Question

Find the pH of 0.1moles of carbonic acid dissolved in 250ml of water

some help would be greatly appreciated!

Answer 1

[H2CO3](initial) = 0.1 mol / 0.25L = 0.4M

H2CO3(aq) <=> H+(aq) + HCO3(aq) (Step 1)
K1 = [H+][HCO3-] / [H2CO3]

HCO3-(aq) <=> H+(aq) + CO3^2-(aq) (Step 2)
K2 = [H+][CO3-2] / [HCO3-]
K2 = 4.7x10^-11

The hydrogen (hydronium) ion concentration depends on step one. Step 2 does not produce many hydrogen ions.

Substituting in the expression for K1, and assuming [H+]=[CO3-2],
[H+]^2 / (0.4-[H+]) = 2.0x10^-4

Since [H+] is much lower than 0.4,
(0.4-[H+]) =0.4 (approximately)

[H+]^2 = (0.4)(2.0x10^-4)
[H+] = sqrt(8.0x10^-5) = 8.9x10^-3
pH = -log(8.9x10^-3) = ______?

<b>PLEASE NOTE</b>: H2CO3 solutions are prepared by dissolving CO2 gas in water under some specified pressure. NOT ALL of the CO2 is converted to H2CO3. That means that the premise of the question is vague or absurd since this hypothetical solution most likely could not exist at normal pressures.