Titration calculations - higher tier
The results of a titration can be used to calculate the concentration of a solution, or the volume of solution needed.
Calculating a concentration
Worked example
In a titration, 25.0 cm3 of 0.1 mol/dm3 sodium hydroxide solution are exactly neutralised by 20.0 cm3 of a dilute solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution in mol/dm3 and g/dm3.
Step 1: Calculate the number of moles of sodium hydroxide in moles
Volume of sodium hydroxide solution = 25.0 Ă· 1000 = 0.025 dm3
Rearrange:
Concentration in mol/dm3 = \(\frac{moles~of~solute}{volume~in~dm^3}\)
Number of moles of solute in mol = concentration in mol/dm3 Ă volume in dm3
Number of moles of sodium hydroxide = 0.1 Ă 0.025 = 0.0025 mol
If you prefer to use volumes in cm3, you can use the equation:
moles of solute = \(\frac{volume (cm^3 ) Ă concentration}{1000}\)
so this calculation becomes:
moles of solute = \(\frac{volume (cm^3 ) Ă concentration}{1000}\) =
\(\frac {25.0 x 0.1}{1000}\)
= 0.0025 mol
This equation can be easier to use as you will mostly be working with volumes in cm3 during a titration.
Step 2: Find the number of moles of hydrochloric acid in moles
The balanced equation is: NaOH(aq) + HCl(aq) â NaCl(aq) + H2O(l)
So the mole ratio NaOH:HCl is 1:1
Therefore 0.0025 mol of NaOH reacts with 0.0025 mol of HCl
Step 3: Calculate the concentration of hydrochloric acid in mol/dm3
Volume of hydrochloric acid = 20.0 Ă· 1000 = 0.02 dm3
Concentration in mol/dm3 = \(\frac{moles~of~solute}{volume~in~dm^3}\)
Concentration in mol/dm3 = \(\frac{0.0025}{0.02}\)
= 0.125 mol/dm3
Again this can be carried out using the equation:
moles of solute = \(\frac{volume (cm^3 ) Ă concentration}{1000}\)
Rearranging this gives:
concentration = \(\frac{moles Ă 1000}{volume}\)
concentration = \(\frac{0.0025 Ă 1000}{20.0}\) = = 0.125 mol/dm3
Step 4:: Calculate the concentration of hydrochloric acid in g/dm3
Relative formula mass (Mr) of HCl = 1 + 35.5 = 36.5
Concentration in g/dm3 = concentration in mol/dm3 Ă Mr
Concentration in g/dm3 = 36.5 Ă 0.125 = 4.56 g/dm3
⎠concentration = 4.56 g/dm3
Worked example
Q: In a titration, 25.0 cm3 of 0.2 mol/dm3 sodium hydroxide solution is exactly neutralised by 22.7 cm3 of a dilute solution of hydrochloric acid.
NaOH(aq) + HCl(aq) â NaCl(aq) + H2O(l)
Calculate the concentration of the hydrochloric acid in mol/dm3.
A: Volume of sodium hydroxide solution = 25.0 Ă· 1000 = 0.025 dm3
Moles of sodium hydroxide = 0.2 Ă 0.0250 = 0.005 mol
From the equation, 0.005 mol of NaOH reacts with 0.005 mol of HCl
Volume of hydrochloric acid = 22.7 Ă· 1000 = 0.0227 dm3
Concentration of hydrochloric acid = 0.005 mol Ă· 0.0227
= 0.220 mol/dm3
Alternatively using the volumes in cm3:
moles of sodium hydroxide\(\frac{volume (cm^3 ) Ă concentration}{1000} = \frac{25.0 Ă 0.2}{1000}\)= 0.005 mol
moles of hydrochloric acid = 0.005 mol (as 1:1 ratio of NaOH:HCl)
concentration of hydrochloric acid\(\frac{moles Ă 1000}{volume} = \frac{0.005 Ă 1000}{22.7}\) =0.220 mol/dm3