Question

Solve this !

⚠️Don't paste copied answer from G*ogle ​

Answer 1

Answer:

(b) 3.32 is the correct option

Explanation:

Let s M be the solubility of silver benzoate in water.

[Ag  +  ]=[PhCOO  −  ]= sM

$K_{sp} = [PhCOO^-][Ag^+]$

s=5.0×10⁻⁷ M. This is solubility of silver benzoate in pure water.

$pH = -log[H^+] = 3.19$

[H⁺] = 6.456×10⁻⁴

Benzoate ions combine with protons to form benzoic acid but hydrogen ion concentration is constant due to buffer solution.

$PhCOOH \rightleftharpoons PhCOO^- + H^+ \\\\K_a = \frac{[PhCOO^-][H^+]}{PhCOOH} \\\\\frac{[PhCOOH]}{[PhCOO^-]} = \frac{[H^+]}{K_a} = \frac{6.456 \times 10^{-4}}{6.46 \times 10^-5} = 10$

Let y M be the solubility in the buffer solution.

$y = [Ag^+] = [PhCOO^-]+ [PhCOOH] = [PhCOO^-] + 10[PhCOO^-] = 11[PhCOO^-]$

$[PhCOO^-] = \frac{y}{11}$

$K_{sp} = [PhCOO^-][Ag^+]\\\\2.5 \times 10^{-13} = \frac{y}{11} \times y\\\\y = 1.66 \times 10^{-6}\\\\\frac{y}{s} = \frac{1.66 \times 10^{-6}}{5 \times 10^{-7}} = 3.32$

Thus, silver benzoate is 3.32 times more soluble in buffer.

Hope it helps

Please mark as brainliest