100ml of 0.1M `H_(3)PO_(4)` is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution
`K_(1)=10^(-3)M`
`K_(2)=10^(-8)M`
`K_(3)=10^(-13)M`
Solubility of compound `A(OH)_(2)` in final solution will be :
A. `10_(10)M`
B. `4xx10^(-10)M`
C. `4xx10^(-22)M`
D. `4xx10^(-14)M`
