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The solubility of AgCl in water at 25^(@) C is found to be 1.06xx10^(-5) moles per litre. Calculate the solubility product of AgCl at this temperature. |
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Answer» Solution :AgCl IONIZES completely in the solution as : `Agcl RARR Ag^(+) + Cl^(-)` i.e., 1 mole of AgCl in the solution gives 1 moles 1 mole of `Ag^(+)`IONS and 1 mole of `Cl^(-)` ions. Now, as the solubility of `AgCl = 1.6 xx 10^(-5)` moles PER litre `:. [Ag^(+)] = 1.06xx10^(-5)` moles/litre and `[Cl^(-)] = 1.06 xx 10^(-5)` moles/litre `:. K_(sp) ` for ` AgCl = [Ag^(+)][Cl^(-)]=1.06xx10^(-5)xx1.06xx10^(_5)=1.1xx10^(-10)` |
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