I'll try my best...
"What volume of HCl was added if 20mL of 1 M NaOH is titrated with 1 M HCl to produce pH=2?
Answer: 20.4 mL"
Lets break it down to two parts:
1) neutralizing the NaOH
2) getting the neutralized solution to a pH of 2.....
1) It takes 20 ml of HCl to neutralize 20 ml of 1M Nacl... which we found out by using M1V1=M2V2... The total solution thus far is 40 ml of the neutralized solution (20ml of NaOH and 20ml of HCl.)
2) To create a pH of 2 of the final solution, we will need to find out how much HCl to add to the total neutralized solution. Which will be...
(1M HCl)(x) = (.01M)(40ml+x)
(40ml +x) = the total volume of the overall solution (40ml of the neutralized solution + the volume of addition HCl needed to get the final solution to pH of 2)
.01M concentration of the final solution. Which comes from pH=[H+] which is 2 = -log[1 x 10^-2]).
x= volume of HCl needed.
1M of HCl given in the question
Then you solve for x, which x=0.4
Then you find out how much of the total volume will contain HCl only, so
First we added 20 ml of HCl to neutralize 20ml of NaOH. Then we need and additional 0.4 ml of HCl to get the neutralized solution to a pH of 2.
Therefore we have 20ml of HCl + 0.4 ml of Hcl = 20.4ml