Problem 68 - Chemistry The Central Science

Question

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: \(\mathrm{Ag}_{2} \mathrm{S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q)\)

Step-by-Step Solution

Answer

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 1: 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 \(K_{f}\)

Step 2: 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: $\mathrm{Ag}_{2} \mathrm{S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q)$

Step 3: 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: \(K = K_{sp} \times K_{a1} \times K_{a2} \times K_{f}^{2}\)

Key Concepts

Solubility Product (Ksp)

The solubility product constant, often abbreviated as \( K_{sp} \), is crucial in predicting the solubility of ionic compounds in water. It specifically applies to sparingly soluble salts, such as \( \text{Ag}_2\text{S} \) in the given exercise.

The formula for the solubility product of a salt \( ext{A} ext{B} ightleftharpoons ext{A}^{+} + ext{B}^{-} \) is given by: \[ K_{sp} = [A^+]^a[B^-]^b \] where \([A^+]\) and \([B^-]\) are the molar concentrations of the ions in the solution at equilibrium and \(a\) and \(b\) are the stoichiometric coefficients.

For \( \text{Ag}_2\text{S} \), which dissociates in water according to: \[ \text{Ag}_2\text{S}(s) \rightleftharpoons 2\text{Ag}^{+}(aq) + \text{S}^{2-}(aq) \] The \( K_{sp} \) expression will be: \[ K_{sp} = [Ag^+]^2[S^{2-}] \] Understanding \( K_{sp} \) helps us determine the extent to which this salt will dissolve in a solution under specified conditions.

Acid Dissociation Constant (Ka)

The acid dissociation constant, represented as \( K_{a} \), measures the strength of an acid in terms of its ability to donate protons (\( H^+ \)) in an aqueous solution.

For a general acid \( HA\) dissociating as: \[ HA(aq) \rightleftharpoons H^+(aq) + A^-(aq) \] The expression for \( K_a \) is: \[ K_{a} = \frac{[H^+][A^-]}{[HA]} \]

In the original problem, \( \text{H}_2\text{S} \) is considered with two dissociation steps and corresponding constants \( K_{a1} \) and \( K_{a2} \):

• \( \text{H}_2\text{S} \rightleftharpoons H^+ + \text{HS}^- \), described by \( K_{a1} \)
• \( \text{HS}^- \rightleftharpoons H^+ + \text{S}^{2-} \), described by \( K_{a2} \)

Each \( K_{a} \) value informs us how readily the proton dissociation occurs, with a higher value indicating a stronger acid.

Formation Constant (Kf)

The formation constant, known as \( K_{f} \), describes the stability of complex ions in solution. Specifically, it refers to the equilibrium between the ions that form a complex ion.

For a reaction where a complex ion \( \text{ML}_n \) is formed: \[ M^{+} + nL^{-} \rightleftharpoons ML_n \] the formation constant expression is: \[ K_{f} = \frac{[ML_n]}{[M^+][L^-]^n} \]
In the problem at hand, the reaction involves forming \( \text{AgCl}_2^- \) from \( \text{Ag}^+ \) and \( \text{Cl}^- \). The given \( K_{f} = 1.1 \times 10^5 \) reveals the high stability and likelihood of this complex ion to form in solution.

Stronger complex formation results in larger \( K_{f} \) values, which indicates that even at low reactant concentrations, the complex ion forms readily.