Question

${\mathrm{CdSO}}_4$ has a solubility in water$8.0\times {10}^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$. What is its solubility $0.01\mathrm{M}$ ${\mathrm{H}}_2{\mathrm{SO}}_4$?(Round off it to the nearest integer)

Answer 1

Step 1:

• $$\begin{gathered} {\mathrm{CdSO}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right) \\ \left(\mathrm{s}\right)\left(\mathrm{s}\right) \end{gathered}$$
• As given $\mathrm{s}$ is $8.0\times {10}^{-4}{\mathrm{molL}}^{-1}$
• $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}& =& {\mathrm{s}}^2\\ & =& {(8.0\times {10}^{-4})}^2\\ & =& 64\times {10}^{-8}\end{array}$

Step 2:

• $$\begin{gathered} {\mathrm{CdSO}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right) \\ \left(\mathrm{s}\right)(\mathrm{s}+{10}^{-2}) \end{gathered}$$
• $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{CdSO}}_4\right)& =& 64\times {10}^{-6}\\ & =& \mathrm{s}(\mathrm{s}+{10}^{-2})\\ & & \end{array}$
• $\begin{array}{rcl}64\times {10}^{-6}& \cong & s(s+{10}^{-2})\\ 64\times {10}^{-6}& \cong & s\times {10}^{-2}\\ & =& 64\times {10}^{-6}\end{array}$

Therefore, $64\times {10}^{-6}$ is its solubility $0.01\mathrm{M}$ ${\mathrm{H}}_2{\mathrm{SO}}_4$.