The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Asked by Pragya Singh | 1 year ago |  276

##### Solution :-

Here, pH = 3.19

$$[H_3O^{+}] = 6.46\times 10^{-4}$$

$$C_6H_5COOH + H_2O$$

$$→C6H5COO^{-} + H_3O$$

$$K_{a}=\dfrac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{C_{6}H_{5}COOH}$$

$$K_{a}=\dfrac{[C_{6}H_{5}COOH]}{C_{6}H_{5}COO^{-}}$$

$$= \dfrac{[H_{3}O^{+}]}{K_{a}}$$

$$= \dfrac{6.46\times 10^{-4}}{6.46\times 10^{-5}} = 10$$

Assuming the solubility of silver benzoate (C6H5COOAg) is y mol/L.

Now,  [Ag+] = y

$$[C_6H_5COOH] = [C_6H_5COO^{-}] = y$$

$$10[C_6H_5COO^{-}] + [C_6H_5COO^{-}]= y$$

$$[C_6H_5COO^{-}] = \dfrac{y}{11}$$

$$Ksp= [Ag^{+}][C_6H_5COO^{-}]2.5\times 10^{-13}$$

$$= y\dfrac{y}{11}$$

y =$$1.66\times 10^{-6}mol/L$$

Hence, solubility of C6H5COOAg in buffer of pH

= 3.19 is $$1.66\times 10^{-6}mol/L$$.

For, water: Assuming the solubility of silver benzoate (C6H5COOAg) is x mol/L.

Now,  [Ag+] = y’ M and [CH3COO] = y’M

Ksp =$$[Ag^{+}][C6H5COO^{-}]$$

Ksp = (y’)2 $$= \sqrt{K_{sp}}$$

= $$\sqrt{2.5\times 10^{-13}}$$

$$= 5\times 10^{-7} mol/L$$

$$\dfrac{y}{x} = \dfrac{1.66\times 10^{-6}}{5\times 10^{-7}}$$

Thus, the solubility of silver benzoate in water is 3.32 times the solubility of silver benzoate in pH = 3.19.

Answered by Abhisek | 1 year ago

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