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Question

100 mL of 0.02 M benzoic acid (pKa=4.20) is titrated using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are:

A
3.50,7
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B
4.2,7
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C
4.2,8.1
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D
4.2,8.25
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Solution

The correct option is D 4.2,7
Given 100ml of 0.02M C6H5COOH with pKa=4.2

So, Ka=104.2=6.31×105
It is titrated using 0.02M NaOH.

(i) After addition of 50ml of NaOH:
Number of moles of C6H5COOH=100×0.02=2mmol
Number of moles of NaOH=50×0.02=1mmol
Here after the addition of 50ml of NaOH,we get basic Buffer solution C6H5COONa and C6H5COOH.
Number of moles of C6H5COOH=21=1mmol
Number of moles of C6H5COONa=1mmol

pH=pKa+log saltacid =4.2+log11
pH=4.2

(ii) After addition of 100ml of NaOH:
Number of moles of C6H5COOH=100×0.02=2mmol
Number of moles of NaOH=100×0.02=2mmol
So, Number of moles of C6H5COONa=2mmol

Therefore the given point is equivalent point. The salt C6H5COONa is weak base, so we will find basic ionization constant.

Now,
Kb=KwKa=10146.31×105=1.58×1010

Concentration of [C6H5COONa]=NumberofmolesofC6H5COONaTotalvolumeofthesolution=2200=0.01

[OH]=Kb×[C6H5COONa]=(1.58×1010)×0.01=1.257×107M

pOH=log[OH]=log(1.257×107)=6.9

pH=14pOH=146.97.0

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