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The `K_(b)` for `AgCl` is `2.8 xx 10^(-10)` at a given temperature. The solubility of `AgCl` in `0.01` molar `HCl` solution at this temperature will be :A. `2.5 xx 10^(-8) molL^(-1)`B. `2.8 xx 10^(-12) molL^(-1)`C. `5.6 xx 10^(-8) molL^(-1)`D. `2.8 xx 10^(-4) molL^(-1)` |
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Answer» Correct Answer - A `K_(sp) = [Ag^(+)][Cl^(-)]` `2.8 xx 10^(-10) = [Ag^(+)]xx 0.01` `[Ag^(+)] = (2.8 xx 10^(-10))/(10^(-2)) = 2.8 xx 10^(-8)"mol" L^(-1)` |
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