Question 6.15: Calculate the solubility of Pb(IO3)2 in a matrix of 0.020 M ...

We begin by calculating the ionic strength of the solution. Since $Pb(IO_3)_2$ is only sparingly soluble, we will assume that its contribution to the ionic strength can be ignored; thus

$\mu =\frac{1}{2} \left[ (0.20  M)(+2)^2 + (0.040  M)(-1)^2\right] =0.060  M$

Activity coefficients for $Pb^{2+}  \text{and}  I^–$ are calculated using equation 6.50

$-\log  \gamma _A =\frac{0.51\times z_A^2\times \sqrt{\mu } }{1+3.3\times \alpha _A\times \sqrt{\mu } }$      6.50

$-\log   \gamma _{pb^{2+}} =\frac{0.51\times (+2)^2\times \sqrt{0.060 } }{1+3.3\times 0.45\times \sqrt{0.060 } } =0.366$

giving an activity coefficient for $Pb^{2+}$ of 0.43. A similar calculation for ${IO_3}^–$ gives its activity coefficient as 0.81. The equilibrium constant expression for the solubility of $PbI_2$ is

$K_{sp} = [Pb^{2+}] + [{IO_3}^-]^2 \gamma _{Pb^{2+}}\gamma_{{IO_3 }^-}=2.5 \times 10^{-13}$

Letting

$[Pb^{2+}] = x        and       [{IO_3}^–] = 2x$

we have

$(x)(2x)^2(0.45)(0.81)^2 = 2.5 × 10^{–13}$

Solving for x gives a value of 6.0 × $10^{–5}$or a solubility of 6.0 × $10^{–5}$ mol/L. This compares to a value of 4.0 × $10^{–5}$ mol/L when activity is ignored. Failing to correct for activity effects underestimates the solubility of $PbI_2$ in this case by 33%.