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Chapter 8: Q31P (page 186)

Ion pairing. As in Problem $8 - 30$ , find the concentration, ionic strength, and ion pair fraction in localid="1654942556135" $0.025 F {Na}_{2} {SO}_{4}$ . Assume that the size of the ${NaSO}^{-}_{4} = 500 pm$

Short Answer

Expert verified

Thus the Ionic Strength and Ion Pair Fraction of $0.025 F {Na}_{2} {SO}_{4}$ is 0.07051M and 9%

Step by step solution

01

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,

$[{Na}^{+}] = 0.04775 M [{SO}^{2 -}_{4}] = 0.02275 M [{NaSO}^{-}_{4}] = 0.002246 M$

Ionic Strength =0.070514M

Ion Pair Fraction =9%

02

Step 2:Ionic Strength and Ion Pair Fraction of 0.025 F Na2SO4

Thus the Ionic Strength and Ion Pair Fraction of $0.025 F {Na}_{2} {SO}_{4}$ is $0.07051 M$ and 9%

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Most popular questions from this chapter

Find the activity coefficient of each ion at the indicated ionic strength:

$(a) {SO}_{4}^{2 -} \to (μ = 0.01 M) (b) {Sc}^{3 +} \to (μ = 0.005 M) (c) {Ec}^{3 +} \to (μ = 0.1 M) (d) {({CH}_{3} {CH}_{2})}_{3} {NH}^{+} \to (μ = 0.05 M)$

Write the charge balance for a solution containing $H^{+}, {OH}^{-}, {Ca}^{2 +}, {HCO}_{3}^{-}, {CO}_{3}^{2 -}, Ca {({HCO}_{3})}^{+}, Ca {(OH)}^{+}, K^{+} and {ClO}_{4}^{-}$

Write the charge balance for a solution of $H_{2} {SO}_{4}$ in water if the $H_{2} {SO}_{4}$ ionizes $H_{2} {SO}^{-}_{4}$ to and ${SO}^{2 -}_{4}$

Systematic treatment of equilibrium for ion pairing. Let’s derive the fraction of ion pairing for the salt in Box $8 - 1$ , which are $0.025 F NaCI, {Na}_{2} {SO}_{4}, {MgCI}_{2}, {MgSO}_{4}$ . Each case is somewhat different. All of the solutions will be near neutral pH because hydrolysis reactions of ${Mg}^{2 +}, {SO}^{2 -}_{4}, {Na}^{+}, {CI}^{-}$ have small equilibrium constants. Therefore, we assume that $[H^{+}] = [{OH}^{-}]$ and omit these species from the calculations. We work ${MgCI}_{2}$ as an example and then you asked to work each of the others. The ion-pair equilibrium constant, $K_{ip}$ comes from Appendix J.

Pertinent reaction:

${Mg}^{2 +} {CI}^{-} ֏ {MgCI}^{+} (aq) K_{ip} = \frac{[{MgCI}^{+} (aq)] γ_{{MgCI}^{+}}}{[{Mg}^{2 +}] γ_{{Mg}^{2 +}} [{CI}^{-}] γ_{{CI}^{-}}} {logK}_{ip} = 0.6 . {pK}_{ip} = - 0.6 \to (A)$

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

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

Mass balance:

$[{Mg}^{2 -}] + [{MgCI}^{+}] = F = 0.025 M \to (C) [{CI}^{-}] + [{MgCI}^{+}] = 2 F = 0.050 M \to (D)$

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 $([{Mg}^{2 +}], [{MgCI}^{+}], [{CI}^{-}])$

Solve: We will use Solver to find

$(number of unknowns) - (number of equiliberia) = 3 - 1 = 2$ unknown concentrations.

The spreadsheet shows the work. Formal concentration $F = 0.0025 M$ appears in cell $G 2$ . We estimate ${pMg}^{2 +}, {pCI}^{-}$ in cell $B 8$ and $B 9$ . The ionic strength in cell $B 5$ is given by the formula in cell $H 24$ . 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" ${Mg}^{2 +}, {CI}^{-}$ are from Table $8 - 1$ and the size of ${MgCI}^{+}$ is a guess. Activity coefficient are computed in columns $E, F$ . Mass balance $b_{1} = F - [{Mg}^{2 +}] - [{MGCI}^{+}], b_{2} = 2 F - [{CI}^{-}] - [{MgCI}^{+}]$ appears in cell $H 14, H 15$ , and the sum of squares ${b^{2}}_{1} + {b^{2}}_{2}$ appears in cell $H 16$ . The charge balance is not used because it is not independentof the two mass balances.

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

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

Ammonia Equilibrium treated by solver. We now use the solve spreadsheet introduced in Figure 8 - 9 for ${TIN}_{3}$ solubility to find the concentration of species in 0.01 M ammonia solution, neglecting activity coefficient. In the systematic treatment of equilibrium of ${NH}_{3}$ hydrolysis, we have four unknowns $([{NH}_{3}], [{NH}_{4}^{+}], [H^{+}], [{OH}^{-}])$ and two equilibrium (8-13) , (8-14). Therefore we will estimate the concentration of 4unknowns - 2equilibirum = 2species, for which I choose localid="1663566766281" ${NH}^{+} and {OH}^{-}$ . Setup the spreadsheet shown below, in which the estimate localid="1663566820791" ${pNH}_{4}^{+} = 3 . {pOH}^{-}$ = 3 appears in B6andB7 . (Estimates comes from the $K_{b}$ equilibrium 8-17 with $[{NH}_{4}^{+}] = [{OH}^{-}] = \sqrt{K_{b} [{NH}_{3}]} \approx \sqrt{10^{- 4.755} [0.01]} \Rightarrow {pNH}_{4}^{+} = {pOH}^{-} \approx 3$ . Estimate donot have to be very good for Solver to work. The formula in cell C8 is $[{NH}_{3}] = [{NH}_{4}^{+}]$ .

$[{OH}^{-}] / K_{b}$ and the formula in the cell C9 is $[H^{+}] = K_{w} / [{OH}^{-}]$ . The mass balance $b_{1}$ appears in cell F6 and the charge balance $b_{2}$ appears in cell F7 . Cell F8 has the sum $b_{1}^{2} + b_{2}^{2}$ . As described for ${TIN}_{3}$ on page 176, open the solver window and set the Solver Option. Then use the Solver to set the target cell F8 Equal to Min by changing cells B6 : B7 . What are the concentrations of the species? What fraction of ammonia $(= [{NH}_{4}^{+}] / [{NH}_{4}^{+}] + [{NH}_{3}])$ is hydrolyzed. Your answer should agree with those from Goal Seek in Figure 8-8

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