The solubility of CaCO3 is pH dependent. (Ka1(H2CO3)=4.3×107,Ka2(H2CO3)=5.6×1011

Question

The solubility of CaCO3 is pH dependent. (Ka1(H2CO3)=4.3×107,Ka2(H2CO3)=5.6×1011.)

**Part A Calculate the molar solubility of CaCO3 ( Ksp = 4.5×109) neglecting the acid-base character of the carbonate ion. Express your answer using two significant figures. S = 6.71105 M (****THIS HAS BEEN ANSWERED, PLACED HERE FOR REFERENCE ONLY****)

Part B-- Use the Kb expression for the CO32 ion to determine the equilibrium constant for the reaction CaCO3(s)+H2O(l)Ca2+(aq)+HCO3(aq)+OH(aq) Express your answer using two significant figures.

K = ??????

Part 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? Express your answer using two significant figures.

S = ????

Part D --What is the molar solubility of CaCO3 at the pH of the ocean (8.3)? Express your answer using two significant figures. S = M

Part E-- If the pH is buffered at 7.5, what is the molar solubility of CaCO3? Express your answer using two significant figures. S = M

Explanation / Answer

A)

CaCO3 --> Ca2+ + CO32-

(C – S) --> S + S

Ksp = [Ca2+][CO3-] = S2 = 4.5 x 10-9

S = 6.71 x 10-5 M

B)

CO32- + H2O --> HCO3- + OH-

Kb = Kw/Ka2 = [OH-][HCO3-]/[CO32-] = 10-14 / 5.6 x 10-11 = 1.79 x 10-4

CaCO3(s)+H2O(l) --> Ca2+(aq)+HCO3?(aq)+OH?(aq)

K = [OH-][HCO3-][Ca2+] = Kb * Ksp

= 8.04 x 10-13

C)

CaCO3(s)+H2O(l) --> Ca2+(aq)+HCO3?(aq)+OH?(aq)

C – S --> S + S + S

K = S3

S = K1/3 = 9.3 x 10-5 M

D)

pH = 8.3

pOH = 14 – 8.3 = 5.7

[OH-] = 10^(-5.7) = 2 x10-6 M

CaCO3 --> Ca2+ + CO32-

(C – S) --> S + (S – x)

Ksp = S (S – x)

CO32- + H2O --> HCO3- + OH-

(S – x) --> x +( x + [OH-])

Kb = x * (x + 2 x 10-6) / (S – x) = 1.79 x 10-4

Ksp = S (S – x) = 4.5 x 10-9 .. eqn (1)

S * x * (x + 2 x 10-6) = 4.5 x 10-9 * 1.79 x 10-4 = 8.04 x10-13   … eqn(2)

Solving eqn (1) & eqn (2) we get,

S = 1.19 x 10-4 M

E)

pH = 7.5

pOH = 14 – 7.5 = 6.5

[OH-] = 10^(-6.5) = 3.16 x10-7 M

CO32- + H2O --> HCO3- + OH-

(S – x) --> x + [OH-]

Kb = x [OH-] / (S – x) = 1.79 x 10-4

x / (S – x) = 1.79 x 10-4 / [OH-] = 5.65 x 102

Ksp = S (S – x) = 4.5 x 10-9   …. eqn(1)

Eqn(1) * eqn (2) = x S = 2.54 x 10-6 … eqn (2)

Solving eqn (1) & eqn (2) we get,

S = 1.6 x 10-3 M

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