(s) ⇌ Ag⁺_(aq) + Br^-_(aq)

Step 2: Calculate the concentration of each ion

$[{\text{Ag}}^{+}]=[{\text{Br}}^{-}]=5.71\times {10}^{-7}{\text{ mol dm}}^{-3}$

Step 3: Write the K_sp expression

K_sp = [Ag⁺][Br^-]

Step 4: Substitution and correct evaluation

${\text{K}}_{\text{sp}}=(5.71\times {10}^{-7})\times (5.71\times {10}^{-7})=3.26\times {10}^{-13}$

Step 5: Determine the units

K_sp will have units of (mol dm^-3)² = mol² dm^-6.

Therefore, the K_sp of silver bromide is 3.26 x 10^-13 mol² dm^-6.

Worked example 2 - Calculating solubility from Ksp

Calculate the solubility of lead chloride (PbCl₂) given that K_sp for PbCl₂ is $1.7\times {10}^{-5}$ mol³ dm^-9 at 298 K. Give your answer to 2 significant figures.

Step 1: Write the equilibrium equation

PbCl_2(s) ⇌ Pb²⁺_(aq) + 2Cl^-_(aq)

Step 2: Write the K_sp expression

Let the solubility of PbCl₂ = $x{\text{ mol dm}}^{-3}$

${\text{K}}_{\text{sp}}=[{\text{Pb}}^{2+}]{[{\text{Cl}}^{-}]}^2=x\times 2x^2=4x^3$

Step 3: Substitution and correct evaluation

$1.7\times {10}^{-5}=4x^3$

$x=\sqrt[3]{\frac{(1.7\times {10}^{-5})}{4}}=1.6\times {10}^{-2}{\text{ mol dm}}^{-3}$

${\text{The solubility of PbCl}}_2\text{ is }1.6\times {10}^{-2}{\text{ mol dm}}^{-3}$

Predicting precipitation using K_sp

K_sp can be used to predict if a precipitate will form when two solutions are mixed.

• If the product of [Mⁿ⁺] and [X^m-] exceeds K_sp, precipitation occurs.
• If the product is less than K_sp, no precipitate forms.

Worked example 3 - Predicting precipitation using Ksp

Determine if a precipitate of CaF₂ will form when 100 cm³ of 1.00 x 10^-3 mol dm^-3 Ca(NO₃)_2(aq) is mixed with 200 cm³ of 1.50 x 10^-3 mol dm^-3 NaF_(aq).

K_sp of CaF₂ = 3.9 x 10^-11 mol³ dm^-9.

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

100 cm³ = 0.100 dm³

200 cm³ = 0.200 dm³

Step 2: Calculate number of moles of Ca²⁺ and F^- before mixing

Ca²⁺: $\text{n }=\text{ c}\times \text{ V}=(1.00\times {10}^{-3})\times 0.100=1.00\times {10}^{-4}\text{ mol}$

F^-:$\text{ n }=\text{ c}\times \text{ V}=(1.50\times {10}^{-3})\times 0.200=3.00\times {10}^{-4}\text{ mol}$

Step 3: Calculate [Ca²⁺] and [F^-]after mixing

Total volume after mixing = 0.300 dm³

$[{\text{Ca}}^{2+}]=\frac{\text{n}}{\text{V}}=\frac{(1.00\times {10}^{-4})}{0.300}=3.33\times {10}^{-4}{\text{ mol dm}}^{-3}$

$[{\text{F}}^{-}]=\frac{\text{n}}{\text{V}}=\frac{(3.00\times {10}^{-4})}{0.300}=1.00\times {10}^{-3}{\text{ mol dm}}^{-3}$

Step 4: Calculate [Ca²⁺][F^-]²

$[{\text{Ca}}^{2+}]{[{\text{F}}^{-}]}^2=(3.33\times {10}^{-4})\times {(1.00\times {10}^{-3})}^2=3.33\times {10}^{-10}{\text{ mol}}^3{\text{ dm}}^{-9}$

Step 5: Compare [Ca²⁺][F^-]²with K_sp

3.3 x 10^-10 > 3.9 x 10^-11

As [Ca²⁺][F^-]² > K_sp, a precipitate of CaF₂ will form when the two solutions are mixed.

The common ion effect

The common ion effect refers to the reduced solubility of an ionic compound when a soluble salt containing one of the same ions is added to the solution. This can lead to precipitation.

For example, consider the equilibrium for AgCl:

AgCl_(s) ⇌ Ag⁺_(aq) + Cl^-_(aq)

Adding NaCl introduces the common ion Cl^-, shifting the equilibrium position to the left and precipitating AgCl. This occurs when [Ag⁺][Cl^-] exceeds the K_sp for AgCl.

The solubility of an ionic compound in a solution with a common ion is always lower than its solubility in pure water.

For instance, the solubility of silver chloride (AgCl) is:

• $1.3\times {10}^{-5}{\text{ mol dm}}^{-3}\text{ in water}$
• $1.8\times {10}^{-9}{\text{ mol dm}}^{-3}\text{ in }0.10{\text{ mol dm}}^{-3}{\text{ NaCl}}_{\text{(aq)}}$

We can explain this using the K_sp expression for AgCl:

${\text{K}}_{\text{sp}}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=1.8\times {10}^{-10}{\text{ mol}}^2{\text{ dm}}^{-6}$

Ignoring the small [Cl^-] from AgCl, the value of [Ag⁺] in 0.10 mol dm^-3 NaCl_(aq) is:

$[{\text{Ag}}^{+}]=\frac{{\text{K}}_{\text{sp}}}{[{\text{Cl}}^{-}]}=\frac{(1.8\times {10}^{-10})}{0.10}=1.8\times {10}^{-9}{\text{ mol dm}}^{-3}$

1.8 x 10^-9 < 1.3 x 10^-5

Therefore, the common chloride ion from NaCl lowers the solubility of AgCl in NaCl_(aq) compared to water.