1.

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `5.0 xx 10^(-7) m^(3)`B. `0.6 xx 10^(-12) M^(3)`C. `4.0 xx 10^(-8)M^(3)`D. `5.0 xx 10^(-9) M^(3)`

Answer» Correct Answer - A
Solubility expressed in terms of molarity
`[OH^(-)] = 10^(-2)`
`[Ba^(+2)] = 0.5 xx 10^(-2)`
`K_(SP) = (0.5 xx 10^(-2)) (10^(-2))^(2)`
`K_(SP) = 5 xx 10^(-7)`


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