Question

Calculate the molar solubility of \(\mathrm{PbCl}_{2}\) in \(0.2 \mathrm{M} \mathrm{NaOH}\). The complex formed is \(\mathrm{Pb}(\mathrm{OH})_{3}^{-}\left(K_{\mathrm{f}}=3.8 \times 10^{14}\right) .\) Ignore any other competing equilibria.

Short answer

Answer: The molar solubility of PbCl2 in a 0.2 M NaOH solution is 2.13 x 10^-5 M.

Step-by-step solution

Write the balanced chemical equations

First, let's write down the balanced chemical equations for the dissociation of \(\mathrm{PbCl}_{2}\) and the reaction forming the complex ion \(\mathrm{Pb(OH)}_{3}^{-}\): (1) \(\mathrm{PbCl}_{2}(s) \longleftrightarrow \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^{-}}(aq)\) (2) \(\mathrm{Pb^{2+}}(aq) + 3\mathrm{OH^{-}}(aq) \longleftrightarrow \mathrm{Pb(OH)}_{3}^{-}\)

Set up the expressions for \(K_\mathrm{sp}\) and \(K_{\mathrm{f}}\)

We need to set up the expressions for the solubility product constant, \(K_\mathrm{sp}\), for the dissociation of \(\mathrm{PbCl}_{2}\) (Equation 1) and the formation constant, \(K_{\mathrm{f}}\), for the formation of \(\mathrm{Pb(OH)}_{3}^{-}\) (Equation 2): \(K_\mathrm{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^{-}}]^2\) \(K_{\mathrm{f}} = \frac{[\mathrm{Pb(OH)}_{3}]}{[\mathrm{Pb^{2+}}][\mathrm{OH^{-}}]^3} = 3.8 \times 10^{14}\)

Calculate the concentration of \(\mathrm{OH^{-}}\)

Since we know that the initial concentration of \(\mathrm{NaOH}\) is \(0.2 \mathrm{M}\), the concentration of \(\mathrm{OH^{-}}\) ions in the solution can be calculated as: \([\mathrm{OH^{-}}] = [\mathrm{NaOH}] = 0.2\mathrm{M}\)

Calculate the concentration of \(\mathrm{Pb(OH)}_{3}^{-}\) using \(K_{\mathrm{f}}\)

We can rearrange the expression for \(K_{\mathrm{f}}\) to determine the concentration of \(\mathrm{Pb(OH)}_{3}^{-}\), given \([\mathrm{OH^{-}}] = 0.2\mathrm{M}\) and \(K_{\mathrm{f}} = 3.8 \times 10^{14}\): \([\mathrm{Pb(OH)}_{3}] = K_{\mathrm{f}}\times[\mathrm{Pb^{2+}}][\mathrm{OH^{-}}]^3\) \([\mathrm{Pb(OH)}_{3}] = (3.8 \times 10^{14}) \times [\mathrm{Pb^{2+}}] (0.2)^3\) \([\mathrm{Pb(OH)}_{3}] = (3.8 \times 10^{14}) \times [\mathrm{Pb^{2+}}] (0.008)\)

Relate \([\mathrm{Pb(OH)}_{3}^{-}]\) and \([\mathrm{Pb^{2+}}]\) using \(K_\mathrm{sp}\)

Since \([\mathrm{Pb(OH)}_{3}^{-}] = [\mathrm{Pb^{2+}}]\) and \([\mathrm{Cl^{-}}] = 2[\mathrm{Pb^{2+}}]\), the expression for \(K_\mathrm{sp}\) can be rewritten as: \(K_\mathrm{sp} = [\mathrm{Pb^{2+}}](2[\mathrm{Pb^{2+}}])^2 = 4[\mathrm{Pb^{2+}}]^3\) Now, we can use this equation to relate the concentration of \(\mathrm{Pb(OH)}_{3}^{-}\) and \(\mathrm{Pb^{2+}}\): \(\frac{[\mathrm{Pb(OH)}_{3}^{-}]}{4} = [\mathrm{Pb^{2+}}]^3\)

Solve for \([\mathrm{Pb^{2+}}]\) and calculate molar solubility

Substituting the expression for \([\mathrm{Pb(OH)}_{3}^{-}]\) from Step 4 and solving for \([\mathrm{Pb^{2+}}]\): \(\frac{(3.8 \times 10^{14})\times [\mathrm{Pb^{2+}}](0.008)}{4} = [\mathrm{Pb^{2+}}]^3\) \(\frac{(3.8 \times 10^{14})(0.008)}{4} = \frac{[\mathrm{Pb^{2+}}]^4}{[\mathrm{Pb^{2+}}]}\) \([\mathrm{Pb^{2+}}] = \sqrt[3]{\frac{(3.8 \times 10^{14})(0.008)}{4}} = 2.13 \times 10^{-5} \mathrm{M}\) Since the molar solubility of \(\mathrm{PbCl}_{2}\) is equal to the concentration of \(\mathrm{Pb^{2+}}\), we can conclude that the molar solubility of \(\mathrm{PbCl}_{2}\) in \(0.2 \mathrm{M} \mathrm{NaOH}\) is: \(2.13 \times 10^{-5} \mathrm{M}\)