dontbam said:

Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.*4* mL

first...you have 20mL of a 1M Strong Base that will take only 1 proton per molecule (NaOH). To neutralize that, add an equal amount of a strong acid (HCl) at 1 M. So to neutralize the 20mL of 1 M NaOH, we add 20 mL of 1 M HCl.

That was step 1, and you do that in your head just by looking at the problem. Then the question is how much more do we have to add of 1M HCl to our now 40mL of neutral solution to get the pH = 2

You need to know this formula below and how to work with it

pH = 2

pH = -log [H+]

2 = -log [H+]

10^-2 = [H+]

or

[H+] = 0.01 M

if we call X the amount of HCl we need to add to our 40ml solution to reach a pH = 2 then you can put it together like this. It's kind of a dilution question. You start with 1 M HCl. You are diluting it by adding some of this 1M HCl to a 40mL neutral solution. When will the solution's pH drop of 0.01 M HCL 9which is pH =2)

(1 M HCl) * (X mL) = (0.01 M HCl) * (40 + X ml)

I'm not gonna do the math here...but the answer comes out right.