a. In such dilute solutions, complete ionsiation of weak acid occurs, but ionisation of water must also be considered.
Let `x [oversetΘ)OH] then [H_(3)O^(o+)] = x +(7.0 xx 10^(-8))`
(From `H_(2)O` and acetic acid)
`:.K_(w) - [H_(3)O^(o+)] [overset(Θ)OH] rArr (x +7.0 xx 10^(-8)) (x) = 10^(-14)`.or
`rArr x^(2) +(7.0 xx 10^(-8))x - 10^(-14) = 0`
`rArr x = 7.1 xx 10^(-8)M = [overset(Θ)OH]`
Hence, `[H_(3)O^(o+)] (7.1 xx 10^(-8) +7.0 xx10^(-8))`
`= 1.4 xx 10^(-7)M`
`pH = - log (1.4 xx 10^(-7)) = 6.85`
b. From charge balance (electroneutrality)
Total negative charge = Total positive charge
`[CH_(3)COO^(Θ)] +[overset(Θ)H] = [H_(3)O^(o+)]`
`:. [CH_(3)COO^(Θ)] = [H_(3)O^(o+)] - [overset(Θ)OH]`
`= (1.4 xx 10^(-7) - (7.1 xx 10^(-8))`
c. `CH_(3)COOH +H_(2)O hArr CH_(3)COO^(Theta) +H_(3)O^(oplus)`
`K_(a) = =([CH_(3)COO^(Θ)][H_(3)O^(o+)])/([CH_(3)COOH]) = 1.8 xx 10^(-15)`
`[CH_(3)COOH] = ([CH_(3)COO^(Θ)][H_(3)O^(o+)])/(K_(a))`
`= ((7 xx 10^(-8))(1.4 xx 10^(-7)))/(1.8xx10^(-5))`
`= 5 xx 10^(-10)M`
