Question

PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.

Answer 1

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

$\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).$

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ii. Given the dissolution of some substance

$xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)}$,

the Ksp, or the solubility product constant, of the preceding equation takes the general form

$K_{sp} = [B]^y [C]^z$.

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

$K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}$.

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iii. Presumably, what we're being asked for here is the molar solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

$1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.$

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.