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

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* 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=([C_9H_8NO_(3)^-][H^+])/([HC_9H_8NO_(3)])`

`K_a=(x* x)/((c -x))`

`2.399* 10^(-4)=(x* 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.