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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 梅 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 脳 volume in dm3

Number of moles of sodium hydroxide = 0.1 脳 0.025 = 0.0025 mol

If you prefer to use volumes in cm3, you can use the equation:

moles of solute = \(\frac{volume (cm^3 ) 脳 concentration}{1000}\)

so this calculation becomes:

moles of solute = \(\frac{volume (cm^3 ) 脳 concentration}{1000}\) =

\(\frac {25.0 x 0.1}{1000}\)

= 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) 鈫 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 梅 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 = \(\frac{volume (cm^3 ) 脳 concentration}{1000}\)

Rearranging this gives:

concentration = \(\frac{moles 脳 1000}{volume}\)

concentration = \(\frac{0.0025 脳 1000}{20.0}\) = = 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 脳 Mr

Concentration in g/dm3 = 36.5 脳 0.125 = 4.56 g/dm3

鈭 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) 鈫 NaCl(aq) + H2O(l)

Calculate the concentration of the hydrochloric acid in mol/dm3.

A: Volume of sodium hydroxide solution = 25.0 梅 1000 = 0.025 dm3

Moles of sodium hydroxide = 0.2 脳 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 梅 1000 = 0.0227 dm3

Concentration of hydrochloric acid = 0.005 mol 梅 0.0227

= 0.220 mol/dm3

Alternatively using the volumes in cm3:

moles of sodium hydroxide\(\frac{volume (cm^3 ) 脳 concentration}{1000} = \frac{25.0 脳 0.2}{1000}\)= 0.005 mol

moles of hydrochloric acid = 0.005 mol (as 1:1 ratio of NaOH:HCl)

concentration of hydrochloric acid\(\frac{moles 脳 1000}{volume} = \frac{0.005 脳 1000}{22.7}\) =0.220 mol/dm3