Question 6.9: Calculate the solubility of Pb(IO3)2 in 1.0 × 10^–4 M Pb(NO3...

Letting x equal the change in the concentration of Pb^{2+} , the equilibrium concentrations are

[Pb^{2+}] = 1.0 × 10^{–4} + x                        [{IO_3}^–] = 2x

and

(1.0 × 10^{–4} + x)(2x)² = 2.5 × 10^{–13}

We start by assuming that

[Pb^{2+}] = 1.0 × 10^{–4} + x ≈ 1.0 × 10^{–4}  M

and solve for x, obtaining a value of 2.50 × 10^{–5}. Substituting back gives the calculated concentration of [Pb^{2+}] at equilibrium as

[Pb^{2+}] = 1.0 × 10^{–4} + 2.50 × 10^{–5} = 1.25 × 10^{–4}  M

a value that differs by 25% from our approximation that the equilibrium
concentration is 1.0 × 10^{–4}  M. This error seems unreasonably large. Rather than shouting in frustration, we make a new assumption. Our first assumption that the concentration of [Pb^{2+}] is 1.0 × 10^{–4}  M was too small. The calculated concentration of 1.25 × 10^{–4}  M, therefore, is probably a little too large. Let us assume that

[Pb^{2+}] = 1.0 × 10^{–4} + x ≈ 1.2 × 10^{–4}  M

Substituting into the solubility product equation and solving for x gives us
x = 2.28 × 10^{–5}

or a concentration of [Pb^{2+}] at equilibrium of

[Pb^{2+}] = 1.0 × 10^{–4} + (2.28 × 10^{–5}) = 1.23 × 10^{–4}  M

which differs from our assumed concentration of 1.2 × 10^{–4}  M by 2.5%. This seems to be a reasonable error since the original concentration of Pb^{2+}  is given to only two significant figures. Our final solution, to two significant figures, is

[Pb^{2+}]  = 1.2 × 10^{–4}  M               [{IO_3}^–] = 4.6 × 10^{–5}  M

and the solubility of Pb(IO_3)_2  \text{is}  2.3 × 10^{–5} mol/L. This iterative approach to solving an equation is known as the method of successive approximations.