How much water must be added to 300 mL of 0.2 M solution of CH3COOH(Ka=1.8×10−5) for the degree of ionisation (α) of the acid to double?
A
600 mL
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B
900 mL
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C
1200 mL
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D
1500 mL
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Solution
The correct option is B900 mL CH3COOH⇋CH3COO−+H+t=0C00t=tC(1−α)CαCα
As α is negligble with respect to 1, then we have Ka=C1α21=C2α22⇒C2=C1(α1α2)2⇒C2=0.2×(14)=0.05
We know that C1V1=C2V2⇒300×0.2=0.05×V2⇒V2=1200 mL
Volume of water added =1200−300=900 mL