Question

When 1.0 mole of acetic acid is diluted with water to a volume of 1.0 L at 25 C, 0.42% of the acetic acid ionizes to form acetate ion and hydronium ion.

CH3CO2H(aq) + H2O(l) = CH3CO2–(aq) + H3O+(aq)

What percentage of the acid ionizes when 0.75 mole of acetic acid is diluted with water to 1.0 L at 25 C?

Can you show the answer step by step so I can figure out how to do the other examples like this one please. Thank you!

Answer 1

molarity of acetic acid = moles/volume of solution in litres = 1/1 = 1 M

Now,

CH₃COOH(aq) <-----> CH₃COO^-(aq) + H⁺(aq)

At eqb., [CH₃COOH] = (1 - 0.0043) M ; [H⁺] = [CH₃COO^-] = 0.0043 M

THus, K_a = {[H⁺]*[CH₃COO^-]}/[CH₃COOH] = 1.857*10^-5

Now, when moles of acetic acid = 0.75

Initial concentration of :-

[CH₃COOH] = 0.75 M ; [H⁺] = [CH₃COO^-] = 0 M

Let at eqb., [CH₃COOH] = (0.75 - x) M ; [H⁺] = [CH₃COO^-] = x M

Thus, 1.857*10^-5 = x²/(0.75-x)

or, x = 0.0037 M

Thus, % ionisation = (0.0037/0.75)*100 = 0.493 %