Question

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?

Answer 1

Answer:

$Ksp=1.07x10^{-8}$

Explanation:

Hello,

In this case, the dissociation reaction is:

$PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)$

For which the equilibrium expression is:

$Ksp=[Pb^{2+}][I^-]^2$

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

$Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M$

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

$[Pb^{2+}]=1.39x10^{-3}M$

$[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M$

Thereby, the solubility product results:

$Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}$

Regards.