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Q.

The dissociation constant of acetic acid at a given temperature is 1.69 x 10-5. The degree of dissociation of 0.01 M acetic acid in presence of 0.01 M HCl is equal to :

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a

0.13

b

0.41

c

0.013 

d

1.69×10−3

answer is C.

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Detailed Solution

H+=H+HCl+H+AcOH1.69×10−5=0.01×0.01α0.01⇒ α=1.69×10−3

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