Citric acid `(H_(3)A)` is a polyprotic acid with `K_(1),K_(2)`, and `K_(3)` equals to `7.4 xx 10^(-4), 1.7 xx 10^(-5)`, and `4.0 xx 10^(-7)`, respectively. Calculate the `[H^(o+)], [H_(2)A^(Theta)], [HA^(2-)]`, and `[A^(3-)]` in `0.01M` citric acid.

Explanation

i. `H_(3)A + H_(2)O hArr H_(2)A^(Theta) + H_(3)O^(o+)`,
`K_(1) = ([H_(2)A^(Theta)][H_(3)O^(o+)])/([H_(3)A]) = 7.4 xx 10^(-4)`
`K_(1) = (x^(2))/(0.01-x) = 7.4 xx 10^(-4)`,
or `x^(2) + (7.4 xx 10^(-4)) x-(7.4 xx 10^(-6)) = 0`
`:. x = (-7.4xx10^(-4)+sqrt((7.4xx10^(-4))^(2)+4(7.4xx10^(-6))))/(2)`
`=2.4 xx 10^(-3) M = [H_(3)O^(o+)] = [H_(2)A^(Theta)]`
ii. `H_(2)A^(Theta) + H_(2)O hArr HA^(2-) + H_(3)O^(o+)`
`K_(2) = ([HA^(2-)][H_(3)O^(o+)])/([H_(2)A])`
`1.7 xx 10^(-5) = ([HA^(2-)][x])/(x) rArr [HA^(2-)] = 1.75 xx 10^(-5)M`
iii. `HA^(2-) + H_(2)O hArr A^(3-) + H_(3)O^(o+)`
`K_(3) = ([A^(3-)][H_(3)O^(o+)])/([HA^(2-)])`
`4.0 xx 10^(-7) = [A^(3-)](2.4 xx 106(-3))/(1.7 xx 10^(-5))`, so `[A^(3-)] = 2.8 xx 10^(-9)M`