Question #73270
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?
Expert's answer
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!
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!
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