The following equation can be used to calculate the solubility of calcium carbon

Question

The following equation can be used to calculate the solubility of calcium carbonate in water: CaCO_3 (s) + CO_2 (g) + H_2O (l) Ca^2+ (aq) + 2 HCO_3^- (aq) K= [Ca^2+][HCO_3^-]^2/P_CO2 K = K_spK_bK_HK_a/K_w At 25 degree C, The K_sp of CaCO_3 is 4.6 times 10^-9 mol^2L^-2, K_w is 1.0 times 10^-14 mol^2L^-2, K_H for CO_2 = 0.047mol/L middot atm, K_b for CO_3^2- is 2.1 times 10^-4, and K_a for H_2CO_3 =3.8 times 10^-7. What is the solubility of calcium carbonate in units of molarity? Assume 386 ppm CO_2 and 1 atm of pressure. (Note ppm in this case means 386 molecules of CO_2 per million gas molecules) What is the solubility if the CO_2 concentration doubles to 772 ppm?

Explanation / Answer

a) CaCO3 (s) + CO2 (g) + H2O (l) <====> Ca2+ (aq) + 2 HCO3- (aq)

Let x moles/L be the molar solubility of CaCO3 in water; then [Ca2+] = x moles/L and [HCO3-] = 2x moles/L.

K = [Ca2+][HCO3-]2/PCO2 …..(1)

K = KspKbKHKa/Kw …..(2)

Given, Ksp = 4.6*10-9 mol2L-2, Kw = 1.0*10-14 mol2L-2, KH for CO2 = 0.047 mol/L.atm, Kb for CO32- = 2.1*10-4 and Ka for H2CO3 = 3.8*10-7; substitute in equation (2) to obtain

K = (4.6*10-9 mol2L-2)*(2.1*10-4)*(0.047 mol/L.atm)*(3.8*10-7)/(1.0*10-14 mol2L-2) = 1.725*10-6 mol/L.atm.

Also, the concentration of CO2 is 386 ppm which, in case of gas means that the mole fraction of CO2 is 386*10-6 mol/1 mol = 386*10-6.

The total pressure is 1 atm; PCO2 = (mole fraction of CO2)*(total pressure) = (386*10-6)*(1 atm) = 386*10-6 atm.

Therefore, we can write for equation (1) as

1.725*10-6 = (x).(2x)2/(3.6*10-6)

====> (1.725*10-6)*(386*10-6) = 4x3

====> 6.6585*10-10 = 4x3

====> x3 = 1.664625*10-10

====> x = 5.501*10-4 5.50*10-4

The molar solubility of CaCO3 is 5.50*10-4 mol/L = 5.50*10-4 M (ans).

b) The concentration of CO2 is 772 ppm; therefore the mole fraction of CO2 is 772*10-6 and PCO2 = 772*10-6 atm.

Let s be the solubility of CaCO3 in water; therefore, [Ca2+] = s and [HCO3-] = 2s.

Put the values in the expression. Note that K remains unchanged.

1.725*10-6 = (s).(2s)2/(772*10-6)

=====> 4s3 = 1.3317*10-9

=====> s3 = 3.32925*10-10

=====> s = 6.931*10-4 6.93*10-4

The molar solubility is 6.93*10-4 mol/L = 6.93*10-4 M (ans).

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