posted by Sophia .
The solubility of CaCO3 at 25 degrees celsius is 6.90*10-5M. The reaction is CaCO3(s)-->Ca 2+(aq) + CO3 2-(aq)
a. Calculate the concentration of Ca +2 and CO3 -2 at equilibrium.
b. Calculate the equilibrium constant for the dissolution reaction. (also known as the solubility product).
The solubility of CaCO3 is quite complex due to the potential reaction of the carbonate ion with H2O to form bicarbonate and carbonic acid. I think the intent of this problem is to ignore that which I shall do.
CaCO3 ==> Ca^+ + CO3^=
Ksp = (Ca^+2)(CO3^=)
(Ca^+2) = 6.90 x 10^-5 M at equilibrium.
(CO3^=) = 6.90 x 10^-5 M at equilibrium.
Substitute into the Ksp expresion and solve for Ksp.