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`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `5.06 xx 10^(-7) M^(3)`B. `4.00xx10^(-6)M^(3)`C. `4.00xx 10^(-7) M^(3)`D. `5.00 xx 10^(-6) M^(3)` |
Answer» Correct Answer - B `pH =12` means `|H^(+)|=10^(-12)M` `:. [OH^(-1)]=10^(-2)` `Ba(OH)_(2) hArr Ba^(2+)+2OH^(-) ` `pH =10^(-2) M :. Ba^(2+)=(10^(-2))/(2)M` `=5 xx 10^(-3)M` `K_(sp)=|Ba^(2+)|" "|OH^(+)|^(2)=(5 xx 10^(-3))(10^(-2))^(2)` `=5 xx 10^(-7) M^(3)` |
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