The ionization constant of benzoic acid is 6.46 × 10^–5 and Ks_p for silver benzoate are 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?

Given:

The ionization constant of benzoic acid (K_a) is 6.46 × 10^–5

Ks_p for silver benzoate is 2.5 × 10^–13

pH = 3.19

Ionization of silver benzoate:

C₆H₅COOAg ↔ C₆H₅COO^- + Ag⁺

Solubility in water: Let solubility in water is x mol/l Then

[C₆H₅COO^- ] = [Ag⁺] = x mol/l

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

^⇒[A⁺] = C₆H₅COO^-

⇒[B^-] = Ag⁺

^⇒∴ K_sp = [C₆H₅COO^-][Ag⁺]

As K_sp = 2.5 × 10^–13 (given)

⇒2.5 × 10^–13 = x²

⇒x = 5 × 10^-7

Solubility in buffer of pH =3.19

As we know that,

pH = –log [H⁺]

∴ –log [H⁺] = 3.19

By taking antilog of both the sides, we get

⇒[H⁺] = antilog -3.19

⇒[H⁺] = 6.457 × 10^-4 M

C₆H₅COO^- ions combine with the H⁼ ions o from benzoic acid but [H⁺] remains almost constant because we have buffer solution. Now,

C₆H₅COOH ↔ C₆H₅COO^- + H⁺

As we know that,

K_a =

⇒K_a =

⇒ =

[H⁺] = 6.457 × 10^-4 M (calculated above)

K_a = 6.46 × 10^–5 (given)

∴ = = = 10

⇒[C₆H₅COOH] = 10 [C₆H₅COO^-]

Suppose solubility in the buffer solution is y mol/l.

Then as most of the benzoate ion is converted into benzoic acid molecules(which remains almost ionized) we have

Y = [Ag⁺] = [C₆H₅COO^-] + [C₆H₅COOH]

As [C₆H₅COOH] = 10 [C₆H₅COO^-] (calculated above)

∴ y = [C₆H₅COO^-] + 10 [C₆H₅COO^-]

⇒y = 11 [C₆H₅COO^-]

⇒[C₆H₅COO^-] =

As we know that,

K_sp = [C₆H₅COO^-][Ag⁺]

As K_sp = 2.5× 10^-13 (given0

[Ag⁺] = y, [C₆H₅COO^-] =

∴ 2.5× 10^-13 = × y

⇒Y² = 2.75 × 10^-12

⇒Y = √ 2.75 × 10^-12

⇒Y = 1.66 × 10^-6

Now,

= = = 3.32

Thus, the silver benzoate is 3.32 times more soluble in a buffer of pH compared to its solubility in pure water

Note: In case of weak acids, the solubility is more in the acidic solution than in water. The reason, in general, may be explained as below:

Taking example of C₆H₅COOAg, we have

C₆H₅COOAg ↔ C₆H₅COO^- + Ag⁺

In acidic solution, the anions (C₆H₅COO^- in the present case) undergo protonation in presence of acid. Thus, C₆H₅COO^- ions are removed. Hence equilibrium shifts forward producing more Ag⁺ ions. Alternatively, as C₆H₅COO^- are removed, Q_sp decreases. In order to maintain solubility product equilibrium (Q_sp= K_sp), Ag⁺ ion concentration must increase. Hence solubility is more.

Note: A buffer solution is defined as a solution which resists any change in its pH value even when small amount of acid or base added to it.