Calculate the molar solubility of \({\bf{CdC}}{{\bf{O}}_{\bf{3}}}\) in a buffer solution containing \({\bf{0}}.{\bf{115}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{N}}{{\bf{a}}_{\bf{2}}}{\bf{C}}{{\bf{O}}_{\bf{3}}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{0}}.{\bf{120}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{NaHC}}{{\bf{O}}_{\bf{3}}}\) .

We have a buffer solution with \(0.115 M N{a_2}C{O_3}{\rm{ }}and 0.120 M NaHC{O_3}\) .

Calculate the molar solubility of CdCO₃

First, let us calculate the concentration of \(\left[ {C{O_3}^{2 - }} \right]\)

\(\begin{array}{*{20}{c}}{{K_a} = \frac{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] \cdot \left[ {{\rm{C}}{{\rm{O}}_3}^{2 - }} \right]}}{{\left[ {HC{O_3} - } \right]}}}\\{4.7 \cdot {{10}^{ - 11}} = \frac{{x \cdot (0.115 + x)}}{{0.120 - x}}}\end{array}\)

Since K_a is lower than\({10^{ - 4}}\), we will assume

That \(0.115 + {\rm{x}} \approx 0.115,{\rm{\;and that \;}}0.120 - {\rm{x}} \approx 0.120\)

\(\begin{array}{*{20}{c}}{4.7 \times {{10}^{ - 11}} = \frac{{x \cdot 0.115}}{{0.120}}}\\{x = 4.9 \times {{10}^{ - 11}}\left[ {C{O_3}^{2 - }} \right]}\end{array}\)

Calculate the molar solubility of \({{\mathop{\rm CdCO}\nolimits} _3}\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {C{d^{2 + }}} \right] \cdot \left[ {CO_3^{2 - }} \right]}\\{\left[ {C{d^{2 + }}} \right] = \frac{{{K_{sp}}}}{{\left[ {CO_3^{2 - }} \right]}}}\\{ = \frac{{5.2 \times {{10}^{ - 12}}}}{{0.115}}}\\{ = 4.52 \times {{10}^{ - 11}}{\rm{M}}}\end{array}\)