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Chapter 16: Problem 109

Calculate the solubility of $\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\( in a solution containing \)1.0 M\( \)\mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)$

Short Answer

Expert verified
The solubility of $\mathrm{AgCN}$ in a solution containing $1.0 \mathrm{M}$ $\mathrm{H}^{+}$ is approximately \(1.48 \times 10^{-6} \mathrm{M}\).

Step by step solution

01

Write the balanced chemical equation for the dissolution of AgCN in water

We will write the balanced chemical equation for the dissolution of silver cyanide in water: \[AgCN(s) \rightleftharpoons Ag^{+}(aq) + CN^{-}(aq)\]
02

Write the Ksp expression for AgCN

Write the Ksp expression for the dissolution of silver cyanide in water: \[K_{sp} = [Ag^{+}][CN^{-}]\]
03

Write the balanced chemical equation for the reaction of HCN with H+ ions

Write the balanced chemical equation for the reaction of hydrogen cyanide with H+ ions in water: \[HCN(aq) + H^+(aq) \rightleftharpoons H_2O(l) + CN^{-}(aq)\]
04

Write the Ka expression for HCN

Write the Ka expression for the reaction of hydrogen cyanide with H+ ions in water: \[K_a = \frac{[CN^{-}]}{[HCN][H^{+}]}\]
05

Set up an ICE table

Set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations of the species. Since we are looking for the solubility of AgCN, we can assume the initial concentrations of Ag+ and CN- are zero: \[ \begin{array}{l|l|l} {\text{Species}} & {\text{Initial (M)}} & {\text{Equilibrium (M)}} \\ \hline {Ag^{+}} & 0 & s \\ CN^{-} & 0 & s \\ {H^{+}} & {1.0} & 1.0 \\ {HCN} & {?} & {?} \end{array} \] Where "s" represents the solubility of AgCN at equilibrium.
06

Substitute equilibrium concentrations into Ksp and Ka expressions

Substituting equilibrium concentrations into the Ksp and Ka expressions, we have: \[K_{sp} = [Ag^{+}][CN^{-}] = (s)(s) = s^2\] and \[K_a = \frac{[CN^{-}]}{[HCN][H^{+}]} = \frac{s}{[HCN](1.0)} =\frac{s}{[HCN]}\]
07

Solve for s using the given values of Ksp and Ka

Now, we can plug in the given values of Ksp and Ka, and solve for s: \[K_{sp} = 2.2 \times 10^{-12} = s^2\] \[K_a = 6.2 \times 10^{-10} = \frac{s}{[HCN]}\] First, solve for "s" in the Ksp equation: \[s = \sqrt{2.2 \times 10^{-12}}\] \[s \approx 1.48 \times 10^{-6}\] Then, substitute this value for "s" into the Ka equation: \[6.2 \times 10^{-10} = \frac{1.48 \times 10^{-6}}{[HCN]}\] \[HCN \approx 2.38 \times 10^{-4}\]
08

Find the solubility of AgCN

Since we found that s ≈ 1.48 × 10^-6, we can now say that the solubility of AgCN in a solution containing 1.0 M H⁺ is approximately 1.48 × 10^-6 M.

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Most popular questions from this chapter

Nitrate salts are generally considered to be soluble salts. One of the least soluble nitrate salts is barium nitrate. Approximately 15 g of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) will dissolve per liter of solution. Calculate the \(K_{\mathrm{sp}}\) value for barium nitrate.

Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of $8.9 \times 10^{-12}\( . If a 10.0 -g sample of \)\mathrm{Mg}(\mathrm{OH})_{2}$ is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10-M \mathrm{KIO}_{3}\) solution is $2.6 \times 10^{-11} \mathrm{mol} / \mathrm{L}\( . Calculate \)K_{\mathrm{sp}}$ for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}.\)

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

What happens to the \(K_{\mathrm{sp}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.
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