Question

Hippuric acid (HC9H8NO3), found in horse urine, has pKa=3.62.

Calculate the pH in 0.150 M hippuric acid.

Answer 1

Solution :-

lets calculate the Ka using the pka

Pka = -log ka

ka = antilog [-pka]

ka= antilog [-3.62]

ka= 2.399*10^-4

now using the Ka lets calculate the pH

When acid dissociate it forms the H3O+ and conjugate base

HA         + H2O ------ > H3O^+ + A^-       (HA is used for the Hippuric acid)

0.150 M                               0            0

-x                                        +x           +x

0.150-x                                x             x

there fore the Ka equation is

Ka= [H3O^+] [A^-]/[HA]

lets put the values in the formula

2.399*10^-4 = [x][x]/[0.150-x ]

neglect the x from the denominator since ka is small

then we get

2.399*10^-4 = [x][x]/[0.150]

2.399*10^-4 * 0.150 = x^2

3.599*10^-5 = x^2

taking square root on both side we get

5.99*10^-3 =x = [H3O+]

now lets calculate the pH

pH= -log[ H3O+]

pH = -log [5.99*10^-3]

pH= 2.22