Question

The concentration of `H^(+)` ion in a `0.2 M` solution of `HCOOH` is `6.4xx10^(-3) "mole" L^(-1)`. To this solution `HCOONa` is added so as to adjust the concentration of `HCOONa` to one mole per litre. What will be the `pH` of this solution? `K_(a)` for `HCOOH` is `2.4xx10^(-4)` and the degree of dissociation of `HCOONa` is `0.75`
A. `3.19`
B. `4.19`
C. `5.19`
D. `6.19`

Answer 1

Correct Answer - B
Assuming that the addition of `HCOONa` suppresses the ionization of `HCOOH`, we can use the expression
`pH=pK_(a)+log``(["Salt"])/([Acid])`
to compute `pH` of the solution, since salt is `75%` dissociated we will get,
`pH= -log(2.4xx10^(-4))+log``(0.75)/(0.2)`
`=3.62+0.57=4.19`