A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:

Options

A. `1.8xx10^(-5)`
B. `8.2xx10^(-6)`
C. `1.8xx10^(-6)`
D. `8.2xx10^(-5)`

Answer

B

Explanation

For neutralisation,
Total Meq.of acid = Meq.of base
`= 26.6xx0.1=2.66`
Now for partial netralisation of acid,
`{:(,HA+,BOHrarr,BA+,H_(2)O),("Meq. before reaction",2.66,1.2,0,0),("Meq. after reaction",1.46,0,1.2,1.2):}`
The rersultant mixture acts as a buffer and, [HA] ans [BA] may be placed in terms of Meq. since volume of mixture is constant.
`pH= -log K_(a)+log([Sal t])/([Acid])`
or `5= -log K_(a)+log (1.2)/(1.46)`
`K_(a)=8.219xx10^(-6)`