Question

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

Answer 1

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`