Titration Dilution

[MDwannabe7]

If 10 mL of 1M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCL was added?

I understand that since they are both strong - 10 mL of HCL will neutralize 10 mL of NaOH. However, I don't quite understand how to get from there to the requested pH of 2. I know that that means you want your final concentration of H+ to be 1.0 x 10^-2, but not sure where to go from there.

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[Phlame217]

Make a table and label top with equation
H+ + OH- = H2O

Get your starting values
H+ = 0
OH- = .1L * 1M = .1 mole
H20 = Extraneous

Your final Values
H+ = 1.0 * 10^-2
OH- = 0
H20 = Extraneous

Quick mental math says you have to add
1.0*10^-2 + .1 moles of H+ to get your pH
= .01 + .1 = .11 moles of H+ added

now you have amount, and now convert to L
.11Moles * 1L/1M = .11 Liters or 11 mL

=11 mL