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The solubility of AgCl in water at 298 K is 1.06 xx 10^(-5) moleper litre. Calculateits solubility product at this temperature. |
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Answer» Solution :The solubility equilibrium in the SATURATED solution is : `AgCI (s) overset(aq)(hArr) Ag^(+) (aq) + CI^(-) (aq)` The solubility of AgCi is `1.06 xx 10^(-5)`mole per litre. `[Ag^(+) (aq)] =1.06 xx 10^(-5) mol L^(-1)` `[CL^(-)(aq)] =1.06 xx 10^(-5) mol L^(-1)` `K_(SP) =[Ag^(+) (aq)] [ Cl^(-) (aq)]` `= (1.06 xx 10^(-5) mol L^(-1)) xx (1.06 xx 10^(-5) mol L^(-1))` `= 1.12 xx 10^(-10) mol^(2)L^(-2)` |
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