If 20.0 ml of glacial acetic acid (pure hc2h3o2 is diluted to 1.70 l with water, what is the ph of the resulting solution? the density of glacial acetic acid is 1.05 g/ml.

Given:

20.0 ml of glacial acetic acid

1.70 L with water

density of the glacial acetic acid is 1.05 g/ml; Ka = 1.8 x 10⁻⁵

1) find out how many grams of HC2H3O2 is in this glacial matter. It says here it has a density of 1.05g/mL.

So we find the grams from that = > 1.05 g/mL * (20.0 mL) = 21 grams of HC2H3O2.

Convert this to moles = > 21 grams (1 mol / 60 g) = 0.35 mol HC2H3O2

We then get the Ka formula from that equation so it is

Ka = [H3O+] [C2H3O2] / [HC2H3O2] (Remember H2O has an activity of 1 so we exclude it)

We know the concentration of HC2H3O2 because we have moles of HC2H3O2

Finding concentration of HC2H3O2 = > 0.35 mol HC2H3O2/1.70 L = 0.20588 M - initial concentration

1.8 x 10 ⁻⁵ = (x) (x) / (0.20588 M - x)

1.8 X 10⁻⁵ (0.20588 M) = x²

3.70584⁻ x 10⁻⁶ = x²

x = 1.9251 x 10⁻³

pH = - log (H₃O⁺) = - log (1.9251 x 10⁻³) = - (-2.7156) = 2.7156

pH of the resulting solution is 2.7156