The chief compound in marble is CaCO3. Marble has been widely used for statues a

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 =4.5×109

H2CO3 Ka1 =4.3×107

H2CO3 Ka2 =5.6×1011

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

Explanation / Answer

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

H2CO3 Ka1 =4.3×107

H2CO3 Ka2 =5.6×1011

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 -----> CO32- + H3O+

I 2.51*10-6                       0          0

C -x                           +x        +x

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

Ka2 = [CO32-][H3O+] /[ HCO3-]

5.6×1011 = x*x / (2.51*10-6-x)

5.6×1011 * 2.51*10-6 = x2

So, x = 1.18*10-8

[CO32-] = 1.18*10-8

We know the Ksp expression for the CaCO3

     CaCO3 <------>Ca2+ + CO32-

Ksp = [Ca2+] [CO32-]

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

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