Calculate `[H^(o+)]` in a soluton that is `0.1M HCOOH` and `0.1 M HOCN. K_(a)(HCOOH) = 1.8 xx 10^(-4), K_(a) (HoCN) = 3.3 xx 10^(-4)`.

Explanation

In the given problem, two weak acids both contribute to `[H^(o+)]`, neither contributing such a big amount that the other share can be neglected.
`{:(,HCOOH,hArr,HCOO^(Theta)+,H^(oplus)),("Initial conc",rArr 0.1,,0,0),("Conc at Eq",rArr 0.1 -x,,x,x+y),(,~~ 0.1,,,):}]`
`K_(a1) = ((x+y)(x))/((0.1-x)) = 1.8 xx 10^(-4)`
`= ((x +y)(x))/(0.1) = 1.8 xx 10^(-4) ..(i)`
`(y M` of `H^(o+)` are contributed from `HOCN`)
`{:(,HOCN,hArr,H^(oplus),+,OCN^(Theta)),("Initial conc",0.1,,0,0,),("Conc at Eq",0.1-y,,(y+x),0,),(,~~ 0.1,,,,):}]`
`(xM` of `H^(o+)` are contributed from `HCOOH)`
`K_(a2) = ((y +x)(y))/(0.1) = 3.3 xx 10^(-4) ...(ii)`
Divide equation (ii) by equaiton (i),
`(y)/(x) = (3.3)/(1.8) = 1.83 or y = 1.83 x`
Subtract equaiton (i) from equaiton (ii)
`((x+y)y-x(x+y))/(0.1)= 1.5 xx 10^(-4)`
or `y^(2) - x^(2) = 1.5 xx 10^(-5) ...(iii)`
Substitutey `= 1.83x` in equaiton (iii) and solve to obtain, ltbRgt `x = 2.5 xx 10^(-3), y = 1.83 xx x = 4.6 xx 10^(-3)`
`[H^(o+)]_(total) x+ y = 7.1 xx 10^(-3)`
Check of assumption. The value of `x` and `y` are slightly less than `10%` of `0.1M`.