Question

The solubility rules outlined in chapter 4 say that are marginally soluble hydroxides. Calculate the pH of a standard solution of each of these marginally soluble hydroxides

Short answer

The pH of$Ba{\left(OH\right)}_2$ is 13.34.

The pH of $Sr{\left(OH\right)}_2$is 12.93.

The pH of $Ca{\left(OH\right)}_2$is 12.38.

Step-by-step solution

Definition of pH

A quantitative measure of the acidity or basicity of aqueous or other liquid solutions is defined as pH.

Calculation of pH

$$\begin{gathered} \mathit{B}\mathit{a}{\left(OH\right)}_{\mathbf{2}}\left(s\right)\mathbf{\iff }\mathit{B}{\mathit{a}}^{\mathbf{+}\mathbf{2}}\left(aq\right)\mathbf{+}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}\left(aq\right) \\ IfthesolubilityofBa{\left(OH\right)}_{2}isxmol/L, \\ {K}_{sp}=\left[B{a}^{2+}\right]\left[O{H}^{-}\right] \\ 5\times {10}^{-3}=x\times \left(2x^2\right) \\ {x}^3=1.25\times {10}^{-3} \\ x={\left(1.25\times {10}^{-3}\right)}^{\frac{1}{3}} \\ x=0.11mol/L \\ \left[O{H}^{-}\right]=0.22M \\ pOH=-\log \left(0.22M\right) \\ pOH=0.65 \\ \mathit{p}\mathit{H}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{34} \end{gathered}$$

$$\begin{gathered} \mathit{S}\mathit{r}{\left(OH\right)}_{\mathbf{2}}\left(s\right)\mathbf{\iff }\mathit{S}{\mathit{r}}^{\mathbf{+}\mathbf{2}}\left(aq\right)\mathbf{+}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}\left(aq\right) \\ IfthesolubilityofSr{\left(OH\right)}_{2}isymol/L \\ {K}_{sp}=\left[S{r}^{+2}\right]{\left[O{H}^{-}\right]}^2 \\ 3.3\times {10}^{-4}=y\times {\left(2y\right)}^2 \\ {y}^3=8\times {10}^{-5} \\ y={\left(8\times {10}^{-5}\right)}^{1/3} \\ y=0.043mol/L \\ \left[O{H}^{-}\right]=0.068M \\ pOH=-\log \left(0.068M\right) \\ pOH=1.07 \\ \mathit{p}\mathit{H}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{93} \end{gathered}$$

$$\begin{gathered} \mathit{C}\mathit{a}{\left(OH\right)}_{\mathbf{2}}\left(s\right)\mathbf{\iff }\mathit{C}{\mathit{a}}^{\mathbf{+}\mathbf{2}}\left(aq\right)\mathbf{+}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}\left(aq\right) \\ IfthesolubilityofCa{\left(OH\right)}_{2}iszmol/L \\ {K}_{sp}=\left[C{a}^{+2}\right]{\left[O{H}^{-}\right]}^2 \\ 6.5\times {10}^{-6}=z\times \left(2z^2\right) \\ {z}^3=1.62\times {10}^{-6} \\ z=0.012mol/L \\ \left[O{H}^{-}\right]=0.024M \\ pOH=-\log \left(0.024M\right) \\ pOH=1.62 \\ pH=12.38 \end{gathered}$$