Question

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.

B) Calculate how many moles of nitric acid react.

C) Calculate the concentration of nitric acid in mol/dm3.

Answer 1

In a titration, the number of moles of sodium hydroxide used is 0.050 mol. The number of moles of nitric acid that react is also 0.050 mol. The concentration of nitric acid is 2.00 mol/dm3.

Step-by-step explanation:

In a titration, the number of moles of sodium hydroxide used can be calculated using the equation:

# moles NaOH = concentration (mol/dm3) x volume (dm3)

= 0.50 mol/dm3 x 0.100 dm3

= 0.050 mol

The number of moles of nitric acid that react can be determined using the balanced chemical equation:

HNO3 + NaOH → NaNO3 + H2O

As the stoichiometric ratio is 1:1, the number of moles of nitric acid is also 0.050 mol.

The concentration of nitric acid can be calculated using the equation:

concentration = moles/volume

= 0.050 mol/0.025 dm3

= 2.00 mol/dm3

Answer 2

Step-by-step 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