Question

The chief compound in marble is CaCO3. Marble has been widely used for statues and ornamental work on buildings, including such structures as the Taj Mahal. However, marble is readily attacked by acids via the following reaction.

CaCO₃(s)+H⁺(aq)⇌Ca²⁺(aq)+HCO₃⁻(aq)

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=4.20?

Answer 1

We knwo that Ksp of CaCO3 can be represented as

CaCO3 (s) ---> Ca^2+ (aq) + CO3^2- (aq).....Ksp= 4.5 X 10^-9.......(1)

The first dissociation of H2CO3 will be:

H2CO3 + H2O (l) ---> HCO3- (aq) + H⁺ (aq).......Ka1= 4.3 X 10^-7........(2)

The second dissociation constant

HCO3- (aq) + H2O (l) ---> CO3^2- (aq) + H⁺ (aq).......Ka2= 5.6 X 10^-11....(3)

We want to get the equation
CaCO₃(s)+H⁺(aq) -->Ca²⁺(aq)+HCO₃⁻(aq) Ksol = Ksp / Ka2 = 80.35

that is we will substract the equation 3 from equation 1

             CaCO₃(s)+H⁺(aq) -->Ca²⁺(aq)+HCO₃⁻(aq)

initial                    a               0             0

Change                 -x             x              x

Equilibrium           a-x            x               x

Ksol = [Ca⁺²] [HCO3^-] / [H⁺]

Ksol = [x]^2 / (a-x)

Given pH = 4.2

pH = -log [H+]

Taking antilog
[H+] = 10^-pH
[H+] = 6.31 X 10^-5

80.35 = [x]^2 / [6.31 X 10^-5-x]

We can ignore x in the denominator
[x]^2 = 80.35 X 6.31 X 10^-5 = 507 X 10^-5

x = 7.12 X 10^-2 M