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.

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)

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 CaCO3using the preceding expression?

Part D

What is the molar solubility of CaCO3 at the pH of the ocean (8.3)?

Part E

If the pH is buffered at 7.5, what is the molar solubility of CaCO3?

Answer 1

Part-A:

CaCO₃(s) + H2O <----> Ca²⁺(aq) + CO₃^2-(aq)

-------------------------------- S ------------- S

Ksp = [Ca²⁺(aq)]*[CO₃^2-(aq)]

=> 4.5*10^-9 = S*S = S²

=> S = square root (4.5*10^-9 )

=> S = 6.71*10^-5 M (Answer)

Part-B:

CaCO₃(s) <----> Ca²⁺(aq) + CO₃^2-(aq) ; Ksp = 4.5*10^-9

CO₃^2-(aq) + H₂O(l) <----> HCO₃^-(aq) + OH^-(aq) ; Kb2 = Kw/Ka2 = 10^-14 / 5.6*10^-11  = 1.79*10^-4  

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Adding the above two equations and cancelling CO₃^2-(aq) from both sides:

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

K = Ksp * Kb2

=> K = 4.5*10^-9 * 1.79*10^-4  

=> K = 8.04*10^-13 (Answer)

Part-C:

--CaCO₃(s) + H₂O(l) <----> Ca²⁺(aq) + HCO₃^-(aq) + OH^-(aq) ; K = 8.04*10^-13

------------------------------------ S ------------ S -------------- S

K = 8.04*10^-13 = S*S*S = S³

=> S = cuberoot (8.04*10^-13)

=> S = 9.30*10^-5 M (Answer)

Part-D:

pOH = 14 - 8.3 = 5.7

=> [OH^-(aq)] = 10^-5.7

=> [OH^-(aq)] = 2.0*10^-6 M

--CaCO₃(s) + H₂O(l) <----> Ca²⁺(aq) + HCO₃^-(aq) + OH^-(aq) ; K = 8.04*10^-13

------------------------------------ S ------------ S -------------- 2.0*10^-6 M

=> K = 8.04*10^-13 = S*S*2.0*10^-6

=> S² = 8.04*10^-13 / 2.0*10^-6 = 4.03*10^-7

=> S = 6.35*10^-4 M (Answer)

PArt-E:

pH = 7.5:

[OH^-] = 10^-6.5 M

K = 8.04*10^-13 = S*S*10^-6.5

=> S² =  8.04*10^-13 / 10^-6.5

=> S = 1.60*10^-3 M (Answer)