Problem 109 - Chemistry

Question

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

Step-by-Step Solution

Answer

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 1: 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: \[AgCN(s) \rightleftharpoons Ag^{+}(aq) + CN^{-}(aq)\]

Step 2: Write the Ksp expression for AgCN

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

Step 3: 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: \[HCN(aq) + H^+(aq) \rightleftharpoons H_2O(l) + CN^{-}(aq)\]

Step 4: Write the Ka expression for HCN

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

Step 5: 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.

Step 6: 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 \[K_a = \frac{[CN^{-}]}{[HCN][H^{+}]} = \frac{s}{[HCN](1.0)} =\frac{s}{[HCN]}\]

Step 7: 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: \[6.2 \times 10^{-10} = \frac{1.48 \times 10^{-6}}{[HCN]}\] \[HCN \approx 2.38 \times 10^{-4}\]

Step 8: 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.

Key Concepts

Ksp expression

The Ksp, or solubility product constant, is crucial in calculating the solubility of sparingly soluble compounds, such as silver cyanide (\(\text{AgCN}\)). For a dissociation reaction like \(\text{AgCN(s) } \rightleftharpoons \text{ Ag}^{+}\text{(aq)} + \text{ CN}^{-}\text{(aq)}\), the \(K_{sp}\) expression is written in terms of the concentrations of the ions produced:

• \[K_{sp} = [\text{Ag}^{+}][\text{CN}^{-}]\]

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This relationship tells us that the product of the molar concentrations of the dissolved ions at equilibrium equals the \(K_{sp}\) value. For silver cyanide, the \(K_{sp}\) is provided as \(2.2 \times 10^{-12}\), showing us it is only slightly soluble.
Understanding the \(K_{sp}\) expression allows us to set up the equilibrium expression necessary to solve for the solubility of \(\text{AgCN}\).

ICE table

An ICE table is a strategic approach to track the Initial concentrations, Changes in concentration, and Equilibrium concentrations in a given chemical reaction.
To determine the solubility of \(\text{AgCN}\), we first establish initial concentrations:

• \(\text{Ag}^{+}\) and \(\text{CN}^{-}\) start at 0 \(\text{M}\), as AgCN has yet to dissolve.
• \(\text{H}^{+}\) begins at \(1.0 \text{ M}\), given a pre-existing condition in the solution.
• The change, denoted as \(s\), represents how much AgCN dissolves.

As dissolution progresses, both \(\text{Ag}^{+}\) and \(\text{CN}^{-}\) increase by \(s\), leading to equilibrium concentrations of \(s\) each. An ICE table effectively simplifies setting up the expressions for the equilibrium constants, aiding the solution process.

Ka expression

The \(K_{a}\) expression represents the acid dissociation constant, offering insight into the extent of acid dissociation in water. For the reaction involving hydrogen cyanide (\(\text{HCN}\) and \(\text{H}^{+}\)), we have:

• \(\text{HCN (aq) } + \text{H}^+(\text{aq}) \rightleftharpoons \text{H}_2\text{O(l) } + \text{CN}^-\text{(aq)}\)
• The \(K_{a}\) expression is:\[K_a = \frac{[\text{CN}^{-}]}{[\text{HCN}][\text{H}^{+}]}\]

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This formula relates the change in concentration of cyanide ions to that of \(\text{HCN}\) and \(\text{H}^+\).
Given a \(K_{a}\) for \(\text{HCN}\) at \(6.2 \times 10^{-10}\), understanding this relationship assists in determining how shifts in ion concentrations affect system equilibrium.
Employing the \(K_{a}\) expression ensures accurate solubility approximations and system evaluations.

Equilibrium concentrations

Achieving equilibrium concentrations involves calculating the end concentrations of all reactants and products in a balanced chemical reaction. In the context of \(\text{AgCN}\) solubility, equilibrium is the state where the rate of dissolution equals the rate of precipitation.
Utilizing the \(K_{sp}\) and \(K_{a}\) values:

• The equilibrium concentrations for \(\text{Ag}^{+}\) and \(\text{CN}^{-}\) are both \(s\), derived from the dissolution of \(\text{AgCN}\).
• Given the established \(K_{sp}\), \(s\) is calculated as \(\sqrt{2.2 \times 10^{-12}}\approx 1.48 \times 10^{-6} \text{ M}\).

Substituting this value into the \(K_{a}\) equation provides the equilibrium concentration of \(\text{HCN}\), ensuring accurate depiction of ionic distribution at system equilibrium.
These concentrations facilitate a holistic understanding of solubility behavior in various chemical environments.