Answer:
Explanation:
1. In a titration, a 25cm3 sample of nitric acid, HNO3, was used to react with 100cm3 of 0.50 mol/dm3 sodium hydroxide (NaOH) solution.
A) Calculate the number of moles of sodium hydroxide used.
the NaOH has 0.5 moles each liter, so 10 ml has 0.05 moles
B) Calculate how many moles of nitric acid react.
1 mole of HNO3 REACTS WITH ONE MOLE OF NAOH
SINCE NAOH WAS 0.05 MOLES, HNO3 WAS 0.05 MOLES
C) Calculate the concentration of nitric acid in mol/dm3.
THESE MOLES OF HNO3 WERE IN 25/1000 L OR 25x10^-3L
SO THE CONCENTRATION OF HNO3 IS
0.05 MOLES/ (25X10^-3) = 2MOLES/LITER
CHECK
2 MOLES/LITER HAS 0.2 MOLES IN 100 ML AND 0.2/4=0.05 MOLES IN 25 ML
