Question

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

Substance	Kc	Value of Kc
CaCO3	Ksp	4.5�10?9
H2CO3	Ka1	4.3�10?7
	Ka2	5.6�10?11

What is the molar solubility of marble (i.e., [Ca2+] in a saturated solution) in normal rainwater, for which pH=5.60?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Given that;

CaCO3 (s) <---> Ca^2+ (aq) + CO3^2- (aq) ----eq.1

Ksp= 4.5*10^-9

H2CO3 + H2O (l) <---> HCO3- (aq) + H3O+ (aq) ----eq.2

Ka1= 4.7*10^-7

HCO3- (aq) + H2O (l) <---> CO3^2- (aq) + H3O+ (aq) ----eq.3

Ka2= 5.6*10^-11

In order to get overall reaction, we add equation 1 to the reverse of equation 3, which is as follows:

( 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

Now add Adding eq(1) to reverse of eq(3),

CaCO3 (s) <---> Ca^2+ (aq) + CO3^2- (aq).

Ksp= 4.5*10^-9
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+]   -----4

according to the problem pH = 5.60 therefore;

pH = -log [H+]
[H+] = 10^-pH
[H+] = 10^-5.6

= 2.51*10^-6 M

putting the value of H+ in equation number 4.

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+] = 1.4×10−2 M