Question

30.0 mL of 0.0200 mol/L NaOH are added to 70.0 mL of 0.0100 mol/L HCI. Calculate the [H3O+] and [OH─] of the resulting solution

My answer: [OH-] = 1.00x10^-10 M; [H3O+] = 1.00x10^-4 M

Teacher's answer: [OH-] = 1.00x10^-11 M; [H3O+] = 1.00x10^-3 M

Am I missing something? I made sure to convert mL to L before my calculations and have checked it over multiple times.

Answer 1

You must be missing something. I obtained 1E-3 for H^+.
millimols NaOH = 30 x 0.02 = 0.6
mmols HCl = 70 x 0.01 = 0.7
mmols HCl in excess = 0.7-0.6 = 0.1

M HCl = mmols/mL = 0.1/100 = 1E-3
So OH must be 1E-14/1E-3 = 1E-11.