Question

How many milliliters of 0.100 M NaOH are required to neutralize the following solutions?

Answer 1

Answers:

• a.) 10.0 mL of 0.0500 M HCl: 5.00 ml

• b.) 25.0 mL of 0.126 M HNO₃: 31.5 ml

• c.) 50.0 mL of 0.215 M H₂SO4: 215. ml

Explanation:

All the reactions are the neutralization of strong acids with the same strong base.

At the neutralization point you have:

• number of equivalents of the base = number of equivalent of the acid

And the number of equivalents (#EQ) may be calculated using the normality (N) concentration and the volume (V)

• # EQ = N × V

Then, at the neutralization point:

• # EQ acid = N acid × V acid

• # EQ base = N base × V base

• # EQ acid = # EQ base

• N acid × V acid = N base × V base

Also, you can use the formula that relates normality with molarity

• N = M × number of hydrogen or hydroxide ions

a.) 10.0 mL of 0.0500 M HCl

• The number of hydrogen ions for HCl is 1 and the number of hydroxide ions for NaOH is 1.

• 10.0 ml × 0.0500 M × 1 = V base × 0.100 M × 1

         ⇒ V base = 10.0 ml ×0.0500 M / 0.100 M = 5.00 ml

b.) 25.0 mL of 0.126 M HNO₃

• The number of hydrogen ions for HNO₃ is 1 and the number of hydroxide ions for NaOH  is 1.

• 25.0 ml × 0.126 M × 1 = V base × 0.100 M × 1

         ⇒ V base = 25.0 ml ×0.126 M / 0.100 M = 31.5 ml

c.) 50.0 mL of 0.215 M H₂SO4

• The number of hydrogen ions for H₂SO4 is 2 and the number of hydroxide ions for NaOH  is 1.

• 50.0 ml × 0.215 M × 2 = V base × 0.100 M × 1

         ⇒ V base = 50.0 ml ×0.215 M × 2 / 0.100 M = 215. ml