Compare the g/L solubility of CaCO3 in pure water, a H3O+ (aq) solution with a PH of 3.00 and a H30+(aq) solution with a pH of 1.00.
Ksp for CaCO3 is 3.36 x 10^-9
CaCO3 (s) + H3O (aq) = Ca (aq) + HCO3 (aq) + H2O (l) (equilibrium equation)
Keq= 71
I knew how to do this for pure water and got
5.8x10-3 g/L
But i don't know how to do this for
Solution with a pH of 3.00 and solution with a pH of 1.00
I would do that this way.
K = 71 = (Ca^2+)(HCO3^-)/(H^+)
71 = (x)(x)/0.001
and solve for x = about 0.27M and convert that to g/L = about 27 g/L.
I just tried that and it told me the answer was incorrect. It gave a hint saying
In acidic solution, use the reaction given for the solvation of CaCO3 in acidic solution. Construct a concentration table that allows for the calculation of final concentration as if the reaction first goes to completion and then returns to equilibrium. The initial hydronium ion concentration is calculated from the pH.
I don't really get what it is telling me to do.
I did that first and didn't like the answer I got. Then I looked in my quant book and found a slightly different method for which I obtain 0.267M solubility. With that as an answer I looked to see how to get it and lo and behold came up with the exact same answer and it looked legit. So I posted.
To compare the g/L solubility of CaCO3 in different solutions, we need to calculate the concentration of H3O+ ions in each solution using the given pH values.
Step 1: Calculate the concentration of H3O+ ions in the solution with a pH of 3.00.
The pH is defined as the negative logarithm (base 10) of the concentration of H3O+ ions in moles per liter (mol/L). Therefore, we can use the formula: pH = -log[H3O+].
pH = 3.00
[H3O+] = 10^(-pH) = 10^(-3.00) = 1.0 x 10^(-3) mol/L
Step 2: Calculate the concentration of H3O+ ions in the solution with a pH of 1.00.
Using the same formula, we have:
pH = 1.00
[H3O+] = 10^(-pH) = 10^(-1.00) = 1.0 x 10^(-1) mol/L
Step 3: Use the equilibrium equation and the equilibrium constant (Keq) to calculate the concentration of Ca2+ ions.
Keq = [Ca2+][HCO3-]/[H3O+]
71 = [Ca2+]/(1.0 x 10^(-3)) (in the solution with pH 3.00)
Therefore, [Ca2+] = 71 x 1.0 x 10^(-3) mol/L
Keq = [Ca2+][HCO3-]/[H3O+]
71 = [Ca2+]/(1.0 x 10^(-1)) (in the solution with pH 1.00)
Therefore, [Ca2+] = 71 x 1.0 x 10^(-1) mol/L
Step 4: Calculate the solubility of CaCO3 in each solution.
Using the stoichiometry of the reaction CaCO3 (s) + H3O (aq) = Ca (aq) + HCO3 (aq) + H2O (l), we know that the molar solubility of CaCO3 is equal to the concentration of Ca2+ ions.
In the solution with pH 3.00:
Molar solubility of CaCO3 = [Ca2+] = 71 x 1.0 x 10^(-3) mol/L
In the solution with pH 1.00:
Molar solubility of CaCO3 = [Ca2+] = 71 x 1.0 x 10^(-1) mol/L
Step 5: Convert the molar solubility to g/L.
To convert from moles per liter (mol/L) to grams per liter (g/L), we need to multiply the molar solubility by the molar mass of CaCO3.
The molar mass of CaCO3 is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O (x3): 16.00 g/mol
Molar mass of CaCO3 = 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol
Finally, multiply the molar solubility by the molar mass to obtain the solubility in g/L.
In the solution with pH 3.00:
Solubility of CaCO3 = molar solubility x molar mass = (71 x 1.0 x 10^(-3)) x 100.09 g/L
In the solution with pH 1.00:
Solubility of CaCO3 = molar solubility x molar mass = (71 x 1.0 x 10^(-1)) x 100.09 g/L
By performing the above calculations, you should be able to determine the g/L solubility of CaCO3 in a solution with a pH of 3.00 and a solution with a pH of 1.00 based on the given data.