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Chapter 8: Q8 TY (page 171)

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

Short Answer

Expert verified

The mass balances are $[{Na}^{+}] = 0.100 M$ and $[{CH}_{3} {CO}_{2}^{-}] + [{CH}_{3} {CO}_{2} H] = 0.100 M$ .

Step by step solution

01

Concept used.

Mass balance:

All species in a solution containing a certain atom (or group of atoms) must be equal to the amount of that atom (or group) given to the solution.

02

Step 2: Calculate the two mass balances.

By definition, mass balance of the solution is given by

For sodium ion:

$[{Na}^{+}] = 0.100 M$

For acetate ion:

$[{CH}_{3} {CO}_{2}^{-}] + [{CH}_{3} {CO}_{2} H] = 0.100 M (dissociated product) (undissociated product) (amount of acetic acid put into the solution)$

Therefore, mass balance equation for 1.0L solution of $0.100 molNa + {CH}_{3} {CO}_{2}^{-}$ was given.

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Most popular questions from this chapter

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.

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

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

14.The temperature-dependent form of the extended Debye-

Hückel equation 8-6 is

$l o g γ = \frac{(- 1.825 \times 10^{6}) {(ε T)}^{- 3 / 2} Z^{2} \sqrt{μ}}{1 + α \sqrt{μ} / (2.00 \sqrt{ε T})}$

where $ε$ is the (dimensionless) dielectric constant* of water, T is

temperature (K), Z is the charge of the ion, $μ$ is ionic strength $(mol / L)$ ,and $α$ is the ion size parameter (pm). The dependence of ´

on temperature is

role="math" localid="1654916203154" $ε = 79.755 e^{(- 4.6 \times 10^{- 3}) (T - 293.15)}$

Calculate the activity coefficient of

{SO}_{4}^{2 -}

50.00 ° C

μ = 0.100 M

. Compare your value with the one in Table 8-1.

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

Write the charge balance for an aqueous solution of arsenic acid, $H_{3} {AsO}_{4}$ , in which the acid can disassociate to role="math" localid="1654936423245" $H_{3} {AsO}^{-}_{4}, H {AsO}_{4}^{2 -}, {andAsO}_{4}^{3 -}$ . Look up the structure of arsenic acid in Appendix G and write the structure of ${HAsO}^{2 -}_{4}$

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