first, add enough HCl to neutralize the sol'n: 20ml (according to equation VaNa = VbNb where a=acid, b=base, N=normality V=volume)...When it is neutralized, a 40 ml NaCl sol'n will be produced.
Next, how much more HCl do you need to add to the neutralized sol'n to produce a pH of 2? pH of 2 = [H+] = 1 x 10^-2 M = .01 M
Let x = amount of HCl to be added. so the final volume of the sol'n w/ph 2 is going to be 40 + x.
set up a dilution, since you are basically adding x amount of HCl to get it to become ph = 2...follow formula M1V1 = M2V2:
(1M)(x ml) = (.01 M)(40 + x ml)
solve for x and we get x = .4
so the FINAL volume of the solution that you titrated/neutralized/whatever is 40.4, but you added 20.4 mL to the original solution in order to get it to be ph = 2 🙂
there must be an easier way to explain this. sorry, i was basically teaching myself while i wrote this up.
