Question

Calculate the pH of a 0.100 M sodium benzoate, NaC6H5CO2, solution. The Ka of acetic acid is 6.3 x 10-5

Answer 1

Consider reaction, C₆H₅COO ^- (aq) + H₂O (l)   OH ^- (aq) + C₆H₅COOH (aq)

Equilibrium constant for above reaction, K _b = [  OH ^- ] [ C₆H₅COOH ] / [C₆H₅COO ^- ] = K _w / K _a

K _b = K _w / K _a = 1.0 10 ^-14 / 6.3 10 ^-05 = 1.59 10 ^-10

Let's use ICE table.

Concentration ( M )	C6H5COO - (aq)    OH - (aq) + C6H5COOH (aq)
I	0.1
C	-X	+X	+X
E	0.1-X	X	X

K _b = ( X) (X) / 0.1 - X = 1.59 10 ^-10

X ² / 0.1 - X = 1.59 10 ^-10

The value of dissociation constant is very small 1.59 10 ^-10 , hence we can assume X is very small as compared to 0.1. We can write 0.1-X 0.1.

X ² / 0.1 = 1.59 10 ^-10

X ² = 0.1 1.59 10 ^-10

X ² = 1.59 10 ^-11

X = 3.98   10 ^-06

We have, X = [  OH ^- ] = [ C₆H₅COOH ] = 3.98   10 ^-06 M

We have relation, pOH = - log [  OH ^- ]

pOH = - log   3.98   10 ^-06

pOH = 5.40

We have relation, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 5.40

pH = 8.6

ANSWER : pH = 8.6