CBSE Class 11-science Answered
Acetic acid has a dissociation constant of 1.8*10^-5 . calculate the pH value of the decinormal solution of acetic acid ???????
Asked by Tushar | 21 Dec, 2013, 11:22: AM
Expert Answer
Given:
Normality of acetic acid = 0.1 N (decinormal solution)
Dissociation constant = Ka = 1.8x10-5
Solution:
CH3COOH + H2O ? CH3COO- +H3O+
Ka = [H3O+][CH3COO-]/[CH3COOH]
1.8x10-5 = x2 / (0.1- x)
x is very negligible compare to initial concentration of acid.
Hence,
1.8x10-5 = x2 / 0.1
x2 = 1.8 x 10-5 x 0.1
x =[H3O+] = 0.001342 N
0.001342 N acetic acid = 0.001342 M acetic acid
As [H+] = 0.001342 M
pH = -log10[H+]
pH= - log10(0.001342)
pH= 2.87
Answered by Prachi Sawant | 21 Dec, 2013, 01:44: PM
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