A solution of 0.312 M KOH is used to neutralize 15.0 mL of an HF solution. If 28.3 mL of KOH is required to reach the end point, what is the molarity of the HF solution?

Answer 1

Chemical eqn: KOH + HF --> H₂O + KF

0.312 M KOH

28.3 mL KOH = 0.0283 L KOH

15.0 mL HF = 0.015 L HF

since Molarity = n / V (n = number of moles of solute)

n = Molarity * Volume in Liters

n = 0.312 M KOH * (0.0283 L KOH)

n = 0.0088296 moles of KOH

Since moles of KOH = moles of HF based on chemical eqn:

use this n and divide by volume of HF because M = n / V

M = 0.0088296 moles of HF / 0.015 L HF

M = 0.588 M HF

Hope this helps :)