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

`Ca(LaC)_(2)+2H_(2)OhArrCa(OH)_(2)+2HLaC`
or `{:(,2LaC^(-)+2H_(2)OhArr,2OH^(-)+,2HLaC),("Before hydrolysis", 1,0,0),("After hydrolysis", (1-h),h,h):}`
`:. [Ca(LaC)_(2)]=(0.13)/(0.5)=0.26M`
`:. [LaC]=0.26xx2=0.52M`
`:. 1mole Ca(LaC)_(2)` gives 2 mol e (LaC).
Now `[OH^(-)]=C.h=C sqrt([(K_(H))/(C )])`
`= sqrt((K_(H).C ))= sqrt([(K_(w)xxC)/(K_(a))])`
where C is conc. of annion which undergoes hydrolysis
`:. 10^(-5.60)= sqrt (((10^(-14)xx0.52)/(K_(a))))`
`K_(a)= 8.25xx10^(-4)`