Question

Systematic treatment of equilibrium for ion pairing. Let’s derive the fraction of ion pairing for the salt in Box $8-1$, which are $0.025\mathrm{F}\mathrm{NaCI},{\mathrm{Na}}_2{\mathrm{SO}}_4,{\mathrm{MgCI}}_2,{\mathrm{MgSO}}_4$. Each case is somewhat different. All of the solutions will be near neutral pH because hydrolysis reactions of ${\mathrm{Mg}}^{2+},{{\mathrm{SO}}^{2-}}_4,{\mathrm{Na}}^{+},{\mathrm{CI}}^{-}$have small equilibrium constants. Therefore, we assume that $\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]$and omit these species from the calculations. We work ${\mathrm{MgCI}}_2$as an example and then you asked to work each of the others. The ion-pair equilibrium constant, ${\mathrm{K}}_{\mathrm{ip}}$comes from Appendix J.

Pertinent reaction:

$$\begin{gathered} {\mathrm{Mg}}^{2+}{\mathrm{CI}}^{-}\mathrm{֏}{\mathrm{MgCI}}^{+}\left(\mathrm{aq}\right) \\ {\mathrm{K}}_{\mathrm{ip}}=\frac{\left[{\mathrm{MgCI}}^{+}\left(\mathrm{aq}\right)\right]{\mathrm{\gamma }}_{{\mathrm{MgCI}}^{+}}}{\left[{\mathrm{Mg}}^{2+}\right]{\mathrm{\gamma }}_{{\mathrm{Mg}}^{2+}}\left[{\mathrm{CI}}^{-}\right]{\mathrm{\gamma }}_{{\mathrm{CI}}^{-}}}{\mathrm{logK}}_{\mathrm{ip}}=0.6.{\mathrm{pK}}_{\mathrm{ip}}=-0.6\to \left(\mathrm{A}\right) \end{gathered}$$

Charge balance (omitting ${\mathrm{H}}^{+},{\mathrm{OH}}^{-}$whose concentrations are both small in comparison with $\left[{\mathrm{Mg}}^{+}\right],\left[{\mathrm{MgCI}}^{+}\right],\left[{\mathrm{CI}}^{-}\right]$:

role="math" localid="1655088043259" $2\left[{\mathrm{Mg}}^{2-}\right]+\left[{\mathrm{MgCI}}^{+}\right]=\left[{\mathrm{CI}}^{-}\right]\to \left(\mathrm{B}\right)$

Mass balance:

$$\begin{gathered} \left[{\mathrm{Mg}}^{2-}\right]+\left[{\mathrm{MgCI}}^{+}\right]=\mathrm{F}=0.025\mathrm{M}\to \left(\mathrm{C}\right) \\ \left[{\mathrm{CI}}^{-}\right]+\left[{\mathrm{MgCI}}^{+}\right]=2\mathrm{F}=0.050\mathrm{M}\to \left(\mathrm{D}\right) \end{gathered}$$

Only two of the three equations (B),(C) and (D) are independent. If you double (c) and subtract (D) , you will produce (B). we choose (C) and (D) as independent equations.

Equilibrium constant expression : Equation (A)

Count : 3 equations (A,C,D) and 3 unknowns $\left(\left[{\mathrm{Mg}}^{2+}\right],\left[{\mathrm{MgCI}}^{+}\right],\left[{\mathrm{CI}}^{-}\right]\right)$

Solve: We will use Solver to find

$\left(\mathrm{number}\mathrm{of}\mathrm{unknowns}\right)-\left(\mathrm{number}\mathrm{of}\mathrm{equiliberia}\right)=3-1=2$unknown concentrations.

The spreadsheet shows the work. Formal concentration $\mathrm{F}=0.0025\mathrm{M}$appears in cell $\mathrm{G}2$. We estimate ${\mathrm{pMg}}^{2+},{\mathrm{pCI}}^{-}$in cell $\mathrm{B}8$and $\mathrm{B}9$. The ionic strength in cell $\mathrm{B}5$is given by the formula in cell $\mathrm{H}24$. Excel must be set to allow for circular definitions as described on page role="math" localid="1655088766279" $179$. The sizes of role="math" localid="1655088853561" ${\mathrm{Mg}}^{2+},{\mathrm{CI}}^{-}$are from Table $8-1$and the size of ${\mathrm{MgCI}}^{+}$is a guess. Activity coefficient are computed in columns $\mathrm{E},\mathrm{F}$. Mass balance ${\mathrm{b}}_1=\mathrm{F}-\left[{\mathrm{Mg}}^{2+}\right]-\left[{\mathrm{MGCI}}^{+}\right],{\mathrm{b}}_2=2\mathrm{F}-\left[{\mathrm{CI}}^{-}\right]-\left[{\mathrm{MgCI}}^{+}\right]$appears in cell $\mathrm{H}14,\mathrm{H}15$, and the sum of squares ${{\mathrm{b}}^2}_1+{{\mathrm{b}}^2}_2$ appears in cell $\mathrm{H}16$. The charge balance is not used because it is not independentof the two mass balances.

Solver is invoked to minimizes ${{\mathrm{b}}^2}_1+{{\mathrm{b}}^2}_2$in cell $\mathrm{H}16$be varying ${\mathrm{pMg}}^{2+},{\mathrm{pCI}}^{-}$in cells $\mathrm{B}8$and $\mathrm{B}9$. From the optimized concentration, the ion-pair fraction $=\raisebox{1ex}{$\left[{\mathrm{MgCI}}^{+}\right]$}\!\left/ \!\raisebox{-1ex}{$\mathrm{F}$}\right.=0.0815$is computed in cell $\mathrm{D}15$.

The problem: Create a spreadsheet like the one for ${\mathrm{MgCI}}^{+}$to find the concentration, ionic strength, and ion pair fraction in $0.025\mathrm{M}\mathrm{NaCI}$. The ion pair formation constant from Appendix J is log ${\mathrm{K}}_{\mathrm{ip}}={10}^{-0.5}$for the reaction ${\mathrm{Na}}^{+}+{\mathrm{CI}}^{-}\mathrm{֏}\mathrm{NaCI}\left(\mathrm{aq}\right)$. The two mass balances are $\left[{\mathrm{Na}}^{+}\right]+\left[\mathrm{NaCI}\left(\mathrm{aq}\right)\right]=\mathrm{F},\left[{\mathrm{Na}}^{+}\right]=\left[{\mathrm{CI}}^{-}\right]$Estimate ${\mathrm{pNa}}^{+},{\mathrm{pCI}}^{-}$ for input and then minimizes the sum of square of the two mass balances.

Short answer

Thus the Ionic strength is 0.02486 M and Ion pair fraction is 0.57%

Step-by-step solution

Step 1:Method which is to be used in the problem.

Systematic treatment of equilibrium is used for ion pairing.

In the given problem, we need to use Excel and need to follow the instructions from the task.

Step 2:Calculating the ionic strength and ion pair fraction.

The values will be as shown in the below,

$$\begin{gathered} \left[{\mathrm{Na}}^{+}\right]=\left[{\mathrm{CI}}^{-}\right]=0.02486\mathrm{M} \\ \left[{\mathrm{NaCI}}^{-}\right]=0.000143\mathrm{M} \end{gathered}$$

Ionic strength$=0.02486\mathrm{M}$

Ion pair fraction $=0.57\%$

Thus the Ionic strength is $0.02486\mathrm{M}$and Ion pair fraction is$0.57\%$ .