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Calculate the pH of 0.1 M solution of acetic acid. If the degree of dissociation of acid is 0.0132.

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Ka =\(\frac{Ca^2}{(1-a)}\)= Ca

= 0.1 × (0.0132)2 

= 1.74 × 10−5 

pKa = −logpKa = −log(1.74 × 10−5

= 5 − 0.2405 = 4.76 

[H+] = Ca 

= 0.1 × 0.0132 

= 1.32 × 10−3M

pH= − log [H+

= −log(1.32 × 10−3

= 3 − 0.1206 = 2.88

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