Question

Calculate the solubility of in each of the following

1. Water
2. A solution buffered at pH = 5.0
3. A solution buffered at pH = 11.0

Short answer

1. The solubility of $Fe{\left(OH\right)}_3$in water is$4\times {10}^{-14}mol/L$
2. The solubility of $Fe{\left(OH\right)}_3$in a solution buffered at pH 5.0 is $4\times {10}^{-11}mol/L$
3. The solubility of $Fe{\left(OH\right)}_3$ in a solution buffered at pH 11.0 is $4\times {10}^{-29}mol/L$

Step-by-step solution

Definition of solubility

The maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature is defined as solubility.

Calculation of solubility

The dissociation is represented as $Fe\left(O{H}_3\right)\iff F{e}^{3+}+3O{H}^{-}$

a.

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ 7.0=-\log \left[H^{+}\right] \\ \left[H^{+}\right]={10}^{-7} \\ Now, \\ \left[H^{+}\right]\left[O{H}^{-}\right]=K_w \\ \left[O{H}^{-}\right]=\frac{{10}^{-14}}{\left[H^{+}\right]} \\ \left[O{H}^{-}\right]={10}^{-7} \\ {K}_{sp}=\left[F{e}^{+3}\right]\left[O{H}^{-}\right]\dots \left(1\right) \\ s=\left[F{e}^{3+}\right],\left[O{H}^{-}\right]={10}^{-7} \\ {K}_{sp}=s\times {\left({10}^{-7}\right)}^3 \\ {K}_{sp}={10}^{-21}s \\ {10}^{-21}s=4\times {10}^{-38} \\ \mathit{s}\mathbf{=}\frac{\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{38}}}{{\mathbf{10}}^{\mathbf{-}\mathbf{27}}} \\ \mathit{s}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L} \end{gathered}$$

b. $$\begin{gathered} pH=-\log \left[H^{+}\right] \\ 5.0=-\log \left[H^{+}\right] \\ \left[H^{+}\right]=1\times {10}^{-5} \\ Now, \\ \left[H^{+}\right]\left[O{H}^{-}\right]=K_w \\ \left[O{H}^{-}\right]=\frac{{10}^{-14}}{\left[H^{+}\right]} \\ \left[O{H}^{-}\right]={10}^{-9} \\ {K}_{sp}=s\times {\left({10}^{-9}\right)}^3 \\ {K}_{sp}={10}^{-27}s \\ s=\frac{4\times {10}^{-38}}{{10}^{-27}} \\ s=4\times {10}^{-11}mol/L \end{gathered}$$

c.

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ 11.0=-\log \left[H^{+}\right] \\ \left[H^{+}\right]=1\times {10}^{-11} \\ Now, \\ \left[H^{+}\right]\left[O{H}^{-}\right]=K_w \\ \left[O{H}^{-}\right]=\frac{{10}^{-14}}{\left[H^{+}\right]} \\ \left[O{H}^{-}\right]={10}^{-3} \\ {K}_{sp}=s\times {\left({10}^{-3}\right)}^3 \\ {\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{s} \\ \mathit{s}\frac{\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{38}}}{{\mathbf{10}}^{\mathbf{-}\mathbf{9}}} \\ \mathit{s}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{29}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L} \end{gathered}$$