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() 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?

Answer 1

0.49 % of the acid ionizes

Step-by-step explanation:

Step 1: data given

number of moles acetic acid = 1.0 moles

Volume = 1.0 L

Temperature = 25°C

0.42% of the acetic acid ionizes to form acetate ion and hydronium ion

Ka of acetic acid = 1.8 * 10^-5

Step 2: The balanced equation

HAc + H2O ⇔ H3O+ + Ac-

Step 3: Calculate concentration

Ka = [H3O+][Ac-] / [Hac]

⇒with [H3O+] = X M

⇒with [Ac-] = X M

⇒with [Hac] = 0.75 M

Ka = 1.8*10^-5 =X*X / 0.75

X = 0.00367 M

⇒with [H3O+] = 0.00367 M

⇒with [Ac-] = 0.00367 M

⇒with [Hac] = 0.75 M

Step 4: Calculate the ionization

% ionization = (0.00367 / 0.75) * 100%

% ionization = 0.49%

0.49 % of the acid ionizes