What volume in mL of 0.50 M NaOH is needed to completely neutralize 8.00 mL of 0.50 M HCl?

Answer 1

Since we have both a strong acid and a strong base, they will both completely dissociate and you'll have H+ and OH- ions in the solution. When we say "completely neutralize" what we mean is that for every H+ produced by the HCl, there is an OH- from the NaOH.

So, we use molarity (mol/L) to determine that there are 0.0040mol of H+ ions (0.50M=n/0.00800L, solve for n). Therefore we need 0.0040mol of OH- ions, which is a volume of 0.0080L or 8.00mL (0.50M=0.0040mol/V, solve for V then convert to mL).

Note, we could have taken a shortcut. Since both have the same concentration, equal number of moles will be the same volume. However, I wanted to show the steps that work even when they're not the same concentration.