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Chapter 8: Q2 TY (page 165)

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

Short Answer

Expert verified

The equilibrium expression is $K = \frac{A_{C a C}}{λ_{C a^{2 +}} X_{C^{-}}^{2}} = \frac{[C a C l_{2}]}{[C a^{2 +}] γ_{C a^{2}} + {[C l^{-}]}^{2} γ_{C l^{-}}}$ .

Step by step solution

01

Concept used.

Equilibrium expression:

The equilibrium equation determines the ratio of products to reactants.

$aA + bB \to cC + dD A_{A}^{2} A_{B}^{D} {| A |}^{2} γ_{A}^{2} {| B |}^{0} γ_{B}^{b}$

where,

-activity of species

[] -concentration of species

$γ$ - activity coefficient of species.

02

Step 2: Calculate the ionic strength of 1mMCaCl2.

Given reaction,

${Ca}^{2 +} + 2 {Cl}^{-} \to {Cacl}_{2} (aq)$

The equilibrium expression by the activity of all species

$K = \frac{A_{C O C}}{A_{C 2} 2 + A_{a}^{2}} = \frac{|C a C h_{2}| γ_{C a C}}{[C a^{2 +}] γ_{C a^{2} 2} + C r | h_{C}^{2}}$

The equilibrium expression for the given reaction was calculated using the species' activity and concentration.

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Most popular questions from this chapter

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.

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

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?

Find the activity coefficient of each ion at the indicated ionic strength:

$(a) {SO}_{4}^{2 -} \to (μ = 0.01 M) (b) {Sc}^{3 +} \to (μ = 0.005 M) (c) {Ec}^{3 +} \to (μ = 0.1 M) (d) {({CH}_{3} {CH}_{2})}_{3} {NH}^{+} \to (μ = 0.05 M)$

What is the ionic strength of $1 {mMCaCl}_{2}$ ?

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