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The solubility of` PbCl_(2)` at `25^(@)C` is `6.3 xx 10^(-3)` mole/litre. Its solubility product at that temperature isA. `(6.3 xx 10^(-3)) xx (6.3 xx 10^(-3))`B. `(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))`C. `(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))^(2)`D. `(12.6 xx 10^(-3)) xx (12.6 xx 10^(-3))` |
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Answer» Correct Answer - C `PbCl_(2) rarr Punderset(S)(b^(++)) + underset(2S)(2Cl^(-))` `K_(sp) = S xx (2S)^(2) = [6.3 xx 10^(-3)] xx [12.6 xx 10^(-3)]^(2)`. |
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