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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,

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

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

Write two mass balances for a 1.0Lsolution containing 0.100molof sodium acetate.

Systematic treatment of equilibrium for ion pairing. Let鈥檚 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:

${Mg}^{2 +} {CI}^{-} 謴 {MgCI}^{+} (aq) K_{ip} = \frac{[{MgCI}^{+} (aq)] 纬_{{MgCI}^{+}}}{[{Mg}^{2 +}] 纬_{{Mg}^{2 +}} [{CI}^{-}] 纬_{{CI}^{-}}} {logK}_{ip} = 0.6 . {pK}_{ip} = - 0.6 鈫� (A)$

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

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

Mass balance:

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

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 ${Na}^{+} + {CI}^{-} 謴 NaCI (aq)$ . 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.

State the meaning of the charge and mass balance equations.

Ammonia equilibrium solved with Goal Seek. Modify Figure $8 - 7$ to find the concentration of species in $0.05 M {NH}_{3}$ . The only change required is the value of F . How do the pH and fraction of ammonia hydrolysis role="math" localid="1654938461639" $(= \frac{[{NH}^{+}_{4}]}{[{NH}^{+}_{4}]} + [{NH}_{3}])$ change when the formal concentration of ${NH}_{3}$ increases from 0.01 to 0.05 M?

Solubility with Activity: Find the concentration of the major species in a saturated aqueous solution of $LiF$ . Consider these reactions:

$LiF (s) 謴 {Li}^{+} + F^{-} K_{sp} = [{Li}^{+}] 纬_{{Li}^{+}} [F^{-}] 纬_{F^{-}} LiF (s) 謴 LiF (aq) K_{ionpair} = [LiF (aq)] 纬尝颈贵 (aq) F^{-} + H_{2} O 謴 HF + {OH}^{-} K_{b} = \frac{K_{w}}{K_{a}} (forHF) H_{2} O \overset{K_{w}}{謴} H^{+} + {OH}^{-} K_{w} = [H^{+}] 纬_{H^{+}} [{OH}^{-}] 纬_{{OH}^{-}}$

Look up the equilibrium constants in the appendixes and write their pK values. The ion pair reaction is the sum of localid="1654945209684" $LiF (s) 謴 {Li}^{+} + F^{-}$ from the Appendix ${FandLi}^{+} 謴 LiF (aq)$ from Appendix J. write the equilibrium constant expressions and the charge and mass balance.
Create a spreadsheet that uses activities to find the concentration of all species and the ionic strength. Use pH and pOH as independent variables to estimate. It does not work to choose pH and pLi because their concentration fixes that of the other through the relation $K_{sp} = [{Li}^{+}] 纬_{{Li}^{+}} [F^{-}] 纬_{F^{-}}$

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