1.

Point out the difference between ionic product and solubility product. (ii) The solubility of AgCl in water at 298 K is 1.06 times 10^-5 mole per litre. Calculate is solubility product at this temperature.

Answer»

Solution :
(ii) The solubility equilibrium in the saturated solution is
`AgCl(s) LEFTRIGHTARROW Ag^(+) (aq)+CL^-(-) (aq)`
The solubility of AgCl is `1.06 TIMES 10^-5` MOLE per litre.
`[Ag^(+)(aq)]=1.06 time 10^-5 mol L^-1`
`[Cl^(-) (aq)]=1.06 times 10^-5 mol L^-1`
`K_(sp)=[Ag^+(aq)][Cl^(-) (aq)]`
`=(1.06 times 10^-5 mol L^-1) times (1.06 times 10^-5 mol L^-1)`
`=1.12 times 10^-2 mol^2 L^-2`


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