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 

They carry out the exchange of gases as we breathe ( oxygen and carbon dioxide).