1.

The solubility of `CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3)` grams per 100 mL of the solution. Calculate solubility product of `CaF_(2).`

Answer» Solubility of `CaF_(2) = 1.7xx10^(-3) g//100mL =17 xx 10^(-3) gL^(-1)`
Molar mass of `CaF_(2) =(40 +2 xx 19)=78 g mol^(-1)`
Solubility of `CaF_(2)=((17xx10^(-3) gL^(-1)))/((78.0 g mol^(-1))) =2.18 xx 10^(-4) mol L^(-1)`
The solubility equibrium for the saturated solution of `CaF_(2)` is :
`CaF_(2)(s)overset(aq)(hArr) Ca^(2+) (aq) +2F^(-) (aq)`
`[Ca^(2+)(aq)] =2.18 xx10^(-4) mol L^(-1)`
`[F^(-)(aq)] =2 xx2.18 xx 10^(-4) mol L^(-1)`
Applying Law of chemical equilibrium :
`K_(sp) =[Ca^(2+) (aq)] [F^(-) (aq)]^(2)`
`=(2.18 xx 10^(-4) mol L^(-1)) xx (2.18 xx 10^(-4) mol L^(-1))^(2) = 4.14 xx 10^(-11) mol^(3).`


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