1.

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

Answer»

Solution :(a) (i)
(ii) The solubility equilibrium in the SATURATED solution is
`AGCL(s) rArr AG^(+) (aq) + Cl^(-) (aq)`
The solubility of AgCl 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^(-2) mol^(2) L^(-2)`


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