Question

(a) Ion Pairing . As in problem $8-30$, find the concentration, ionic strength, and ion pair fraction in $0.025\mathrm{F}{\mathrm{MgSO}}_4$

(b) Two possibly important reactions that we did not consider are acid hydrolysis of ${\mathrm{Mg}}^{+}\left({\mathrm{Mg}}^{2+}+{\mathrm{H}}_2\mathrm{O}\mathrm{֏}{\mathrm{MgOH}}^{+}+{\mathrm{H}}^{+}\right)$and base hydrolysis of ${{\mathrm{SO}}^{2-}}_4$. Write these two reactions and find their equilibrium constants in Appendices I and G. With the assumed pH near $7.20$ and neglecting activity coefficient, show that both reactions are negligible.

Short answer

Thus the Ionic Strength and Ion Pair Fraction of $0.025\mathrm{F}{\mathrm{MgSO}}_4$ is $0.06463\mathrm{M}$and $35.4\%$ and the equilibrium constants are$6\times {10}^{-7}\mathrm{M}\mathrm{and}2\times {10}^{-7}\mathrm{M}$

Step-by-step solution

Step 1:Calculating the Ionic Strength and Ion Pair Fraction.

This is also a task that need to be performed in the Excel. And the values will be,

$$\begin{gathered} \left[{\mathrm{Mg}}^{2+}\right]=\left[{{\mathrm{SO}}^{2-}}_4\right]=0.01616\mathrm{M} \\ \left[{{\mathrm{MgSO}}^{-}}_4\right]=0.008844\mathrm{M} \end{gathered}$$

Ionic Strength $=0.06463\mathrm{M}$

Ion Pair Fraction$=35.4\%$

Step 2:Calculating the equilibrium Constants.

In the part (B) , we need to find the equilibrium constant.

$$\begin{gathered} {\mathrm{Mg}}^{2+}+{\mathrm{OH}}^{-}\mathrm{֏}{\mathrm{MgOH}}^{+}+{\mathrm{K}}_1={10}^{2.6} \\ \left[{\mathrm{MgOH}}^{+}\right]{\mathrm{֏K}}_1\left[{\mathrm{Mg}}^{2+}\right]\left[{\mathrm{OH}}^{-}\right]={10}^{2.6}\times 0.016\times {10}^{-7}=6\times {10}^{-7}\mathrm{M} \end{gathered}$$

This is negligible in comparison to $\left[{\mathrm{Mg}}^{2+}\right]=0.016\mathrm{M}$

$$\begin{gathered} {{\mathrm{SO}}_4}^{2-}+{\mathrm{H}}_2\mathrm{O}\mathrm{֏}{{\mathrm{HSO}}^{-}}_4+{\mathrm{OH}}^{-}{\mathrm{pK}}_{\mathrm{b}}=12.01 \\ \left[{{\mathrm{HSO}}^{-}}_4\right]\mathrm{֏}{\mathrm{K}}_{\mathrm{b}}\left[{{\mathrm{SO}}^{2-}}_4\right]/\left[{\mathrm{OH}}^{-}\right]={10}^{-12.01}\times \frac{0.016}{{10}^{-7}}=2\times {10}^{-7}\mathrm{M} \end{gathered}$$

This is negligible in comparison to$\left[{{\mathrm{SO}}_4}^{2-}\right]=0.016\mathrm{M}$

Thus the Ionic Strength and Ion Pair Fraction of $0.025\mathrm{F}{\mathrm{MgSO}}_4\mathrm{is}0.06463\mathrm{M}$and $35.4\%$and the equilibrium constants are $6\times {10}^{-7}\mathrm{M}\mathrm{and}2\times {10}^{-7}\mathrm{M}$