What is the molar solubility of CaCO3 in ocean water at pH 8.3, given that the solubility is pH dependent and the dissociation constants are Ka1(H2CO3) = 4.3x10^-7 and Ka2(H2CO3) = 5.6x10^-11?
A
1.3 x 10^-6 M
B
2.5 x 10^-5 M
C
4.8 x 10^-7 M
D
3.2 x 10^-4 M
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1
Identify the relevant chemical equilibrium: The dissolution of calcium carbonate (CaCO3) in water can be represented by the equation: CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq).
Consider the effect of pH on carbonate species: At pH 8.3, the carbonate ion (CO₃²⁻) is in equilibrium with bicarbonate (HCO₃⁻) and carbonic acid (H₂CO₃). The relevant equilibria are: H₂CO₃ ⇌ H⁺ + HCO₃⁻ and HCO₃⁻ ⇌ H⁺ + CO₃²⁻.
Use the given dissociation constants: The dissociation constants Ka1 and Ka2 for carbonic acid are provided. These will help determine the concentrations of carbonate species at the given pH.
Calculate the concentration of carbonate ions: Use the pH to find the concentration of H⁺ ions, then apply the equilibrium expressions and dissociation constants to find the concentration of CO₃²⁻.
Determine the molar solubility of CaCO3: Use the concentration of CO₃²⁻ to find the molar solubility of CaCO3, as the solubility product (Ksp) is related to the concentrations of Ca²⁺ and CO₃²⁻ at equilibrium.