43.8 mL of 0.246 M hydrobromic acid is added to 19.2 mL of sodium hydroxide, and the resulting solution is found to be acidic. 29.7 mL of 0.190 M potassium hydroxide is required to reach neutrality. What is the molarity of the original sodium hydroxide solution

0.267M of NaOH

Explanation:

Hydrobromic acid, HBr, reacts with NaOH thus:

HBr + NaOH → H₂O + NaBr

1 mole of acid reacts per mole of base

The resulting solution is titrated with KOH, that means in the initial reaction HBr is in excess.

The reaction of HBr with KOH, potassium hydroxide, is:

HBr + KOH → H₂O + KBr

Also 1 mole of acid reacts per mole of base

Moles of KOH added to neutralize HBr are:

0.0297L ₓ (0.190mol / L) = 5.643x10⁻³ moles of KOH = moles in excess of HBr

And initial moles of HBr are:

0.0438L ₓ (0.246mol / L) = 0.01077 moles of HBr

That means moles of HBr thar react with the NaOH:

0.01077 moles HBr - 5.643x10⁻³ moles =

5.1318x10⁻³ moles of HBr = moles of NaOH

As the volume of the NaOH solution is 19.2mL = 0.0192L, molarity of the NaOH solution is:

5.1318x10⁻³ moles NaOH / 0.0192L =

0.267M of NaOH