Given:
pH = 2
[H+] = 10^(-2)
[H+] = 0.01 M
-> This is where the 0.01 M came from
Since HCl + NaOH -> H2O + NaCl, it's a neutralization reaction.
20 mL 1M HCl + 20 mL 1M NaOH = 40 mL of water
If we keep adding HCl, the pH goes down as HCl will dissociate into H+ and Cl-.
How much extra HCl do we add?
Given:
Volume ~ 40 mL
Concentration = 0.01 M
moles = ?
c = moles/volume
moles = (0.040L)(0.01M)
moles = 0.0004 mol
We know the concentration of HCl is 1M.
So the amount of 1M HCl needed to add to the solution is..
c=moles/volume
volume = moles/concentration
volume = 0.0004 mol/1.0M
volume = 0.0004 L = 0.4 mL
Total Volume of HCl added = neutralizing volume + volume needed to lower pH
= 20 mL + 0.4 mL
= 20.4 mL
Hope this helps...