Chemical equilibrium — Question 123

Answer
$[P{b}^{2+}]=1.20\times {10}^{-3}$ and $[I^{-}]=1.60\times {10}^{-3}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$. $Q_{sp}=3.07\times {10}^{-9}<K_{sp}$, so no precipitate forms. Added iodide lowers dissolved lead-ion concentration; Ksp is unchanged.

Working
The total volume is 0.0500 L. Divide each original ion amount by this volume: $[P{b}^{2+}]=0.0300(0.00200)/0.0500$ and $[I^{-}]=0.0200(0.00400)/0.0500$. The ion product is $[P{b}^{2+}][I^{-}{]}^2=3.072\times {10}^{-9}$. In the separate saturated solution, $\mathrm{P}\mathrm{b}{\mathrm{I}}_2(\mathrm{s})\rightleftharpoons \mathrm{P}{\mathrm{b}}^{2+}(\mathrm{a}\mathrm{q})+2{\mathrm{I}}^{-}(\mathrm{a}\mathrm{q})$. Added iodide drives precipitation and reduces [Pb²⁺] until the ion product returns to the unchanged Ksp.

Marking criteria

• Calculates both diluted ion concentrations correctly. [1 mark]
• Calculates Qsp = 3.07 × 10⁻⁹ using the squared iodide concentration. [1 mark]
• Compares Qsp with Ksp and concludes no precipitation in the mixed sample. [1 mark]
• Explains that added iodide drives precipitation and decreases dissolved Pb²⁺. [1 mark]
• States Ksp is unchanged at constant temperature. [1 mark]