The solubility of CaCO3 is pH dependent. B) Use the Kb expression for the CO3^2-

Question

The solubility of CaCO3 is pH dependent.

B) Use the Kb expression for the CO3^2- ion to determine the equilibrium constant for the reaction

CaCO3 (s) + H2O (l) <==> Ca^2+ (aq) + HCO3^- (aq) + OH^- (aq)

C) If we assume that the only sources of Ca^2+, HCO3^-, and OH^- ions are from the dissolution of CaCO3, what is the molar solubility of CaCO3 using the preceding expression?

E) If the pH is buffered at 8.2 (as is historically typical for the ocean), what is the molar solubility of CaCO3?

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

G) How much does this drop in pH increase solubility?

Explanation / Answer

Kb= [ca+2][Hco3-][OH-]

Let us assume the concentration of calciumcarbonate to be 1 ,at equilibrium the concentrartions of ions be x and (1-x)caco3.

now x^3/1-x =Kb =>x=Kb^(1/3)

the molar solubilty of caco3 is Kb^(1/3)

E) [H+][oH-]=10^-14

[H+]=(10^-14)/x

-log([H+])=8.2

[H+]=10^-8.2

(10^-14)/x =10^-8.2

x=10^-5.8

F) similarly x= 10^(-6.5)

G)increase in solubility =10^-5.8 -10^-6.5

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.