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When 10ml of 0.1 M acetic acid (pK(a)=50...

When `10ml` of `0.1 M` acetic acid `(pK_(a)=50)` is titrated against `10 ml` of `0.1 M` ammonia solution `(pK_(b)=5.0)`, the equivalence point occurs at `pH`

A

`5.0`

B

`6.0`

C

`7.0`

D

`9.0`

Text Solution

Verified by Experts

The correct Answer is:
C
`pK_(a)=-logK_(a),pK_(b)=-logK_(b)`
`pH=(1)/(2)[logK_(a)+logK_(w)-logK_(b)]`
`=-(1)/(2)[-5+log(1xx10^(-14))-(-5)]`
`=-(1)/(2)[-5-14+5]=-(1)/(2)(-14)=7`
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