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The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

Grade:11

1 Answers

Arun
25763 Points
3 years ago
Dear student
 
[H30]+ = 0.01 + Ca = 0.01 M
[CH3COO]- = Ca = 0.01 a
[CH3COOH] = 0.01
 
1.69 * 10-5 = (0.01)(0.01 a)/(0.01)
0.01 a = 1.69 * 10-5
a = 1.69 * 10-3
a = 0.169 * 10-2
Hence option C is correct.
Regards
Arun (askIITians forum expert)

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