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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. `5.0 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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