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Chapter 8: Q9P (page 183)

The equilibrium constant for dissolution of a non ionic compound, such as diethyl ether $({CH}_{3} {CH}_{2} {OCH}_{2} {CH}_{3})$ , in water can be written
$ether (I) ֏ ether (aq) K = [ether (aq)] γ_{ether}$
At low ionic strength, role="math" localid="1654837696819" $γ \approx 1$ for neutral compounds. At high ionic strength, most neutral molecules can be salted out of aqueous solution. That is, when a high concentration typically $(> 1 M)$ of a salt such as $NaCI$ is added to aqueous solution, neutral molecules usually becomes less soluble. Does the activity coefficient, $γ_{ether}$ increases or decreases at high ionic strength?

Short Answer

Expert verified

Thus, the neutral compound in ether would preferentially be extracted in the ether layer and can be easily separated by salting out the layer.

Step by step solution

01

Step 1:Activity Coefficient of Ether at high NaCI concentration

At high concentrations, the activity coefficient value of ether would be very low, and the value would decrease at high ionic strength. Which makes the value of as,

$K = [ether (aq)] \times activity coefficient$

The value is too small to extract from the aqueous layer.

02

Neutral molecules of Ether at high NaCI concentration

As a result, the neutral compound in ether will preferentially extract in the ether layer and can be easily separated by salting out the layer.

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Most popular questions from this chapter

Find $γ$ for ${Cl}^{-}$ in 0.33mM ${CaCl}_{2}$ .

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.

24. Consider the dissolution of the compound,which gives $X_{2} Y_{2}^{2 +}, X_{2} Y^{4 +}, X_{2} Y_{3} (aq)$ and $Y^{2 -}$ . Use the mass balance to find an

expression for $[Y^{2 -}]$ in terms of the other concentrations. Simplify

your answer as much as possible.

Write the charge and mass balances for dissolving ${CaF}_{2}$ in water if the reactions are

role="math" localid="1654770961556" ${CaF}_{2} (s) ٟ {Ca}^{2 +} + 2 F {Ca}^{2 +} + H_{2} O ٟ {CaOH}^{+} + H^{+} {Ca}^{2 +} F ؚ {CaF}^{+} {CaF}_{2} (s) ؚ {CaF}_{2} (aq) F + H^{+} ؚ HF (aq) HF (aq) + F ؚ {HF}_{2}$

(a) Ion Pairing . As in problem $8 - 30$ , find the concentration, ionic strength, and ion pair fraction in $0.025 F {MgSO}_{4}$

(b) Two possibly important reactions that we did not consider are acid hydrolysis of ${Mg}^{+} ({Mg}^{2 +} + H_{2} O ֏ {MgOH}^{+} + H^{+})$ and base hydrolysis of ${SO}^{2 -}_{4}$ . 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.

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