Question

${\mathrm{PbI}}_2$ has a solubility product of $8.0\times {10}^{-9}$. Lead Iodide has a solubility in $0.1\mathrm{molar}$ Lead Nitrate is $\mathrm{x}\times {10}^{-6}\mathrm{mol}/\mathrm{L}$. What is the value of $\mathrm{X}$? (Round off to the nearest integer)

Answer 1

Step 1:

• Given,
• ${\left[{\mathrm{K}}_{\mathrm{sp}}\right]}_{{\mathrm{PbI}}_2}=8\times {10}^{-9}$
• $$\begin{gathered} \mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{2+}\left(\mathrm{s}\right)+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right) \\ 0.1\mathrm{M}0.1\mathrm{M}0.2\mathrm{M} \end{gathered}$$
• $$\begin{gathered} {\mathrm{PbI}}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right) \\ \left(\mathrm{s}\right)\left(2\mathrm{s}\right) \end{gathered}$$
• ${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{I}}^{-}\right]}^2$
• $\begin{array}{rcl}{\mathrm{Pb}}^{2+}& =& \mathrm{s}+0.1\\ & \cong & 0.1\end{array}$

Step 2:

• $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}& =& 8\times {10}^{-9}\\ & =& \left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{I}}^{-}\right]}^2\end{array}$
• $\begin{array}{rcl}8\times {10}^{-9}& =& \left(0.1\right){\left(2s\right)}^2\\ 8\times {10}^{-8}& =& 4s^2\\ {\mathrm{s}}^2& =& \frac{8\times {10}^{-9}}{{10}^{-1}\times 4}\\ s^2& =& 2\times {10}^{-8}\\ s& =& \frac{{10}^{-2}\times \sqrt{2}\times {10}^{-4}}{{10}^{-2}}\\ \mathrm{s}& =& \sqrt{2}\times {10}^{-4}\\ & =& 141\times {10}^{-6}\mathrm{M}\\ & =& 141\end{array}$

Therefore, is the value of $\mathrm{X}$ is $141$