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?

Explanation

To solve the problem of how many times silver benzoate is more soluble in a buffer of pH 3.19 compared to its solubility in pure water, we will follow these steps:

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

Silver benzoate dissociates in water as follows:
\[ \text{PhCOOAg} \rightleftharpoons \text{PhCOO}^- + \text{Ag}^+ \]
Let the solubility of silver benzoate in pure water be \( S \). Therefore, at equilibrium:
- \([\text{PhCOO}^-] = S\)
- \([\text{Ag}^+] = S\)

The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{PhCOO}^-][\text{Ag}^+] = S \cdot S = S^2 \]
Given that \( K_{sp} = 2.5 \times 10^{-3} \), we can solve for \( S \):
\[ S^2 = 2.5 \times 10^{-3} \]
\[ S = \sqrt{2.5 \times 10^{-3}} \approx 0.050 \text{ M} = 5.0 \times 10^{-7} \text{ M} \]

### Step 2: Determine the solubility of silver benzoate in a buffer of pH 3.19.

In a buffer solution, the benzoate ion \([\text{PhCOO}^-]\) can react with \( \text{H}^+ \) ions from the buffer:
\[ \text{PhCOO}^- + \text{H}^+ \rightleftharpoons \text{PhCOOH} \]
Let the solubility of silver benzoate in the buffer be \( S_2 \). The equilibrium concentrations will be:
- \([\text{PhCOO}^-] = S_2 - x\)
- \([\text{Ag}^+] = S_2\)
- \([\text{H}^+] = 10^{-3.19} \approx 6.46 \times 10^{-4} \text{ M}\)

The \( K_a \) for benzoic acid is given as:
\[ K_a = 6.46 \times 10^{-5} \]

Using the relationship:
\[ K_a = \frac{[\text{PhCOOH}]}{[\text{PhCOO}^-][\text{H}^+]} \]
We can express this as:
\[ K_a = \frac{x}{(S_2 - x)(6.46 \times 10^{-4})} \]
Assuming \( S_2 \) is much larger than \( x \), we can simplify to:
\[ K_a \approx \frac{x}{S_2 \cdot (6.46 \times 10^{-4})} \]

### Step 3: Relate \( K_{sp} \) and \( K_a \).

The solubility product can be expressed in terms of \( S_2 \):
\[ K_{sp} = [\text{PhCOO}^-][\text{Ag}^+] = (S_2 - x)S_2 \approx S_2^2 \]
From the previous equations, we can relate \( K_{sp} \) and \( K_a \):
\[ \frac{K_w}{K_a} = [\text{OH}^-] = 10^{-14} / (6.46 \times 10^{-5}) \approx 1.55 \times 10^{-10} \]

### Step 4: Calculate the new solubility \( S_2 \).

Using the equilibrium expression:
\[ K_{sp} = S_2^2 \]
We can substitute \( S_2 \) into the equation:
\[ K_{sp} = (S_2)(S_2 - x) \approx S_2^2 \]
Where \( x \) is negligible in comparison to \( S_2 \).

### Step 5: Calculate the ratio of solubilities.

Finally, we can find how many times more soluble silver benzoate is in the buffer compared to pure water:
\[ \text{Ratio} = \frac{S_2}{S_1} = \frac{15.8 \times 10^{-7}}{5.0 \times 10^{-7}} \approx 3.16 \]

### Conclusion:
Silver benzoate is approximately 3 times more soluble in a buffer of pH 3.19 compared to its solubility in pure water.