Chemistry

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).

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  1. 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.

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