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Chapter 8: Q3 TY (page 166)

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

Short Answer

Expert verified

The activity coefficient value for ${Cl}^{-}$ in 0.33mM ${CaCl}_{2}$ is 0.964.

Step by step solution

01

Concept used.

Extended Debye-Huckel equation:

The extended Debye-Huckel equation describes the link between activity coefficient and ionic strength.

$γ$ -activity coefficient of ion

$μ$ -ionic strength of solution

$α$ -size of ion

02

Step 2: Calculate the value of γ.

Given information,

$0.33 mM {CaCl}_{2}$

Size of ${Cl}^{-}$ ion is 300pm

Ionic strength calculation formula is given by

$\begin{array}{l} μ = \frac{1}{2} (c_{1} z_{1}^{2} + c_{2} z_{2}^{2} + . . . . . . . . . . . . .) \\ = \frac{1}{2} \sum_{i} c_{i} z_{i}^{2} \end{array}$

where,

$c_{i}$ -concentration of the $i^{t h}$ species and $z_{i}$ is its charge.

Calculate the ionic strength of the solution by substituting the numbers into the equation above.

$μ = \frac{1}{2} [[{Ca}^{2 +}] {(+ 2)}^{2} + 2 [{Cl}^{-}] {(- 1)}^{2}] = (\frac{1}{2} [0.33] {(+ 2)}^{2} + 2 [0.33] {(- 1)}^{2}) mM = 0.99 mM = 0.00099 M$

Using the extended Debye-Huckel equation, find the value of $γ$ for ${Cl}^{-}$ .

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Most popular questions from this chapter

Using activities, calculate the pH of a solution containing 0.010 M NaOH plus 0.012 0 M LiNO₃ . What would be the pH if you neglected activities?

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.

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 equilibrium expression for ${Ca}^{2 +} + 2 {Cl}^{-} \to {Cacl}_{2} (aq)$ with activity coefficients.

(a) Following the example of Mg(OH)₂ in Section 8-5, write the equations needed to find the solubility of Ca(OH)₂. Include activity coefficients where appropriate. Look up the equilibrium constants in Appendixes F and I.

(b) Suppose that the size of CaOH⁺= Ca(H₂O)₅(OH)⁺ is 500 pm. Including activity coefficients, compute the concentrations of all species, the fraction of hydrolysis (= [CaOH⁺]/{[Ca²⁺] + [CaOH⁺]}), and the solubility of Ca(OH)₂ in g/L. The Handbook of Chemistry and Physics lists the solubility of Ca(OH)₂ as 1.85 g/L at 0⁰C and 0.77 g/L at 100⁰C

See all solutions

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