Titration calculations  

In a nutshell

The results of a titration can be used to calculate the concentrations of unknown solutions. This can also be done for diprotic acids. 

Equations

Below are general equations you need to know. You will need to be able to use them for specific calculations.

Variable definitions

Titration results to calculate concentrations

Titration results can be used to calculate the concentration of an unknown acid or base.

Make sure to carry out the titration to the nearest $$0.05cm^3$$ and repeat until concordant results are achieved. 

Example

$$12.50 \ cm^3$$ of $$1.20\ mol\ dm^{-3}\ HCl$$  was required to neutralise $$20 \ cm^3$$ of $$KOH$$ solution. Calculate the concentration of the $$KOH$$ solution.

$$HCl\ +\ KOH\ \rightarrow\ KCl\ +\ H_2O$$

Convert $$cm^3$$ to $$dm^3$$.​

$$\frac{12.50\ cm^3}{1000}\ =\ 0.0125\ dm^3$$ 

$$\frac{20\ cm^3}{1000}\ =\ 0.020\ dm^3$$​​

Calculate the number of moles of $$HCl$$:

$$n \ (HCl)\ =\ c\ \times\ v \\ n \ (HCl) = 1.20\ \times\ 0.0125\ =\ 0.015\ mol$$​​​

From the equation, $$1$$​ mol of $$HCl$$ is neutralised by $$1$$​ mol of $$KOH$$. Therefore the moles of $$HCl$$ and $$KOH$$ are equal. Now the concentration can be calculated.

​$$c\ (KOH)\ =\ \frac{n \ (KOH)}{v}\ =\ \frac{0.015}{0.020}\ =\ \underline{0.75\ mol\ dm^{-3}}$$

Diprotic acids release two protons

A diprotic acid is an acid that releases two protons in solution. Diprotic acids neutralise a base in two stages.

Example

Some examples of diprotic acids include; sulfuric acid $$(\underline{H_2}SO_4)$$​, carbonic acid $$(\underline{H_2}CO_3)$$​, oxalic acid $$(\underline{H_2}C_2O_4)$$​.

$$H_2SO_4\ \rightarrow\ HSO_4^-\ +\ H^+\ \rightleftharpoons\ SO_4^{2-}\ +\ H^+$$​​

Calculating concentrations of diprotic acids

Calculating the concentration of a diprotic acid is almost the same as calculating the concentration for a monoprotic acid. 

Example

$$20 \ cm^3$$ of $$H_2SO_4$$ is neutralised by $$15.50 \ cm^3$$ of $$1.6\ mol\ dm^{-3}\ NaOH$$. Determine the concentration of the sulfuric acid. 

Firstly, the volumes must be converted from $$cm^3$$​ to $$dm^3$$: 

$$\frac{20\ cm^3}{1000}\ =\ 0.020\ dm^3$$ 

$$\frac{15.50\ cm^3}{1000}\ =\ 0.0155\ dm^3$$ 

​

Calculate the moles of $$NaOH$$:

$$n \ (NaOH)\ =\ c\ \times\ v \\ n \ (NaOH) = 1.6 \times\ 0.0155\ =\ 0.0248\ mol$$​​

Calculate the moles of $$H_2SO_4$$ from the moles of $$NaOH$$:

$$H_2SO_4\ +\ 2NaOH\ \rightarrow\ Na_2SO_4\ +\ 2H_2O$$​​

$$H_2SO_4:NaOH \\1:2$$​​

You can see from the ratio of sulphuric acid to sodium hydroxide, the moles of sulphuric acid is half of that of sodium hydroxide.

$$\frac{0.0248\ mol}{2}\ =\ 0.0124\ mol$$​​

Since the moles of sulphuric acid is known, the concentration can be calculated.

$$c \ (H_2SO4)\ =\ \frac{n\ (H_2SO_4)}{v}\ =\ \frac{0.0124}{0.020}\ =\ \underline{0.62\ mol\ dm^{-3}}$$ 

​

The concentration of sulphuric acid is $$\underline{0.62\ mol\ dm^{-3}}$$.​