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

Answer 1

Ellie Andrews

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

We evaluate the reaction at equilibrium.

#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#

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Answer 2

Luna Chen

The volume of water required to dissolve a 6.7 g mass of calcium sulfate can't be determined solely based on the solubility product constant (Ksp) of calcium sulfate. The solubility of calcium sulfate depends on various factors such as temperature, pressure, and the presence of other ions. To calculate the volume of water required, additional information about the solubility of calcium sulfate under specific conditions is needed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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