In topscore test # 2:

What volume of HCL was added if 20 ml of 1 M NaOH is titrated with 1 M HCL to produce a pH = 2?

The answer is 20.4 ml

I don't understand their solution...what equation is being used. I get the whole idea of there being 40 ml of a .5 M solution but from there I'm lost. Please someone calrify. Taking the DAT Monday at noon.

okay it is actually very easy when thinking qualitatively.

when you titrate 1M NaOH of 20 ml with 1 M of HCL, what do you see from this ? this is actually a neutralization rxn, in which the strong base reacts with the strong acid to produce salt and water.

so if you add the equal volume of HCL into the NaOH, you will get pH of 7 right? b/c they are all strong acid and base. so in order to get pH 2, you need to add more HCL into the solution. a pH of 2 would produce a concentration of 10e-2 M by using the log.

so by using titration equation to tirate the whole volume:

(20ml NaOH + 20ml HCL )*( 1 M ) = ( 0.1 M )*( # ml)

==> 0.4 ml

so additional of 0.4 ml into 20ml of HCL => 20.4 ml

if you know how to do destroyer, you will be fine!