What volume of water is required to dissolve a #6.7*g# mass of #"calcium sulfate"# if #K_(sp)*CaSO_4=2.4xx10^-5#?

1 Answer
May 29, 2017

Answer:

A volume of #10*L# is required........

Explanation:

We assess the equilibrium reaction.......

#CaSO_4(s) rightleftharpoonsCa^(2+) + SO_4^(2-)#,

Where #K_"eq"=K_"sp"=[Ca^(2+)][SO_4^(2-)]#.

And if we put #[Ca^(2+)]=[SO_4^(2-)]=x#..............

.......then #x=sqrt(K_"sp")=sqrt(2.4xx10^-5)=0.005*mol*L^-1#

And so #((6.7*g)/(136.14*g*mol^-1))/(??*L)=0.005*mol*L^-1#

i.e. #((6.7*g)/(136.14*g*mol^-1))/(0.005*mol*L^-1)~=10*L#