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When 10ml of 0.1 M acetic acid (pK(a)=5....

When `10ml` of `0.1 M` acetic acid `(pK_(a)=5.0)` 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)= - log K_(u),pK_(b) = - log K_(b)`
`pH = -1/2 [log K_(a)+ log K_(w) - log K_(b)] `
`=-1/2 [-5 +log (1xx10^(-14))-(-5)]`
`=-1/2 [-5 -14+5] =-1/2 (-14)=7`
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