The ionisation constant of benzoic acid `(PhCOOH)` is `6.46 xx 10^(-5)` and `K_(sp)` for silver benzoate is `2.5 xx 10^(-3)`. How many times is silver benzoate more soluble in a buffer of `pH 3.19` compared to its solubility is pure water?

Options

A. a. 4
B. b. 3.32
C. c. 3.01
D. d. 2.5

Explanation

To solve the problem, we need to determine how many times silver benzoate is more soluble in a buffer solution with a pH of 3.19 compared to its solubility in pure water. We'll follow these steps:

### Step 1: Determine the solubility of silver benzoate in pure water.

The dissociation of silver benzoate (AgC₆H₅O₂) in water can be represented as:
\[ \text{AgC}_6\text{H}_5\text{O}_2 (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{C}_6\text{H}_5\text{O}_2^- (aq) \]

Let the solubility of silver benzoate in pure water be \( C \). Then, at equilibrium:
- \([Ag^+] = C\)
- \([C_6H_5O_2^-] = C\)

The solubility product constant (\( K_{sp} \)) is given by:
\[ K_{sp} = [Ag^+][C_6H_5O_2^-] = C \cdot C = C^2 \]

Given \( K_{sp} = 2.5 \times 10^{-3} \):
\[ C^2 = 2.5 \times 10^{-3} \]
\[ C = \sqrt{2.5 \times 10^{-3}} \]
\[ C = \sqrt{2.5} \times 10^{-1.5} \]
\[ C \approx 1.58 \times 10^{-1.5} \approx 5.0 \times 10^{-2} \text{ mol/L} \]

### Step 2: Determine the hydrogen ion concentration in the buffer solution.

The pH of the buffer solution is given as 3.19. We can find the hydrogen ion concentration (\([H^+]\)) using:
\[ [H^+] = 10^{-\text{pH}} \]
\[ [H^+] = 10^{-3.19} \approx 6.46 \times 10^{-4} \text{ mol/L} \]

### Step 3: Use the ionization constant of benzoic acid to find the solubility in the buffer solution.

The dissociation of benzoic acid (PhCOOH) can be represented as:
\[ \text{PhCOOH} \rightleftharpoons \text{PhCOO}^- + H^+ \]

The ionization constant (\( K_a \)) is given by:
\[ K_a = \frac{[\text{PhCOO}^-][H^+]}{[\text{PhCOOH}]} \]

Let \( x \) be the solubility of silver benzoate in the buffer solution. The concentration of benzoate ion will be approximately \( x \) and the concentration of benzoic acid will be \( [H^+] = 6.46 \times 10^{-4} \).

Using \( K_a = 6.46 \times 10^{-5} \):
\[ K_a = \frac{x \cdot [H^+]}{[\text{PhCOOH}]} \]

Assuming the concentration of benzoic acid is much larger than \( x \):
\[ K_a \approx \frac{x \cdot 6.46 \times 10^{-4}}{[H^+] + x} \]

Since \( x \) is small compared to \( [H^+] \):
\[ K_a = \frac{x \cdot 6.46 \times 10^{-4}}{[H^+]} \]
\[ 6.46 \times 10^{-5} = \frac{x \cdot 6.46 \times 10^{-4}}{6.46 \times 10^{-4}} \]
\[ x = \frac{6.46 \times 10^{-5} \cdot 6.46 \times 10^{-4}}{6.46 \times 10^{-4}} \]
\[ x = 10 \cdot 6.46 \times 10^{-4} \]
\[ x \approx 6.46 \times 10^{-3} \text{ mol/L} \]

### Step 4: Calculate the ratio of solubility in buffer to solubility in pure water.

Now, we can find the ratio of solubility in the buffer solution to that in pure water:
\[ \text{Ratio} = \frac{x}{C} = \frac{6.46 \times 10^{-3}}{5.0 \times 10^{-2}} \]
\[ \text{Ratio} = \frac{6.46}{5.0} \approx 1.292 \]

### Conclusion

Silver benzoate is approximately **1.29 times** more soluble in a buffer of pH 3.19 compared to its solubility in pure water.

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