The ionization constant of benzoic acid is 6.46× 10- 5 and Ks p for silver benzoic 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?

Question

The ionization constant of benzoic acid is 6.46× 10- 5 and Ks p for silver benzoic 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? (A) 4 (B) 3.32 (C) 3.01 (D) 2.5

Tardigrade Admin · Accepted Answer

Suppose S is the molar solubility of silver benzoate in water, then

C_{6} H_{5} COO A g_{s} ⇌ C_{6} H_{5} CO O_{a q}^{-} + A g_{a q}^{+}

So

S = 2.5 \times 1 0^{- 13} = 5.0 \times 1 0^{- 7}

M
If the solubility of salt of weak acid of ionization constant

K_{a}

is S, then

K_{s p}, K_{a}

and S' are related to each other at

p H = 3.19.

So

[H^{+}] = 6.46 \times 1 0^{- 4} M

(∵ pH = 3.19)

K_{s p} = S'^{2} [\frac{K _{a}}{K _{a} + [ H ^{+} ]}]

S' = {\frac{2.5 \times 1 0 ^{- 13}}{[ \frac{6.46 \times 1 0 ^{- 5}}{6.46 \times 1 0 ^{- 5} + 6.46 \times 1 0 ^{- 4}} ]}}^{\frac{1}{2}}

S' = {\frac{2.5 \times 1 0 ^{- 13} \times 7.106 \times 1 0 ^{- 4}}{6.46 \times 1 0 ^{- 5}}}^{\frac{1}{2}}

= (2.75 \times 1 0^{- 12})^{\frac{1}{2}}

= 1.658 \times 1 0^{- 6} M

So the ratio of

\frac{S '}{S} = \frac{1.658 \times 1 0 ^{- 6}}{5.0 \times 1 0 ^{- 7}} = 3.32

Silver benzoate is

3.32

times more soluble in buffer of

p H = 3.19

than in pure water.

Q. The ionization constant of benzoic acid is 6.46×10−5 and Ksp​ for silver benzoic 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?

4

3.32

3.01

2.5

Q. The ionization constant of benzoic acid is $6.46 \times 1 0^{- 5}$ and $K_{s p}$ for silver benzoic is $2.5 \times 1 0^{- 13}$ . How many times is silver benzoate more soluble in a buffer of $p H = 3.19$ compared to its solubility in pure water?