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A volume of 50.00 mL of a weak acid of u...

A volume of `50.00 mL` of a weak acid of unknown concentration is titrated with `0.10 M` solution of `NaOH`. The equivalence point is reached after `39.30 mL` of `NaOH` solution has been added. At the half-equivalence point `(19.65 mL)`, the `pH` is `4.85`. Thus, initial concentration of the acid and its `pK_(a)` values are

A

`[HA]` initial `=0.1 M`, `pK_(a)=4.85`

B

`[HA]` initial `=0.079 M`, `pK_(a)=2.93`

C

`[HA]` initial `=0.1 M`, `pK_(a)=3.70`

D

`[HA]` initial `=0.079 M`, `pK_(a)=4.85`

Text Solution

Verified by Experts

The correct Answer is:
D
`underset(Acid)(M_(1)V_(1))=underset(Base)(M_(2)V_(2)`
`M_(1)xx50.0=0.10xx39.30`
`:. M_(1)=(0.10xx39.30)/(50)=0.0786 M`
When half-equivalence point is reached
`[HA]=[A^(-)]`
`pH=pK_(a)+log``([A^(-)])/([HA])`
`:. pH=pK_(a)=4.85`
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