Question

Compare the molar solubility of AgOH in pure water with that in a solution buffered at pH 7.00. Note the difference between the two: When AgOH is dissolved in pure water, the pH does not remain at 7.

Short answer

The solubility of AgOH is more in a solution that is buffered with pH 7.00 compared to pure water.

Step-by-step solution

The concentration of hydroxide ion in solution buffered at pH 7.00

Given pH of the solution is 7.00.

Hence, $\left[O{H}^{-}\right]=1.0\times {10}^{-7}M$

Molar solubility of AgOH in this solution:

localid="1661941464750" $$\begin{gathered} \mathrm{AgOH}\left(\mathrm{s}\right)\to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ -\mathrm{s}\mathrm{s} \end{gathered}$$

$\begin{array}{rcl}{K}_{sp}& =& \left[A{g}^{+}\right]\left[O{H}^{-}\right]\\ 1.5\times {10}^{-8}& =& \left[A{g}^{+}\right]\times 1.0\times {10}^{-7}\\ \left[A{g}^{+}\right]& =& 0.15mol/L\end{array}$

Thus, the molar solubility of silver (I) is 0.15 mol/L.----------(1)

Molar solubility of AgOH

The molar solubility of AgOH can be calculated as shown below:

$$\begin{gathered} AgOH\left(s\right)\to A{g}^{+}\left(aq\right)+O{H}^{-}\left(aq\right) \\ -ss \\ {K}_{sp}=s^2 \\ 1.5\times {10}^{-8}=s^2 \\ s=1.22\times {10}^{-4}mol/L \end{gathered}$$----------------(2)

Comparison

If the solubilities of (1) and (2) are compared, the solubility of AgOH is more in a solution that is buffered with pH 7.00.