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Question

A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) solution has a pH of 5. At the end point, the volume of the same base solution required is 27 mL.Ka of acid is:

A
1.8×105
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B
8×106
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C
1.8×106
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D
8×105
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Solution

The correct option is B 8×106
At end point, M1V1=M2V2
M1V1=27×0.1=2.7 millimoles
Millimoles of acid present= 2.7
Millimoles of salt formed= 1.2
[Salt]=1.2V,[acid]=2.71.2=1.5V
pH=pKa+log[Salt][acid]
5=pKa+log1.2/V1.5/V
pKa=5log1.21.5=5.0969
logKa=5.0969
Ka=8×106

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