The solubility of CaCO3 at 25 degrees celsius is 6.90*10-5M. The reaction is CaCO3(s)-->Ca 2+(aq) + CO3 2-(aq)
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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.
To calculate the concentration of Ca2+ and CO32- at equilibrium, we need to use the solubility product expression (Ksp) for the dissolution reaction.
a. The solubility product expression for CaCO3 is:
Ksp = [Ca2+][CO32-]
Given that the solubility of CaCO3 is 6.90 * 10-5 M, we can assume that the concentration of Ca2+ and CO32- at equilibrium will be equal to the solubility value. Therefore:
[Ca2+] = 6.90 * 10-5 M
[CO32-] = 6.90 * 10-5 M
b. The equilibrium constant, also known as the solubility product (Ksp), is calculated by multiplying the concentrations of the ions at equilibrium:
Ksp = [Ca2+][CO32-] = (6.90 * 10-5 M) * (6.90 * 10-5 M)
To find the actual value of the equilibrium constant, we need to calculate this expression using the given values:
Ksp = (6.90 * 10-5 M) * (6.90 * 10-5 M) = 4.76 * 10-9
Therefore, the equilibrium constant (solubility product) for the dissolution reaction of CaCO3 is 4.76 * 10-9.