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100 mL of 0.02 M benzoic acid (pKa= 4.2)...

100 mL of 0.02 M benzoic acid `(pK_a= 4.2)` is titrated using 0.02 M NaOH. PH after 50 ml and 100 mL of NaOH have been added are

A

3.50, 7

B

4.2, 7

C

4.2, 8.1

D

4.2, 8.25

Text Solution

Verified by Experts

The correct Answer is:
D
When 50mL of 0.02 M NaOH is added inside, we will have [salt]=[acid] at half equivalence point,
`pH =pK_a +log (["Salt"])/(["Acid"])=4.2`
When 100mL of NaOH is added we will have equivalence point so pH will be calculated according to hydrolysis of salt of weak acid and strong base.
`pH =1/2 (pK_(w)+pK_(a)+log C)`
`=1/2 (14+4.2 +log 0.02)=8.25`
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