How many milliliters of 0.100 m naoh are required to neutralize the following solutions?

a.) 10.0 ml of 0.0500 m hcl
b.) 25.0 ml of 0.126 m hno3
c.) 50.0 ml of 0.215 m h2so4

Answers:

a.) 10.0 mL of 0.0500 M HCl: 5.00 mlb.) 25.0 mL of 0.126 M HNO₃: 31.5 mlc.) 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 baseN 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

They carry out the exchange of gases as we breathe ( oxygen and carbon dioxide).