Problem 49 - Chemistry: Principles and Reactions

Question

Calculate the molar solubility of gold(I) chloride \(\left(K_{s p}=2.0 \times 10^{-13}\right)\) in \(0.10 \mathrm{M} \mathrm{NaCN}\). The complex ion formed is \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) with \(K_{\mathrm{f}}=2 \times 10^{38}\). Ignore any other competing equilibrium systems.

Step-by-Step Solution

Answer

Answer: The molar solubility of gold(I) chloride in 0.10 M sodium cyanide solution is approximately 2.0 x 10^-6 M.

Step 1: Write the balanced chemical equations for the solubility of AuCl and the formation of the complex ion

For the solubility of AuCl: \(\mathrm{AuCl} \rightleftharpoons \mathrm{Au}^{+} + \mathrm{Cl}^{-}\) For the formation of the complex ion: \(\mathrm{Au}^{+} + 2\mathrm{CN}^{-} \rightleftharpoons \left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\)

Step 2: Calculate the initial concentration of cyanide ions

Since NaCN completely dissociates in water, the concentration of cyanide ions is equal to the concentration of NaCN, which is given as 0.10 M.

Step 3: Set up the equilibrium expressions

For the solubility of AuCl, we have the expression: \(K_{sp} = [\mathrm{Au}^{+}][\mathrm{Cl}^{-}]\) For the formation of the complex ion, we have the expression: \(K_{f} = \frac{[\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}]}{[\mathrm{Au}^{+}][\mathrm{CN}^{-}]^2}\)

Step 4: Determine the solubility of AuCl in the presence of the complex ion

Let the solubility of AuCl be represented by S. Then, the concentration of Au+ ions and Cl- ions in the solution will also be S, since it forms in a 1:1 ratio from the dissociation of AuCl. Now, the formation of a complex ion \([\mathrm{Au}(\mathrm{CN})_{2}]^-\) reduces the concentration of Au+ ions. Let the concentration of the complex ion be x. Hence, the concentration of Au+ ions left in the solution will be (S - x). From the above equilibrium expressions, we can write: \(K_{sp} = S(S - x)\) ...(1) \(K_{f} = \frac{x}{(S - x)(0.10)^2}\) ...(2)

Step 5: Solve for the concentration of complex ion and the molar solubility of AuCl

From equation (2), we can write the expression for x (complex ion concentration) as, \(x = K_{f}(S - x)(0.10)^2\) As \(K_{f}\) is very large, we can consider the concentration of complex ion x to be approximately equal to S. Now we can solve for S: \(x \approx S = K_{f}(S - x)(0.10)^2\) \(S = K_{f}(S - S)(0.10)^2\) \(S = K_{f} \times 0.01 \times S\) Now, we can find the value of S: \(\frac{S}{0.01 K_{f}} = 1\) \(S = \frac{0.01}{K_{f}}\) But we have not taken \(K_{sp}\) into consideration yet. As in the previous analysis, the concentration of complex ion, x, is approximately equal to S. Therefore, we can write equation (1) as: \(K_{sp} = S^2\) Plugging the value of S from above, \(K_{sp} = \left(\frac{0.01}{K_{f}}\right)^2\) Now, we can solve for S: \(S = \sqrt{K_{sp} \times K_{f} \times 0.01^2}\) Plugging the given values of \(K_{sp}\) and \(K_{f}\): \(S = \sqrt{2.0\times 10^{-13} \times 2.0 \times 10^{38} \times 0.0001}\) Finally, we can calculate the solubility S: \(S \approx 2.0 \times 10^{-6}\, M\) The molar solubility of gold(I) chloride in \(0.10\,M\, \mathrm{NaCN}\) is approximately \(2.0 \times 10^{-6}\, M\).

Key Concepts

Gold(I) Chloride

Gold(I) chloride, often represented as AuCl, is a compound consisting of gold and chloride ions. It's known for having low solubility in water. In pure water, AuCl dissociates slightly, producing gold ions (\(\mathrm{Au}^+\)) and chloride ions (\(\mathrm{Cl}^-\)). This process is important in calculating its solubility, often expressed through the solubility product constant (\(K_{sp}\)).

The dissociation equilibrium can be seen as:

• \(\mathrm{AuCl} \rightleftharpoons \mathrm{Au}^+ + \mathrm{Cl}^-\)

Understanding this reaction is essential when calculating how factors like the presence of other ions influence the solubility of gold(I) chloride.

Complex Ion Formation

The formation of complex ions can drastically change the solubility of a compound. In the case of gold(I) chloride, the presence of cyanide ions (\(\mathrm{CN}^-\)) leads to the formation of a stable complex ion: \([\mathrm{Au}(\mathrm{CN})_2]^-\).

This occurs through the following equilibrium reaction:

• \(\mathrm{Au}^+ + 2\mathrm{CN}^- \rightleftharpoons [\mathrm{Au}(\mathrm{CN})_2]^-\)

The formation constant (\(K_f\)) is extremely large, indicating a strong preference for forming the complex, which effectively reduces the concentration of free \(\mathrm{Au}^+\) ions. As a result, the solubility of gold(I) chloride increases in such solutions.

Solubility Product Constant

The solubility product constant, \(K_{sp}\), represents the extent to which a compound can dissolve in water. For gold(I) chloride, this value helps us understand its low natural solubility.

The expression for \(K_{sp}\) in terms of the ions is:

• \(K_{sp} = [\mathrm{Au}^+][\mathrm{Cl}^-]\)

In this equation, \([\mathrm{Au}^+]\) and \([\mathrm{Cl}^-]\) are the molar concentrations of the dissociated ions. The \(K_{sp}\) value reflects how much of the compound can dissolve before the solution becomes saturated. In the presence of complex formation, this equilibrium shifts, often leading to higher solubility.

Equilibrium Calculations

Equilibrium calculations allow us to determine the solubility of compounds under specific conditions. In the given exercise, we first wrote the equilibrium equations for the dissolution of gold(I) chloride and the complex ion formation.

For gold(I) chloride, \(K_{sp}\) gives:

• \(K_{sp} = S(S - x)\)

And for complex formation, \(K_f\) is:

• \(K_f = \frac{[\mathrm{Au}(\mathrm{CN})_2]^-}{(S - x)(0.10)^2}\)

Assuming significant complex formation, the concentration of \([\mathrm{Au}(\mathrm{CN})_2]^-\) approaches \(S\). Solving these equations with the given \(K_{sp}\) and \(K_f\) values produces the molar solubility. These calculations show how equilibrium constants and ion concentrations define the solubility in complex systems.