The information in the table shows how the solubility of lead chloride is affected by temperature.
Using a graph, calculate the solubility product \((K_{sp})\) of the dissolution of lead chloride at 50°C. Include a fully labelled graph and a relevant chemical equation in your answer (6 marks)
--- 12 WORK AREA LINES (style=lined) ---
\(\ce{PbCl2(s) \rightleftharpoons Pb^2+(aq) + 2Cl^-(aq)}\)
\(\ce{Using the graph:}\)
\(\ce{Solubility (50°) = 0.7 g/100 g water = 7 g/L}\)
\(\ce{Converting to mol L^{-1}:}\)
\[\ce{MM(PbCl2) = 207.2 + 2 \times 35.45 = 278.1}\]
\[\ce{n = \frac{m}{MM} = \frac{7}{278.1} = 0.0252 mol L^{-1}}\]
\(\ce{[Pb^2+(aq)] = 0.0252 mol L^{-1}}\)
\(\ce{Mole ratio \ Pb^2+ : Cl^- = 1:2}\)
\(\Rightarrow \ce{[Cl^-] = 2 \times 0.0252 = 0.0504 mol L^{-1}}\)
\begin{aligned}
\ce{$K_{sp}$} & \ce{= [Pb^2+][Cl^-]^{2}} \\
& \ce{=0.0252 \times (0.0504)^{2}} \\
& \ce{= 6.4 \times 10^{-5}} \\
\end{aligned}
