Question

Equilibrium constants at 25 ?C are listed in the table below.

Substance	Kc	Value of Kc
CaCO3	Ksp	4.5

Answer 1

We were given Ksp values for the following equations.

1) CaCO3 (s) <---> Ca^2+ (aq) + CO3^2- (aq).....Ksp= 4.5*10^-9
2) H2CO3 + H2O (l) <---> HCO3- (aq) + H3O+ (aq).......Ka1= 4.7*10^-7
3) HCO3- (aq) + H2O (l) <---> CO3^2- (aq) + H3O+ (aq).......Ka2= 5.6*10^-11

In order to get overall reaction, we add equation 1 to the inverse of equation 3.
*The inverse of equation 3*
( HCO3- (aq) + H2O (l) <---> CO3^2- (aq) + H3O+ (aq).......Ka2= 5.6*10^-11)^-1
=CO3^2- (aq) + H3O+ (aq) <---> HCO3- (aq) + H2O (l).....Ka2= 1.7857*10^10

Adding eq(1) to inverse of eq(3),

1) CaCO3 (s) <---> Ca^2+ (aq) + CO3^2- (aq).....Ksp= 4.5*10^-9
3) CO3^2- (aq) + H3O+ (aq) <---> HCO3- (aq) + H2O (l).....Ka2^-1= 1.7857*10^10
________________________________________...
CaCO3 (s) + H+ (aq) <---> Ca^2+ (aq) + HCO3- (aq).....K = Ksp * Ka2^-1 = 80.357

K = [Ca^2+ (aq)] [HCO3- (aq)] / [H+]
K = [X]^2 / [H+]

We know our new K value and we know our [H+] because we were given the pH.
pH = -log [H+]
[H+] = 10^-pH
[H+] = 10^-5.6 = 2.51*10^-6 M

Plugging in the values into K expression
80.357 = [X]^2 / [2.51*10^-6]
X= square root of 80.357 * [2.51*10^-6]
X = [Ca^2+] = 0.0142

Thus, [Ca^2+] = 0.0142 M