1.

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.

Answer» 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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