The concentration of hydrogen ions in a `0.2M` solution of formic acid is `6.4 xx 10^(-3) mol L^(-1)`. To this solution, sodium formate is added so as to adjust the concentration of sodium formate to `1 mol L^(-1)`. What will be the `pH` of this solution? The dissociation constant of formic acid is `2.4 xx 10^(-4)` and the degree of dissociation fo sodium formate is `0.75`.

Answer

A, D

Explanation

For weak acid,
`[H^(o+)] = Calpha`
`[H^(o+)] =6.4 xx 10^(-3)` (given)
`:. Calpha = 6.4 xx 10^(-3)`
or `0.2 alpha = 3.2 xx 10^(-2)`
When sodium formate is added to formic acid solution, its dissociation is supressed sue to common ion effect.
`{:(HCOONahArr,HCOO^(Theta)+,Na^(o+)),(1,0,0),((1-0.75),0.75,0.75):}`
Due to suppression in dissociation, the dissociation of formic acid is so small that formic acid concentration is practically constant. The combination of formic acid and sodium formate forms an acidic buffer.
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
`=- log 2.4 xx 10^(-4) + "log" (0.75)/(0.2) = 4.19`