I'm working on this problem as well. I have calculated the ratios of acid and base, but am not sure what you mean by
Then assume whatever was added reacted stoichiometrically with whatever was in the solution.
Lactic acid disassociates as follows:
CH3CH(OH)COOH (aq) + H2O (l) <---------> CH3CH(OH)COO- (aq)+ H2O (l)
It has a Ka of 1.37 x 10-4 and a pKa of 3.86.
pH = pKa + log [conjugate base]
[acid]
3.50 = 3.86 + log [NaOH]
[CH3CH(OH)COOH]
-0.36 = log [NaOH]
[CH3CH(OH)COOH]
10-0.36 = 0.44
So a mole ratio of acid:base would be 0.44:1. Therefore, 4.4 L of 0.10 M lactic acid would react with 10 L of 0.10 M NaOH to produce 14.4 L of solution.
14.4 x 0.800 L = 18
4.4 ÷ 18 = 0.244 L
10 ÷ 18 = 0.556 L
To produce an 800mL buffer solution with a pH of 3.50, we would require 244mL of lactic acid and 556 mL of NaOH.
Now, adding in 0.10 g of NaOH is obviously going to increase the pH. NaOH is 40 g/mol, so .10 g is 4 moles. Does this mean that the new ratio of acid: base is now 0.44:5?