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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_(sp)pK_(b) = -logK_(b)`
`pH = -(1)/(2)[log K_(a) + log K_(w) - log K_(b)]`
`= (1)/(2)[-5 + log(1 xx 10^(-14)) - (-5)]`
`= -(1)/(2)[-5-14+5] = -(1)/(2)(-14) =7`
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