Question

The solubility of CaCO_3 is pH dependent, Calculate the molar solubility of CaCO_3 (K_sp = 4.5 times 10^-9) neglecting the acid-base character of the carbonate Ion. Use the K_b expression for the CO_3^2- ion to determine the equilibrium constant for the reaction CaCO_3(s) + H_2O(l) equilibrium Ca^2 (aq) + HCO_3^-(aq) + OH^- (aq) If we assume that the only sources of Ca^2+, HCO_3^-, and OH^- ions are from the dissolution of CaCO_3, what is the molar solubility of CaCO_3 using the equilibrium expression from part (b)? What is the molar solubility of CaCO_3 at the pH of the ocean (8.3)? If the pH is buffered at 7.5, what is the molar solubility of CaCO_3?

Answer 1

CO_{3^2- (aq)} + H₂O $\rightleftharpoons$ HCO₃^- + OH^- Carbonate ion hydrolyses in water to form a basic solution.

K_b =[HCO₃^-] [OH^-]/[CO₃^-] Therefore, [HCO₃^-][OH^-] = K_b[CO₃^-]

K_b(CO₃^2-) = 2.1 x10^-4

For this equation, CaCO3(s)+H2O(l)⇌Ca2+(aq)+HCO3−(aq)+OH−(aq)

equilibrium constant K = [Ca²⁺][HCO₃^-][OH^-] CaCO₃ is in the solid state, and water is in excess. Hence their concentration is not considered in the equilibrium expression

K = [Ca²⁺][HCO₃^-][OH^-] =    K = [Ca²⁺] K_b[CO₃^-] = [Ca²⁺][CO₃^-] x2.1x10^-4  = K_spx2.1x10^-4

⁼ 4.5x10^-9x 2.1x10^-4 = 10.29x10^-13 = 1.029x10^-12  = 1.0x10^-12

D. CaCO3(s) → ← Ca²⁺ + CO3 ^2-

Ksp = [Ca2+] [CO3 2-] (1) If you don't take into account the hydrolysis of carbonate ion

S = [Ca²⁺] = [CO3 ^2-] = K_sp = 6.71x10^-5 (Has been given correct)

Taking into account K_a2 K2 = [H+ ] [CO3 ^2-]/ [HCO]3^-

S = [Ca²⁺] = [CO3 ^2-] + [HCO3 ^-] = [CO3 ^2-] + [H⁺ ] [CO3 ^2-]/ K2

S = [Ca²⁺] = [CO3 2-] ( 1 + [H⁺ ] /K2 )

Multiply both sides by Ca⁺² ions

S²= [Ca²⁺] ² = Ksp ( 1 + [H⁺ ] /K2) Substituting the values of K_sp and K₂ = 5.0x10^-11 and[ H⁺]

We can get the solubility of CaCO3 at any Desired pH

For instance at 8.3 pH

-log[H⁺] = 8.3 [H⁺] = 5.01x10^-9

S² = 45.45x10^-8   S = 6.74x10^-4 mol L^-1