Calcium lactate is a salt of weak acid and represented as `Ca(LaC)_(2)`. A saturated solution of `Ca(LaC)_(2)` contains `0.13` mole of salt in `0.50` litre solution. The pOH of this is `5.60`. Assuming complete dissociation os salt, calculate `K_(a)` of lactic acid.

Explanation

In solution, `Ca(Lac)_(2)` is hydrolysed as follows :
`Ca(Lac)_(2)+2H_(2)O hArr Ca(OH)_(2) + underset(underset("(weak)")("lactic acid"))(2HLac)`
or `Ca^(2)+2 Lac^(-1) + 2 H_(2)O hArr Ca^(2+) + 2 OH^(-) + 2 Hlac`
or `2 Lac^(-) + 2H_(2)O hArr 2 OH^(-) + 2 HLac`
or `Lac^(-) + H_(2)O hArr OH^(-) + HLac`
Hydrolysis constant,
`K_(h) = ([OH^(-)][HLac])/([Lac^(-)])`
But `[Ca(Lac)_(2)]=0.26 ` mol `L^-1)` so that `[Lac^(-)]=0.52 ` mol `L^(-)`
and `pOH = 5.60 ` so that
`- log [OH^(-) ] = 5.60`
or `[OH^(-)] = ` antilog `(-5.6)=2.51xx10^(-6)M`
`:. K_(h) = ((2.51xx10^(-6))^(2))/(0.52)=1.21xx10^(-11)`
Further, `HLac hArr H^(+) + Lac^(-)`
`K_(a) = ([H^(+)][Lac^(-)])/([HLac]) " " ...(ii)`
Also `K_(w) = [H^(+)][OH^(-)]` ...(iii)
From (i), (ii) and (iii), `K_(a)=(K_(w))/(K_(h))=(10^(-14))/(1.21xx10^(-11))=8.26xx10^(-4)`