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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` |
Answer» Correct Answer - b |
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