Question

Fill in the blanks in the following table. $$ \begin{array}{llll} \hline & \text { Compound } & \text { [cation] } & \text { [anion] } & K_{\mathrm{sp}} \\ \hline \text { (a) } & \mathrm{BaC}_{2} \mathrm{O}_{4} &\text {_________} & \text {_______}& 1.6 \times 10^{-6} \\ \text {(b) } & \mathrm{Cr}(\mathrm{OH})_{3} & 2.7 \times 10^{-8} &\text {_______} & 6.3 \times 10^{-31} \\ \text {(c) } & \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2} &\text {_________} & 8 \times 10^{-6} & 1 \times 10^{-54} \\ \hline \end{array} $$

Short answer

Answer: The missing cation and anion concentrations for the compounds are: (a) [Ba^2+] = 1.0 × 10^-2 M and [C2O4^2-] = 2.0 × 10^-2 M (b) [OH^-] = 5.0 × 10^-8 M (c) [Pb^2+] = 7.9 × 10^-19 M

Step-by-step solution

Write the balanced chemical equation for dissociation of each compound

Write the balanced chemical equation for each compound as they dissociate in solution: (a) BaC2O4(s) -> Ba^2+(aq) + 2C2O4^2- (aq) (b) Cr(OH)3(s) -> Cr^3+(aq) + 3OH^-(aq) (c) Pb3(PO4)2(s) -> 3Pb^2+(aq) + 2PO4^3- (aq)

Relate the dissociation stoichiometry with Ksp

For each compound, we can write the expression for Ksp using the given reaction stoichiometry. Let x, y, z be the concentration of cation in (a), (b), and (c), respectively. Let x', y', z' be the concentration of anion in (a), (b), and (c), respectively. (a) Ksp = [Ba^2+][C2O4^2-]^2 = x * (2x)^2 (b) Ksp = [Cr^3+][OH^-]^3 = y * (3y)^3 (c) Ksp = [Pb^2+]^3[PO4^3-]^2 = z^3 * (z'/2)^2

Solve for the missing concentrations

Use the given Ksp values and the expressions derived in Step 2 to find the missing concentrations. (a) 1.6 × 10^-6 = x * (2x)^2 (b) 6.3 × 10^-31 = 2.7 × 10^-8 * (3y)^3 (c) 1 × 10^-54 = z^3 * (4 × 10^-6)^2 Solve for each missing concentration and fill in the blanks: (a) x = [Ba^2+] = 1.0 × 10^-2 M [C2O4^2-] = 2x = 2.0 × 10^-2 M (b) [Cr^3+] = 2.7 × 10^-8 M (already given) y' = [OH^-] = 5.0 × 10^-8 M (c) z = [Pb^2+] = 7.9 × 10^-19 M [PO4^3-] = 8 × 10^-6 M (already given) The completed table looks like: $$ \begin{array}{llll} \hline & \text { Compound } & \text { [cation] } & \text { [anion] } & K_{\mathrm{sp}} \\ \hline \text { (a) } & \mathrm{BaC}_{2} \mathrm{O}_{4} & {1.0 x 10^{-2}} & {2.0 x 10^{-2}} & 1.6 \times 10^{-6} \\ \text {(b) } & \mathrm{Cr}(\mathrm{OH})_{3} & 2.7 \times 10^{-8} & {5.0 x 10^{-8}} & 6.3 \times 10^{-31} \\ \text {(c) } & \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2} & {7.9 x 10^{-19}} & 8 \times 10^{-6} & 1 \times 10^{-54} \\ \hline \end{array} $$