Answer to Question #82227 in Inorganic Chemistry for Raza Ahmed

Question

Calculate the pH of a 0.01moldm-3 solution of propanoic acid which has a constant of 1.34x10-5 moldm-3

Answer 1

Calculate the pH of a 0.01moldm-3 solution of propanoic acid which has a constant of 1.34x10-5 moldm-3

C(C2H5COOH) = 0.01 mol/l

Ka= 1.34x10-5

pH- ?

Solution

C2H5COOH = C2H5COO- + H+

Ka = [C2H5COO-][H+]/[C2H5COOH]

[C2H5COOH]= C(C2H5COOH), because Ka is small.

[C2H5COO-]=[H+]= x

x^2 = Ka*C= 1.34x10-5 * 0.01= 1.34x10-7

x= 0.000366. pH = - lg[H]= -lg(0.000366)= 3.43

Answer: pH = 3.43