The solubility of CaCO3 is pH-dependent.
Ka1(H2CO3)=4.3×10^−7
Ka2(H2CO3)=5.6×10^−11
a. Calculate the molar solubility of CaCO3 (Ksp = 4.5×10^−9) neglecting the acid-base character of the carbonate ion.
b. Use the Kb expression for the CO3^2− ion to determine the equilibrium constant for the reaction
CaCO3(s) + H2O(l) ⇌ Ca2^+(aq) + HCO3^−(aq) + OH^−(aq)
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?
d. What is the molar solubility of CaCO3 at the pH of the ocean (8.3)?
f. If the pH is buffered at 7.5, what is the molar solubility of CaCO3?
See you other post above. I assume for parts c and f that we consider only k1 for H2CO3 since k1 and k2 are widely separated.
a. To calculate the molar solubility of CaCO3 neglecting the acid-base character of the carbonate ion, we can use the solubility product constant (Ksp) equation. The equation is:
CaCO3 (s) ⇌ Ca2+ (aq) + CO3^2- (aq)
The Ksp for CaCO3 is given as 4.5×10^-9. Since the stoichiometry of the reaction is 1:1 for CaCO3 to Ca2+ and CO3^2-, we can assume that x mol/L of CaCO3 dissolves to form x mol/L of Ca2+ and x mol/L of CO3^2-. Thus, the equilibrium expression is:
Ksp = [Ca2+][CO3^2-] = x * x = x^2
Substituting the given value of Ksp (4.5×10^-9) into the equation:
4.5×10^-9 = x^2
Taking the square root of both sides:
x = √(4.5×10^-9)
x ≈ 6.71 × 10^-5 mol/L
Therefore, the molar solubility of CaCO3 (neglecting the acid-base character of the carbonate ion) is approximately 6.71 × 10^-5 mol/L.
b. To determine the equilibrium constant for the reaction CaCO3(s) + H2O(l) ⇌ Ca2+(aq) + HCO3^-(aq) + OH^-(aq), we need to find the Kb expression for the CO3^2- ion. Since Kb = Kw/Ka, where Kw is the autoionization constant of water (1.0×10^-14) and Ka is the acid dissociation constant, we can use the equation:
Kb(CO3^2-) = Kw/Ka2
Substituting the given values:
Kb(CO3^2-) = (1.0×10^-14)/(5.6×10^-11)
Kb(CO3^2-) ≈ 1.79×10^-4
Therefore, the equilibrium constant for the reaction CaCO3(s) + H2O(l) ⇌ Ca2+(aq) + HCO3^-(aq) + OH^-(aq) is approximately 1.79×10^-4.
c. If we assume that the only sources of Ca2+, HCO3^-, and OH^- ions are from the dissolution of CaCO3, we can use the equilibrium constant expression to find the molar solubility of CaCO3. The equilibrium expression is:
Kb(CO3^2-) = [Ca2+][HCO3^-][OH^-]
Since the stoichiometry of the reaction is 1:1:1 for CaCO3 to Ca2+, HCO3^-, and OH^-, the molar solubility of CaCO3 is equal to the concentration of Ca2+ or HCO3^- or OH^-. Therefore, the molar solubility of CaCO3 using the preceding expression is approximately 1.79×10^-4 mol/L.
d. To calculate the molar solubility of CaCO3 at the pH of the ocean (pH 8.3), we need to consider the carbonate equilibrium reactions and the related acid-base chemistry.
At pH 8.3, we can assume that all the H2CO3 molecules dissociate into HCO3^- and H+ ions, while all the CO3^2- ions are generated by the dissociation of HCO3^-.
The concentrations of H2CO3, HCO3^-, and CO3^2- are related by the equilibrium reactions and the given Ka values. By applying the Henderson-Hasselbalch equation for carbonic acid, we have:
pH = pKa + log([HCO3^-]/[H2CO3])
Rearranging the equation, we get:
[HCO3^-] = [H2CO3] * 10^(pH - pKa)
Substituting the given Ka values and pH = 8.3:
[HCO3^-] = [H2CO3] * 10^(8.3 - 4.3)
[HCO3^-] ≈ [H2CO3] * 5623
The concentration of CO3^2- can be calculated as:
[CO3^2-] = Ka2 / [HCO3^-]
Substituting the given Ka2 value and the calculated [HCO3^-]:
[CO3^2-] ≈ (5.6×10^-11) / ([H2CO3] * 5623)
Since the stoichiometry of the CaCO3 dissolution reaction is 1:1 for Ca2+ and CO3^2-, the molar solubility of CaCO3 is equal to the concentration of the CO3^2- ion. Therefore, the molar solubility of CaCO3 at the pH of the ocean (8.3) can be estimated by calculating the [CO3^2-] concentration using the above equations.
f. To determine the molar solubility of CaCO3 at a buffered pH of 7.5, we can follow the same process as in part (d), but now using a different pH value (7.5). Following the steps explained in part (d) and substituting the pH = 7.5, you can calculate the molar solubility of CaCO3.