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# the dissociation constant of a substituted benzoic acid at 25 degree C is 1 * 10^(-4). the pH of a 0.01 M solution of its sodium salt is ...

Sunil Kumar FP
6 years ago
here
BH------B- +H+
let final concentration of H+ =x
final concentration of BH=.01-x
final concentration of B- =x
Ka =x^2/(.01-x)
10^-4 =x^2 /(.01-x)
x^2 +10^-4x -10^-6 =0
x=9.5*10^-4 M
pH=-log(9.5*10^-4)
=3.02
NOOR ASHIQUE
37 Points
3 years ago

Ka (C6H5COOH) = 1 × 10-4

pH of 0.01 M C6H5COONa

C6H5COO- + H2O -------> C6H5COOH + OH-1

0.01(1-h)                                 0.01 h            0.01 h

Kh = Kw/Ka =(0.01 h2)/(1-h)

10-14/10-4 = 10-2 h2/(1-h) (1 – h ≈ 1)

[OH-] = 0.01 h = 0.01 × 10-4 = 10-6

[H+] = 10-8

pH = 8