Question

A) The Ka of a monoprotic weak acid is 2.29 × 10-3. What is the percent ionization of a 0.129 M solution of this acid? I got 104.8% which I know is impossible.

B) The Ka value for acetic acid, CH3COOH(aq), is 1.8× 10–5. Calculate the pH of a 2.40 M acetic acid solution.

C) Calculate the pH of the resulting solution when 4.00 mL of the 2.40 M acetic acid is diluted to make a 250.0 mL solution.

I also got these incorrect, but am not sure why and would like to understand. Please help me understand these before my upcoming test! Thanks!

Answer 1

Solution.

A) The percent ionization can be found from Ostwald's dilution law:

$K_a=\frac{\alpha ^{2}c}{1-\alpha };$

$2.29\times 10^{-3}=\frac{\alpha ^{2}\cdot 0.129}{1-\alpha },$

$\alpha = 0.125, \ or \ 12.5\%.$

B) As acetic acid is a weak acid, the simplified formula can be used:

$pH=\frac{pK_a+pc}{2};$

$pK_a = -\log K_a = -\log 1.8\times 10^{-5} = 4.745;$

$pc = -\log c = -\log 2.4 = -0.38;$

$pH = \frac{4.745-0.38}{2} = 2.18.$

C) The new concentration of an acetic acid is

$c = \frac{4\times 2.4}{250} = 0.0384;$

$pH = \frac{4.745+1.416}{2} = 3.08.$