A 25.0 milliliter sample of #HNO_3(aq)# is neutralized by 32.1 milliliters of 0.150 M #KOH(aq)#. What is the molarity of the #HNO_3(aq)#?

1 Answer
Apr 12, 2017

Answer:

0.1926 M

Explanation:

We would normally us Normality instead of Molarity for a titration equivalency calculation, but because both are monoprotic the values for N and M are the same.

#25.0 mL_(HNO_3) * ?M_(HNO_3) = 32.1 mL_(KOH) * 0.150 M_(KOH)#

#?M_(HNO_3) = [32.1 mL_(KOH) * 0.150 M_(KOH)]/(25.0 mL_(HNO_3))#

#M_(HNO_3) = 0.1926#