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 (Figure 1) . 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.

CaCO3 Ksp =4.5×10−9

H2CO3 Ka1 =4.3×10−7

H2CO3 Ka2 =5.6×10−11

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

Answer 1

Answer – We are given, pH = 5.60 , Ksp of CaCO₃ = 4.5*10^-9

H₂CO₃ Ka1 =4.3×10⁻⁷

H₂CO₃ Ka2 =5.6×10⁻¹¹

We know the Ka2 is the very low, so the first dissociation is give the pH

So,

[H₃O⁺] = 10^-pH

            = 10^-5.60

            = 2.51*10^-6 M

[HCO₃^-] = [H₃O⁺] = 2.51*10^-6 M

We know second dissociation,

   HCO₃^- + H₂O -----> CO₃^2- + H₃O⁺

I 2.51*10^-6                       0          0

C -x                           +x        +x

E 2.51*10^-6-x               +x        +x

Ka2 = [CO₃^2-][H₃O⁺] /[ HCO₃^-]

5.6×10⁻¹¹ = x*x / (2.51*10^-6-x)

5.6×10⁻¹¹ * 2.51*10^-6 = x²

So, x = 1.18*10^-8

[CO₃^2-] = 1.18*10^-8

We know the Ksp expression for the CaCO₃

     CaCO₃ <------>Ca²⁺ + CO₃^2-

Ksp = [Ca²⁺] [CO₃^2-]

4.5*10^-9 = x * 1.18*10^-8

So, [Ca²⁺] = 4.5*10^-9 / 1.18*10^-8

                 = 0.379 M

So, the molar solubility of marble is 0.379 M