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Chapter 18: Problem 80

The solubility of \(\mathrm{AgCN}(\mathrm{s})\) in \(0.200 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) is \(8.8 \times 10^{-6} \mathrm{mol} / \mathrm{L} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{AgCN}\).

Short Answer

Expert verified
The \(K_{sp}\) for \(AgCN\) is \(7.744 \times 10^{-11}\).

Step by step solution

01

Write the balanced chemical equation

First, we need to write out the balanced chemical equation. The chemical equation for the dissolution of \(AgCN\) in \(NH_3\) is: \[AgCN(s) + 2NH_3(aq) \leftrightarrow [Ag(NH_3)_2]^+ (aq) + CN^{-} (aq)\]
02

Write the expression for \(K_{sp}\)

Now, express the equilibrium constant for the reaction. In this case, it is called the solubility product constant (\(K_{sp}\)): \[K_{sp} = [[Ag(NH_3)_2]^+] \cdot [CN^-]\]
03

Substitution

As the solubility of \(AgCN\) is given to be \(8.8 \times 10^{-6}\) mol/L, this is the concentration of both [Ag(NH3)2]+ and CN- at equilibrium. Substitute this into our \(K_{sp}\) expression: \[K_{sp} = (8.8 \times 10^{-6}) \cdot (8.8 \times 10^{-6})\]
04

Calculate the value of \(K_{sp}\)

Finally, calculate the value of \(K_{sp}\) by multiplying these concentrations: \[K_{sp} = (8.8 \times 10^{-6})^2 = 7.744 \times 10^{-11}\]

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Most popular questions from this chapter

Write solubility equilibrium equations that are described by the following \(K_{\mathrm{sp}}\) expressions. For example, \(K_{\mathrm{sp}}=\) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \quad\) represents \(\quad \mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq})\). (a) \(K_{\mathrm{sp}}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}\) (b) \(K_{\mathrm{sp}}=\left[\mathrm{BiO}^{+}\right]\left[\mathrm{OH}^{-}\right]\) (c) \(K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\) (d) \(K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]^{3}\left[\mathrm{AsO}_{4}^{3-}\right]^{2}\)

Saturated solutions of sodium phosphate, copper(II) chloride, and ammonium acetate are mixed together. The precipitate is (a) copper(II) acetate; (b) copper(II) phosphate; (c) sodium chloride; (d) ammonium phosphate; (e) nothing precipitates.

To increase the molar solubility of \(\mathrm{CaCO}_{3}(\mathrm{s})\) in a saturated aqueous solution, add (a) ammonium chloride; (b) sodium carbonate; (c) ammonia; (d) more water.

A solution is \(0.05 \mathrm{M}\) in \(\mathrm{Cu}^{2+},\) in \(\mathrm{Hg}^{2+},\) and in \(\mathrm{Mn}^{2+}\). Which sulfides will precipitate if the solution is made to be \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) and \(0.010 \mathrm{M} \mathrm{HCl}(\mathrm{aq}) ?\) For \(\mathrm{CuS}\), \(K_{\mathrm{spa}}=6 \times 10^{-16} ;\) for \(\mathrm{HgS}, K_{\mathrm{spa}}=2 \times 10^{-32} ;\) for \(\mathrm{MnS}\), \(K_{\mathrm{spa}}=3 \times 10^{7}\).

Which one of the following solutions can be used to separate the cations in an aqueous solution in which \(\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.050 \mathrm{M}: 0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq}), 0.05 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\) \(0.001 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}),\) or \(0.50 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\text { aq }) ?\) Explain why.
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