Two weak monobasic organic acids `HA` and `HB` have dissociation constans as `3.0 xx 10^(-5)` and `1.5 xx 10^(-5)`, respectively, at `25^(@)C`. If `500mL` of `1M` solutions of each of these two acids are mixed top product `1L` of mixed acid solution, what iws the `pH` of the resulting solutions?

Explanation

Note that `K_(a)` of two acids is nearly same. In such cases, we have to consider `H^(o+)` from both `hA` and `HB` simulatenously. The concentration of `HA` and `HB` in the mixture `= 0.5M` [equal volumes are mixed] `=C M`
`SrF_(2)(s) rArr Sr^(2+)(aq)+2F^(Theta)(aq)`
Now, `[H^(o+)]_("final") = [H^(o+)]_("from HA") + [H^(o+)]_("from HB")`
`rArr K_(a(HA)) = ((x+y)x)/(C-x)`and`K_(a(HB)) =((x+y)y)/(C-y)`
As `K_(a)` of both acid `~~ 10^(-5)` and `H^(o+)` from one acid acts as common ion for other's dissocitation, `x` and `y` are very less as compared to `C`.
`rArr C -x ~~ C` and `C-y ~~C`
`rArrK_(a(HA)) = ((x+y)y)/(C-y)` and `K_(a(HB)) = ((x+y)y)/(C)`
Divide the two expressions to get:
`(1)/(2) = (x)/(y) rArr x = 2y`
Substitute for `y = (1)/(2)x` in `K_(a(HA)) = (x^(2)+xy)/(c)`
`rArr 3.0 xx 10^(-5) = (x^(2) + 0.5x^(2))/(0.5)`
`rArr x = sqrt(10) xx 10^(-3)M` and `y = (sqrt(10))/(2) xx10^(-3)M`
`[H^(o+)] = x+y = (3sqrt(10))/(2) xx 10^(-3)M`
`pH =- log ((3sqrt(10))/(2)xx10^(-3))rArrpH = 2.32`