The solubility of a compound can be affected by the pH of the solution it is in. In the case of silver benzoate, its solubility can be compared in a buffer solution of pH = 3.19 and in pure water.

To determine the solubility in the buffer solution, we need to consider the ionization of benzoic acid and the formation of silver benzoate.

1. Ionization of benzoic acid:
The ionization constant (Ka) of benzoic acid is given as 6.46 x 10^-5. In water, benzoic acid partially dissociates into its conjugate base, benzoate ion (C6H5COO-) and a hydrogen ion (H+).
C6H5COOH ⇌ C6H5COO- + H+

2. Formation of silver benzoate:
When benzoate ions react with silver ions (Ag+), silver benzoate (AgC6H5COO) is formed.
C6H5COO- + Ag+ → AgC6H5COO

The solubility product constant (Ksp) for silver benzoate is given as 2.5 x 10^-13. Ksp is a measure of the solubility of a compound in a solution. It relates the concentrations of the ions in a saturated solution of the compound.

Now, let's compare the solubility of silver benzoate in the buffer solution of pH = 3.19 and in pure water.

In a buffer solution of pH = 3.19, the concentration of hydrogen ions (H+) is higher than in pure water. This higher concentration of H+ ions can shift the equilibrium of the ionization of benzoic acid to the left, decreasing the concentration of benzoate ions (C6H5COO-).

In pure water, the concentration of hydrogen ions (H+) is lower compared to the buffer solution, allowing the ionization of benzoic acid to proceed to a greater extent, resulting in a higher concentration of benzoate ions (C6H5COO-).

Since the solubility of silver benzoate is directly proportional to the concentration of benzoate ions, it will be higher in the buffer solution of pH = 3.19 compared to pure water.

To find the ratio of solubility, we need to calculate the concentration of benzoate ions in both solutions using the ionization constant of benzoic acid.

The concentration of benzoate ions in the buffer solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
3.19 = -log(6.46 x 10^-5) + log([A-]/[HA])
log([A-]/[HA]) = 3.19 + log(6.46 x 10^-5)
[A-]/[HA] = antilog(3.19 + log(6.46 x 10^-5))

Similarly, the concentration of benzoate ions in pure water can be calculated using the ionization constant of benzoic acid at pH = 7 (neutral pH).

The solubility ratio can then be calculated as the concentration of benzoate ions in the buffer solution divided by the concentration of benzoate ions in pure water.

Given