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Acids and Bases - Titration Example Problem

Worked Chemistry Problems

By Todd Helmenstine

Titration is used to determine concentration of an acid.

Titration is used to determine concentration of an acid.

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Question

A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?

Solution

Step 1 - Determine [OH-]

Every mole of NaOH will have one mole of OH-. Therefore [OH-] = 0.5 M.

Step 2 - Determine the number of moles of OH-

Molarity = # of moles/volume

# of moles = Molarity x Volume

# of moles OH- = (0.5 M)(.025 L)
# of moles OH- = 0.0125 mol

Step 3 - Determine the number of moles of H+

When the base neutralizes the acid, the number of moles of H+ = the number of moles of OH-. Therefore the number of moles of H+ = 0.0125 moles.

Step 4 - Determine concentration of HCl

Every mole of HCl will produce one mole of H+, therefore number of moles of HCl = number of moles of H+.

Molarity = # of moles/volume

Molarity of HCl = (0.0125 mol)/(0.050 L)
Molarity of HCl = 0.25 M

Answer

The concentration of the HCl is 0.25 M.

Another Solution Method

The above steps can be reduced to one equation

MacidVacid = MbaseVbase

where

Macid = concentration of the acid
Vacid = volume of the acid
Mbase = concentration of the base
Vbase = volume of the base

This equation works for acid/base reactions where the mole ratio between acid and base is 1:1. If the ratio were different as in Ca(OH)2 and HCl, the ratio would be 1 mole acid to 2 moles base. The equation would now be

MacidVacid = 2MbaseVbase

For the example problem, the ratio is 1:1

MacidVacid = MbaseVbase

Macid(50 ml)= (0.5 M)(25 ml)
Macid = 12.5 MmL/50 ml
Macid = 0.25 M

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