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

a) The solubility of salts of weak acids is pH dependent.pH dependence of the solubility can be expalined because when CaCO3 dissolves :

CaCO3------>Ca^+2 + CO3^2-

orCaCO3 (s) + H2O (l) <==> Ca^2+ (aq) + HCO3^- (aq) + OH^- (aq)

Hydrolysis decreases the conc of CO3^2- which pulls the solubility quilibrium to the right making CaCO3 more soluble .

b)Keq =[Ca^2+ (aq)]*[HCO3^- (aq)]*[OH^- (aq)]/[CaCO3-(s)]*[H2O(l)]

=[Ca^2+ (aq)]*[HCO3^- (aq)]*[OH^- (aq)] Assuming the conc of pure solid and liquid as unity.

c)molar solubility of CaCO3 :

CaCO3------>Ca^+2 + CO3^2-

c-x x x

Ksp =x*x

so, [Ca^+2]=[CO3^2-]=x=(Ksp)^1/2

e)solubility is a function of pH as :

s=[Ca^+2]=[CO3^2-]=sqroot(Ksp(1+ [H+]/K2) )

for pH =8 [H+] =10^-8 M,K2=4.8*10^-11, Ksp =5*10^-9

So, we can calculate the molar solubilty s=sqroot(5*10^-9(1+ {10^-8/4.8*10^-11}) )=answer

f)Same as (e)

s=sqroot(5*10^-9(1+ {3.16*10^-8 /4.8*10^-11}) )=answer

g) drop in pH increases solubility as it is clear form the solubility relation

s=[Ca^+2]=[CO3^2-]=sqroot(Ksp(1+ [H+]/K2) ) lower the pH higher the molar solubility CaCO3.

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