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Chapter 15: Problem 103

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 AgCN in the given solution containing 1.0 M H⁺ is \(1.48 \times 10^{-6}\) M.

Step by step solution

01

1. Write the dissolution and acid-base equilibrium reactions.

First, let's write the chemical equilibrium equation for the dissolution of AgCN: \[\mathrm{AgCN}(s) \rightleftharpoons \mathrm{Ag}^{+}(aq) + \mathrm{CN}^{-}(aq)\] Next, we write the chemical equilibrium equation for the acid-base reaction involving H⁺ and CN⁻: \[\mathrm{H}^{+}(aq) + \mathrm{CN}^{-}(aq) \rightleftharpoons \mathrm{HCN}(aq)\]
02

2. Write the Ksp and Ka expressions.

Now, let's write the equilibrium expressions corresponding to the Ksp and Ka values given. For the dissolution of AgCN: \[K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{CN}^{-}]\] For the acid-base reaction between H⁺ and CN⁻: \[K_{\mathrm{a}} = \frac{[\mathrm{HCN}]}{[\mathrm{H}^{+}][\mathrm{CN}^{-}]}\]
03

3. Define the terms in the equilibrium expressions.

Let x be the solubility of AgCN in moles per liter. Then at equilibrium: \[[\mathrm{Ag}^{+}] = x\] \[[\mathrm{CN}^{-}] = x\] \[[\mathrm{HCN}] = x\] \[[\mathrm{H}^{+}] = 1.0\,\mathrm{M}\text, since it is given that the solution contains 1.0 M H⁺)
04

4. Substitute the equilibrium concentrations into the Ksp and Ka expressions.

Now, we substitute the equilibrium concentrations into the Ksp and Ka expressions: For Ksp: \[2.2 \times 10^{-12} = x^2\] For Ka: \[6.2 \times 10^{-10} = \frac{x}{1.0\,\mathrm{M} \times x}\]
05

5. Solve the equations to find the solubility of AgCN.

First, solve for x using the Ka expression: \[6.2 \times 10^{-10} = \frac{x}{x}\] \[\text{Since the value of HCN is negligible, we have:}\] \[[\mathrm{HCN}] \approx 0\] Now, solve for x using the Ksp expression: \[2.2 \times 10^{-12} = x^2\] \[x = \sqrt{2.2 \times 10^{-12}}\] \[x = \mathrm{1.48 \times 10^{-6}\, M}\] Therefore, the solubility of AgCN in the given solution is \(1.48 \times 10^{-6}\) M.

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Most popular questions from this chapter

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} M\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \quad K_{\mathrm{overall}}=?$$

On a hot day, a 200.0 -mL sample of a saturated solution of \(\mathrm{PbI}_{2}\) was allowed to evaporate until dry. If \(240 \mathrm{mg}\) of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{PbI}_{2}\) on this hot day.

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7},\) or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2}\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

Order the following solids (a-d) from least soluble to most soluble. Ignore any potential reactions of the ions with water. a. \(\mathrm{AgCl} \quad K_{\mathrm{sp}}=1.6 \times 10^{-10}\) b. \(\mathrm{Ag}_{2} \mathrm{S} \quad K_{\mathrm{sp}}=1.6 \times 10^{-49}\) c. \(\mathrm{CaF}_{2} \quad K_{\mathrm{sp}}=4.0 \times 10^{-11}\) d. CuS \(\quad K_{\mathrm{sp}}=8.5 \times 10^{-45}\)

Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.
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