Acids and Bases - Titration Example Problem

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

M_acidV_acid = M_baseV_base

where

M_acid = concentration of the acid
V_acid = volume of the acid
M_base = concentration of the base
V_base = 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

M_acidV_acid = 2M_baseV_base

For the example problem, the ratio is 1:1

M_acidV_acid = M_baseV_base

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