The following equation can be used to calculate the solubility of calcium carbon
ID: 1050188 • Letter: T
Question
The following equation can be used to calculate the solubility of calcium carbonate in water: CaCO_3 (s) + CO_2 (g) + H_2 O (1) rightarrow left arrow Ca^2+ (aq) + 2 HCO_3 (aq) K=[Ca^2+][HCO_3^-]^2/P_CO_2 K = K_sp K_b K_H K_a/K_w At 25^degree C. The K_sp of CaCO_3 is 4.6x10^-9 mol^2L^-2, K_w is 1.OxlO^-l4 mol^-2 L^-2, K_H for CO_2=0.047mol/L.atm, K_b for CO^-2_3 is 2.1xl0^-4. and K_a for H_2CO_3, =3.8x10^-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): For the given reaction in equilibrium,
K = KspxKbxKHxKa / Kw
=> K = (4.6x10-9 mol2.L-2x2.1x10-4 x0.047 mol.L-1atm-1x 3.8x10-7) / 1.0x10-14
=> K = 1.725x10-6 mol3.L-3atm-1 ---------- (1)
Again for the given reaction equilibrium constant,
K = [Ca2+(aq)]x[HCO3-(aq)]2 / Pco2 -------- (2)
From equation (1) and (2)
K = 1.725x10-6 mol3.L-3atm-1 = [Ca2+(aq)]x[HCO3-(aq)]2 / Pco2
Given the concentration of CO2 = 386 ppm
Hence partial pressure of CO2, Pco2 = (386 / 106) x 1 atm = 3.86x10-4 atm
=> K = 1.725x10-6 mol3.L-3atm-1 = [S]x[2S]2 / 3.86x10-4 atm
=> 1.725x10-6 mol3.L-3atm-1 = 4S3 / 3.86x10-4 atm
=> S3 = 1.665x10-10 mol3.L-3
=> S = cuberoot (1.665x10-10 mol3.L-3) = 5.5x10-4 mol/L or 5.5x10-4 M
Hence solubility of calcium carbonate is 5.5x10-4 M (answer)
(b): Given the concentration of CO2 = 772 ppm
Hence partial pressure of CO2, Pco2 = (772 / 106) x 1 atm = 7.72x10-4 atm
=> K = 1.725x10-6 mol3.L-3atm-1 = [S]x[2S]2 / 7.72x10-4 atm
=> 1.725x10-6 mol3.L-3atm-1 = 4S3 / 7.72x10-4 atm
=> S3 = 3.33x10-10 mol3.L-3
=> S = cuberoot (3.33x10-10 mol3.L-3) = 5.5x10-4 mol/L or 5.5x10-4 M
Hence solubility of calcium carbonate is 6.93x10-4 M (answer)
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