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Question: Fixed volume of 0.1 M benzoic acid (pK<sub>a</sub> = 4.2) solution is added into 0.2 M sodium benzoa...

Fixed volume of 0.1 M benzoic acid (pKa = 4.2) solution is added into 0.2 M sodium benzoate solution and formed a 300 ml, resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid –

A

100 ml

B

150 ml

C

200 ml

D

None of these

Answer

200 ml

Explanation

Solution

Ph = pKa­ + log[C6H5COO][C6H5COOH]\frac { \log \left[ \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { COO } ^ { - } \right] } { \left[ \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { COOH } \right] }

[C6H5COO][C6H5COOH]\therefore \frac { \left[ \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { COO } ^ { - } \right] } { \left[ \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { COOH } \right] } = 2

Let volume of acid is V ml.

= 2

Ž V = 200 ml.