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Question

A weak acid HA after treatment with 12 ml of 0.1 M strong base BOH has pH=5. At the end point, the volume of same base required is 13.2 ml. Ka of acid is:

A
106
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B
105
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C
104
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D
107
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Solution

The correct option is C 104
According to the question, 12 mL of 0.1 M strong base is used. Thus, at the end point, the volume of same base required is 13.2 mL
HA+BOHBA+H2Ot=0 a 12×0.1 0 0t=eq (a1.2) 0 1.2 1.2

Given, at the end point, the volume of same base required is 13.2 ml
Thus, Meq of HA = Meq of BOH
Meq of BOH at the end point =13.2×0.1=1.32
Meq of HA left =1.321.2=0.12
Now, pH=pKa+log[Salt][acid]
5=logKa+log1.2(0.12)5=logKa+1Ka=104

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