Hello, I'm currently working on solubility, and I'm not quite sure if I got this question correct...
The Ksp of CaCO3(s) is 8.7 x 10-9 at 25°C. Write the dissociation equation, then
calculate its solubility?
I know that the dissociation eq. would be CaCO3 <==> Ca + CO3, but I'm not sure where to go from there. If you could show how you get the final answer, 10 wonderful points are yours. THanks a bunch :D
The Ksp of CaCO3(s) is 8.7 x 10-9 at 25°C. Write the dissociation equation, then
calculate its solubility?
I know that the dissociation eq. would be CaCO3 <==> Ca + CO3, but I'm not sure where to go from there. If you could show how you get the final answer, 10 wonderful points are yours. THanks a bunch :D
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Good job with the equation!
Now, write the solublity expression exactly as you wold for an equilibrium expression. Don't forget to not include the reactant since it is a solid!
So...
Ksp = (Ca)(CO3)
To calculate solubility, it is the same as an equilibrium problem. Set Ca and CO3 equal to x.
Ksp = X^2
And just solve for x!
9.33x10^-5
Now, write the solublity expression exactly as you wold for an equilibrium expression. Don't forget to not include the reactant since it is a solid!
So...
Ksp = (Ca)(CO3)
To calculate solubility, it is the same as an equilibrium problem. Set Ca and CO3 equal to x.
Ksp = X^2
And just solve for x!
9.33x10^-5