Question

Solubility of $\mathrm{AgCN}$ buffer solution has a $\mathrm{pH}=3$ is $\mathrm{X}$.What is the value of $\mathrm{X}$? Assume no Cyano is formed ${\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{AgCN}\right)=2.2\times {10}^{-16}$ and ${\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HCN}\right)=6.2\times {10}^{-10}$

Answer 1

Step 1:

• Assume solubility of $\mathrm{AgCN}$ is $\mathrm{M}$ in a buffer of $\mathrm{pH}=3$
• 
• ,

Step 2:

For weak $\mathrm{HCN}$,

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}& =& \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{CN}}^{-}\right]}{\left[\mathrm{HCN}\right]}\\ 6.2\times {10}^{-10}& =& \frac{{10}^{-3}\left[{\mathrm{CN}}^{-}\right]}{\mathrm{a}}\\ \left[{\mathrm{CN}}^{-}\right]& =& 6.2\times {10}^{-7}\times \mathrm{a}\end{array}$

Step 3:

For $\mathrm{AgCN}$

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}& =& \left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{CN}}^{-}\right]\\ 2.2\times {10}^{-16}& =& \mathrm{a}\times 6.2\times {10}^{-10}\times \mathrm{a}\\ a^2& =& \frac{2.2\times {10}^{-16}}{6.2\times {10}^{-10}}\\ a^2& =& 3.55\times {10}^{-10}\\ a& =& 1.88\times {10}^{-5}\\ & =& 1.9\times {10}^{-5}\end{array}$

Therefore, the solubility of $\mathrm{AgCN}$ is $1.9\times {10}^{-5}$.