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 | 21st 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 | 21st Dec, 2013, 01:44: PM

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