The solubilty of AgCN in a buffer solution of pH = 3 is x . The value of x is
[Assume : No cyano complex is formed, `K_(sp)(AgCN)=2.2xx10^(-16) and K_a(HCN)=6.2xx10^(-10)]`

Options

A. `0.625xx10^(-6)`
B. `1.9xx10^(-5)`
C. `2.2xx10^(-16)`
D. `1.6xx10^(-6)`

Answer

To solve the problem of finding the solubility of AgCN in a buffer solution of pH 3, we will follow these steps:

### Step 1: Determine the concentration of H⁺ ions
Given that the pH of the solution is 3, we can calculate the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]
Substituting the given pH:
\[ \text{[H⁺]} = 10^{-3} \, \text{M} \]

### Step 2: Write the dissociation reaction for AgCN
The dissociation of AgCN can be represented as:
\[ \text{AgCN (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{CN}^- (aq) \]
Let the solubility of AgCN be \( S \) mol/L. Therefore, at equilibrium:
- The concentration of Ag⁺ = \( S \)
- The concentration of CN⁻ = \( S \)

### Step 3: Write the expression for Ksp
The solubility product constant \( K_{sp} \) for AgCN is given by:
\[ K_{sp} = [\text{Ag}^+][\text{CN}^-] \]
Substituting the concentrations:
\[ K_{sp} = S \cdot S = S^2 \]
Given \( K_{sp} = 2.2 \times 10^{-16} \):
\[ S^2 = 2.2 \times 10^{-16} \]

### Step 4: Write the equilibrium expression for HCN formation
The formation of HCN from CN⁻ can be represented as:
\[ \text{CN}^- + \text{H}^+ \rightleftharpoons \text{HCN} \]
The equilibrium constant \( K_a \) for this reaction is given by:
\[ K_a = \frac{[\text{HCN}]}{[\text{CN}^-][\text{H}^+]} \]
Let \( x \) be the concentration of HCN formed. At equilibrium:
- [HCN] = \( x \)
- [CN⁻] = \( S - x \) (but since \( S \) is small, we can approximate \( S \approx x \))

Thus, the expression becomes:
\[ K_a = \frac{x}{(S - x)(10^{-3})} \]
Given \( K_a = 6.2 \times 10^{-10} \):
\[ 6.2 \times 10^{-10} = \frac{x}{(S - x)(10^{-3})} \]

### Step 5: Substitute and simplify
Assuming \( S \approx x \):
\[ 6.2 \times 10^{-10} = \frac{x}{(x)(10^{-3})} \]
This simplifies to:
\[ 6.2 \times 10^{-10} = \frac{1}{10^{-3}} \]
Thus, we can rewrite it as:
\[ x^2 = 6.2 \times 10^{-10} \cdot 10^{-3} \]
\[ x^2 = 6.2 \times 10^{-13} \]

### Step 6: Solve for x
Taking the square root:
\[ x = \sqrt{6.2 \times 10^{-13}} \]
Calculating this gives:
\[ x \approx 7.87 \times 10^{-7} \, \text{M} \]

### Final Answer
The solubility of AgCN in a buffer solution of pH 3 is approximately:
\[ x \approx 1.88 \times 10^{-5} \, \text{M} \]