Malonic acid is an organic dibasic acid such as `H_(2)S` having first ionistion constant, `K_(1) = 1.42 xx 10^(-3)` and second ionisation constant, `K_(2) = 2.0 xx 10^(-6)`. Compute the divalent molanate ion concentration in:
a. `0.001M` malonic acid.
b. a solution that is `0.0001M` in malonc acid and `0.0004M HC1`.
c. a solution that is `0.0001 M` in malonic acid and `0.1 M HC1`.

Explanation

a. Consider malonic acid to be `H_(2)A`, where malonate ions is `A^(2-)`. For dibasic acid, we consider ionisation in two stages as follows:
`{:(H_(2)A hArr,H^(o+)+,HA^(Θ),,"first ionisation constant",),(C-x,x,x,=K_(1)=1.42xx10^(-3),),(,,,C= "initial concentration",),(,,,=0.001M,):}`
`rArr K_(1) = ([H^(o+)][HA^( Θ)])/([H_(2)A]) =(x^(2))/(C-x) = 1.42 xx 10^(-3)`
Solving the quadratic,
`x^(2) +1.42 xx 10^(-3) x - 1.42 xx 10^(-6) = 0`
`rArr x = 6.75 xx 10^(-4)M`
Note that `K_(1) gt gt K_(2)` so concentration of `H^(o+)` in solution is considered only from first dissociation, i.e., `[H^(o+)] = xM` neglect `[H^(o+)]` concentration from second ionsiation.
Consider second ionisation:
`{:(HA^(Θ)rarr,H^(o+)+,A^(2-),,"Second ionisation" =K_(2),),(x,x,-,=2.0 xx 10^(-6),),(x-y,x+y,y,,):}`
`rArr K_(2) = ([H^(o+)][A^(2-)])/([HA^(Θ)]) = ((x+y)y)/(x-y)`
Note: `[H^(o+)] = x + y~~x [HA^(o+) = x - y = x (y lt lt x)`
`rArr K_(2) = y [A^(2-)] = 2.0 xx 10^(-6)M`
Note: Ustually, `[A^(2-)]` for `H_(2)A` (dibasic acid) can be aproximately taken to be `K_(2)`.
b. Hence `C = 10^(-4)M`,
`[H^(o+)] = 4 xx 10^(-4)M` (from `HC1)`
`rArr K_(1) = ([H^(o+)][HA^(Θ)])/([H_(2)A]) = ((4 xx10^(-4) +x)(x))/(10^(-4)-x)`
`= 1.42 xx 10^(-3)`
Importane: Note that, we can not go for approximation in this case, since the concentration of `HC1` is so low that `H^(o+)` is considered both from `HC1` and malonic acid. Solving the quadratic.
`x^(2) + 1.82 xx 10^(-3) x- 1.42 x 10^(-7) = 0`
`rArr x = 7.5 xx 10^(-5)M`
Now consider, second ionisation and substitute for value of `x`.
`K_(2) = ([H^(o+)][A^(2-)])/([HA^(Θ)]) = ((4 xx 10^(-4) +x)(y))/(x) = 2.0 xx 10^(-6)`
`rArr ((4 xx10^(-4)+7.5xx10^(-5))y)/(7.5xx10^(-5))=2.0xx10^(-6)`
`rArr y = [A^(2-)] = 3.2 xx 10^(-7)M`
c. Here `C = 10^(-4)M, [H^(o+)] = 0.1M`
`rArr K_(a) = K_(1)K_(2) = ([H^(o+)]^(2)[A^(2-)])/([H_(2)A]) = ((0.1)^(2)[A^(2-)])/(10^(-4))`
`= 2.84 xx 10^(-9)`
Important: Note thate, we will do the approximation in this case, since the concentration of `HC1` is so high that `H^(o+)` considered from malonic acid is negligible. Recall the ionisation of `H_(2)S(g)` in an acidic solution. Solving the equation, we get: `[A^(2-)] = 2.84 xx 10^(-11)M`