Question

The solubility of CaCO3 is pH dependent. (Ka1(H2CO3)=4.3×10−7,Ka2(H2CO3)=5.6×10−11.)

**Part A Calculate the molar solubility of CaCO3 ( Ksp = 4.5×10−9) neglecting the acid-base character of the carbonate ion. Express your answer using two significant figures. S = 6.71⋅10−5 M (****THIS HAS BEEN ANSWERED, PLACED HERE FOR REFERENCE ONLY****)

Part B-- Use the Kb expression for the CO32− ion to determine the equilibrium constant for the reaction CaCO3(s)+H2O(l)⇌Ca2+(aq)+HCO3−(aq)+OH−(aq) Express your answer using two significant figures.

K = ??????

Part C --- If we assume that the only sources of Ca2+, HCO3−, and OH− ions are from the dissolution of CaCO3, what is the molar solubility of CaCO3 using the preceding expression? Express your answer using two significant figures.

S = ????

Part D --What is the molar solubility of CaCO3 at the pH of the ocean (8.3)? Express your answer using two significant figures. S = M

Part E-- If the pH is buffered at 7.5, what is the molar solubility of CaCO3? Express your answer using two significant figures. S = M

Answer 1

A)

CaCO₃ --> Ca²⁺ + CO₃^2-

(C – S) --> S + S

Ksp = [Ca²⁺][CO₃^-] = S² = 4.5 x 10^-9

S = 6.71 x 10^-5 M

B)

CO₃^2- + H₂O --> HCO₃^- + OH-

Kb = Kw/Ka2 = [OH-][HCO₃^-]/[CO₃^2-] = 10^-14 / 5.6 x 10^-11 = 1.79 x 10^-4

CaCO₃(s)+H₂O(l) --> Ca²⁺(aq)+HCO₃^?(aq)+OH^?(aq)

K = [OH-][HCO₃-][Ca²⁺] = Kb * Ksp

= 8.04 x 10^-13

C)

CaCO₃(s)+H₂O(l) --> Ca²⁺(aq)+HCO₃^?(aq)+OH^?(aq)

C – S --> S + S + S

K = S³

S = K^1/3 = 9.3 x 10^-5 M

D)

pH = 8.3

pOH = 14 – 8.3 = 5.7

[OH-] = 10^(-5.7) = 2 x10^-6 M

CaCO₃ --> Ca²⁺ + CO₃^2-

(C – S) --> S + (S – x)

Ksp = S (S – x)

CO₃^2- + H₂O --> HCO₃^- + OH-

(S – x) --> x +( x + [OH-])

Kb = x * (x + 2 x 10^-6) / (S – x) = 1.79 x 10-4

Ksp = S (S – x) = 4.5 x 10^-9 .. eqn (1)

S * x * (x + 2 x 10^-6) = 4.5 x 10^-9 * 1.79 x 10^-4 = 8.04 x10^-13   … eqn(2)

Solving eqn (1) & eqn (2) we get,

S = 1.19 x 10^-4 M

E)

pH = 7.5

pOH = 14 – 7.5 = 6.5

[OH-] = 10^(-6.5) = 3.16 x10-7 M

CO₃^2- + H₂O --> HCO₃^- + OH-

(S – x) --> x + [OH-]

Kb = x [OH-] / (S – x) = 1.79 x 10-4

x / (S – x) = 1.79 x 10-4 / [OH-] = 5.65 x 10²

Ksp = S (S – x) = 4.5 x 10^-9   …. eqn(1)

Eqn(1) * eqn (2) = x S = 2.54 x 10^-6 … eqn (2)

Solving eqn (1) & eqn (2) we get,

S = 1.6 x 10^-3 M