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Chapter 8: Q4 TY (page 167)

By interpolation, find $γ$ for $H^{+}$ when $μ = 0.06 M$ .

Short Answer

Expert verified

The activity coefficient value for $H^{+}$ ion is 0.854.

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

$σ$ -size of ion

02

Step 2: Calculate the value of γ.

The activity coefficient and ionic strength values for ion:

The linear interpolation:

$\frac{U n k n o w n i n t e r v a l}{∆ γ} = \frac{k n o w n i n t e r v a l}{∆ γ} \frac{0.83 - 1}{0.83 - 0.86} = \frac{0.1 - 0.06}{0.1 - 0.05} 0.83 - γ = \frac{0.04}{0.05} (- 0.03) - γ = - 0.024 - 0.83 = - 0.854 γ = 0.854$

The activity coefficient value for $H^{+}$ ion is 0.854.

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Most popular questions from this chapter

Find $[{Hg}_{2}^{2 +}]$ in equilibrium with 0.010MKCl saturated with ${Hg}_{2} {Cl}_{2}$ .

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.

Calculate the activity coefficient of ${AI}^{3 +}$ when role="math" localid="1654836623327" $μ = 0.083 M$ by using linear interpolation in Table $8 - 1$ .

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 ${NaSO}^{-}_{4} = 500 pm$

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

See all solutions

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