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

Explanation

`{:(,Ca(LaC)_(2),+2H_(2)OhArr,Ca(OH)_(2)+,2HLaC),(or,2LaC^(Theta)+,2H_(2)OhArr,2overset(Theta)(O)H,+2HLaC),("Before 1 hydrolysis",rArr1,,0,0),("After hydrolysis",rArr1-h,,h,h):}`
`:. [Ca(LaC)_(2)] = 0.13//0.5 = 0.26M`
`:. [LaC^(Θ)] = 0.26 xx2xx 0.52M`
[because 1 mole `Ca(LaC)_(2)` gives 2 mole `[LaC^(Θ)]`
Now, `[overset(Θ)OH] = C.h = sqrt(((K_(h))/(C))) = sqrt((K_(h).C)) = sqrt((K_(w)xxC))/(K_(a))`
where `C` is concentration of anion which undergoes hydrolysis.
`:. 10^(-5.60) = sqrt(((10^(-14)xx0.52))/(K_(a))) K_(a) = 8.25 xx 10^(-4)`
Second Method
Use direct formula for the `pH` of salt of `W_(A)//S_(B)`
`pH = (1)/(2) (pK_(w) + pK_(a) + log C)`
(concentration of salt `=2 xx 0.26 M`)
`8.4 =(1)/(2) (14 + pK_(a) + log 2 xx 0.26)`
Solver for `pK_(a), :. K_(a) = 8.25 xx 10^(-4)`