Question

The chief compound in marble is CaCO3. However, marble is readily attacked by acids.

[Ca2+] in normal rainwater of pH 5.0 = 0.024 M

a) Determine the molar solubility of marble (that is, [Ca2+] in a saturated solution).

b) Determine the equilibrium constant for the overall reaction that occurs when marble reacts with acid

CaCO3 (s) + H3O+ ↔ Ca2+ (aq) + HCO3- (aq) + H2O (l)

My problem is that I have been given no more information other than what is written above. I'm not sure how to go about solving it with this amount of information.

Answer 1

Answer

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