Answer to Question #73270 in General Chemistry for Angelo

Question

If I add 1000 mg of CaCO3 to 25 ml of water, I achieve an experimental pH around 9, how can I prove this mathematically? If I then add 25 ml of 0.1 N HCl, I achieve a pH of about 7.5, how can I prove this mathematically?

Answer 1

CaCO3 = Сa2+ + CO32-
Ksp (25C) = 3.36x10-9 solubility product constant for CaCO3

The possible concentration of CO32- is 5.8x10-5 mol/L .
CO32- + HOH = HCO3- + OH-
[OH-] = 5.8x10-5 mol/L
pOH = - lg[OH-] ≈ 4.3. pH = 14 – pOH = 14 – 4.3 ≈ 9.7

n (HCl) = 0.1 N ∙ 0.025 L = 0.0025 mol
This quantity reacts completely with 1g of CaCO3. The salt is in excess! So if you wait till the end of the reaction the pH won’t change!