Calculate `[H^(o+)], [C_(3)H_(5)O_(3)^(Θ)]`, and `[PhO^(Θ)]` in a solution that is `0.03M (C_(2)H_(5)overset(O)overset(||)(C)-O-O-H)` and `0.1MPhOH? K_(a)` values for `C_(3)H_(5)O_(3)H` and `PhOH` are `1.48 xx 10^(-4)` nad `1.05 xx 10^(-10)` respectively.

Explanation

The ionisation of weaker acid `(PhOH)` is neglected in calculating `[H_(3)O^(o+)]` since its `K_(a)` value is very less.
a. `{:(,CH_(3)H_(5)O_(3)H+,H_(2)O,hArr,C_(3)H_(5)O_(3)^(Theta)+,H_(3)O^(oplus)),("Initial conc",rArr 0.03,-,,0,0),("Eq conc",rArr(0.03-x)M,-,,xM,xM),(,~~ 0.03 M,,,,):}`
`:. K_(a1) = (x^(2))/(0.03) = 1.38xx10^(-4)`, thus `x = 2.0 xx 10^(-3)M`
Hence, `[H_(3)O^(o+)] = [C_(3)H_(5)O_(3)^(Θ)] = 2.0 xx 10^(-3)M`
b. `{:(,PhOH+,H_(2)O,hArr,PhO^(Theta) +,H_(3)O^(oplus)),("Initial conc",0.1,-,,0,0),("Eq conc",(0.1-y),-,,y(y+x)~~x,),(,~~ 0.1,,,=2.0xx 10^(-3),):}`
Since both acids are in the same solution, the concentration of `[H_(3)O^(o+)]` is contributed from the ionisation of more stronger acid `(C_(3)H_(5)O_(3)H)`
`K_(a_(2)) = ([H_(3)O^(o+)][PhO^(Θ)])/(0.1) = ((2.0xx10^(-3))(PhO^(Θ)))/(0.1)`
So, `[PhO^(Θ)] = 5.2 xx 10^(-9)M`