Calculate the equilibrium constants for the reactions with water of `H_(2)PO_(4)^(Theta), HPO_(4)^(2-)`, and `PO_(4)^(3-)` as ase. Comparing the relative values of two equilibrium constants of `H_(2)PO_(4)^(Theta)` with water, deduce whether solutions of this ion in water are acidic or base, Deduce whether solutions of `HPO_(4)^(2-)` are acidic or bases. Given `K_(1),K_(2)`,and `K_(3)` for `H_(3)PO_(4)` are `7.1 xx 10^(-3), 6.3 xx 10^(-8)`, and `4.5 xx 10^(-13)` respectively.

Explanation

i. `H_(2)PO_(4)^(Theta) +H_(2)O hArr H_(3)PO_(4) + overset(Theta)OH`
`K_(h) = (K_(w))/(K_(1)) = (10^(-14))/(7.1 xx 10^(-3)) = 1.4 xx 10^(-12)`
ii. `HPO_(4)^(2-) + H_(2)O hArr H_(2)PO_(4)^(Theta) + overset(Theta)OH`
`K_(h) = (K_(w))/(K_(2)) = (10^(-14))/(6.3xx10^(-8)) = 1.6 xx 10^(-7)`
iii. `PO_(4)^(3-) + H_(2)O hArr HPO_(4)^(2-) + overset(Theta)OH`
`K_(h) = (K_(w))/(K_(3)) = (10^(-14))/(4.5 xx 10^(-13)) = 2.2 xx 10^(-2)`
`H_(2)PO_(4)^(Theta)` can react with water to yield `HPO_(4)^(2-)` and `H_(3)O^(o+)` with a constant `6.3 xx 10^(-8)`.
It reaction with water to yield `H_(3)PO_(4)` and `overset(Theta)OH` has a much smaller equilibrium constant, `1.4 xx 10^(-12)`, and so this ion in water is acidic comparision of the constants for `HPO_(4)^(2-) (4.5 xx 10^(-3)` as an acid, `1.6 xx 10^(-7)` as a base ) leads to the conclusion that this ion is solution is basic.