Question

Hippuric acid (HC9H8NO3)(HC9H8NO3), found in horse urine, has pKa=3.62pKa=3.62. Part A Calculate the pHpH in 0.140 MM hippuric acid. Express your answer to two decimal places.

Answer 1

Answer:

The pH in 0.140 M hippuric acid solution is 2.2.

Explanation :

Dissociation constant of the acid = $K_a$

$pK_a=-\log[K_a]$

$3.62=-\log[K_a]$

$K_a=2.399\times 10^{-4}$

Concentration of hippuric acid = c = 0.140 M

$HC_9H_8NO_3\rightleftharpoons C_9H_8NO_{3}^-+H^+$

Initially

c           0     0

At equilibrium

(c-x)     x      x

Concentration of acid = c $[HC_9H_8NO_3]=0.140 M$

Dissociation constant of an acid is given by:

$K_a=\frac{[C_9H_8NO_{3}^-][H^+]}{[HC_9H_8NO_{3}]}$

$K_a=\frac{x\times x}{(c -x)}$

$2.399\times 10^{-4}=\frac{x\times x}{(0.140 -x)}$

Solving for x:

x = 0.005677 M

$[H^+]=x = 0.005677 M$

The pH of the solution :

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

$pH=-\log[0.005677 M]=2.246\approx 2.2$

The pH in 0.140 M hippuric acid solution is 2.2.