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Chapter 17: Problem 68

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{S}, K_{a 1}\) and \(K_{a 2}\) for \(\mathrm{H}_{2} \mathrm{S},\) and \(K_{f}=1.1 \times 10^{5}\) for \(\mathrm{AgCl}_{2}^{-}\) , calculate the equilibrium constant for the following reaction: \(\mathrm{Ag}_{2} \mathrm{S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q)\)

Short Answer

Expert verified
To find the equilibrium constant for the overall reaction, multiply the individual equilibrium constants: \(K = K_{sp} \times K_{a1} \times K_{a2} \times K_{f}^{2}\).

Step by step solution

01

Determine the initial reaction equations and their respective equilibrium constants.

We have the following reactions and their equilibrium constants: 1. \(\mathrm{Ag}_{2} \mathrm{S}(s) \rightleftharpoons 2\mathrm{Ag}^{+}(a q)+\mathrm{S}^{2-}(a q)\) with \(K_{sp}\) 2. \(\mathrm{H}_{2} \mathrm{S}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HS}^{-}(a q)\) with \(K_{a1}\) 3. \(\mathrm{HS}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{S}^{2-}(a q)\) with \(K_{a2}\) 4. 2 \(\mathrm{Ag}^{+}(a q) + 2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q)\) with \(K_{f}\)
02

Determine the overall reaction.

Now we'll combine the given reactions in such a way that they add up to the desired reaction without changing the \(K_{sp}, K_{a1}, K_{a2}\), and \(K_{f}\) constants nor the stoichiometric coefficients. The overall reaction is: $\mathrm{Ag}_{2} \mathrm{S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q)$
03

Calculate the equilibrium constant for the overall reaction.

When we multiply or divide the reactions, we can simply do the same with the equilibrium constants (multiply, divide, raise to powers). In this case, we need to combine reaction 1, reaction 2, reaction 3, and reaction 4 (multiplied by 2) such that: 1 * 2 * 3 * (4^2) = overall reaction Hence, the equilibrium constant for the given reaction can be calculated by multiplying the individual equilibrium constants: \(K = K_{sp} \times K_{a1} \times K_{a2} \times K_{f}^{2}\)

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Most popular questions from this chapter

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) \((\mathbf{b} ) \)The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

Assume that 30.0 \(\mathrm{mL}\) of a 0.10 \(\mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a 0.10\(M\) solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (a) Is the pH \(7,\) less than \(7,\) or more than 7 at the equivalence point?\( (\mathbf{d} )\) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating 0.200 \(\mathrm{M}\) solutions of each of the following bases with 0.200 \(M \mathrm{HBr} :(\mathbf{a})\) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\)

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

A biochemist needs 750 \(\mathrm{mL}\) of an acetic acid-sodium acetate buffer with \(\mathrm{pH} 4.50 .\) Solid sodium acetate \((\mathrm{CH}_{3}$$ \mathrm{COONa}\) and glacial acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) are \right. available. Glacial acetic acid is 99\(\% \mathrm{CH}_{3} \mathrm{COOH}\) by mass and has a density of 1.05 \(\mathrm{g} / \mathrm{mL}\) . If the buffer is to be 0.15 \(\mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COOH}\) , how many grams of \(\mathrm{CH}_{3} \mathrm{COONa}\) and how many milliliters of glacial acetic acid must be used?
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