Titration calculations

In a nutshell

The purpose of a titration is to determine the concentration of an unknown sample. In this summary you will learn how to carry out calculations from titration data.

Equations

​$$concentration\ =\ \frac{amount\space of\space solute}{volume}\\ \ \\ \\ \ \\ amount\space of \space solute\ =\ concentration\ \times\ volume\\ \ \\ \\ \ \\ mass\ =\ molecular\space mass\ \times\ amount\space of \space solute\\ \ \\ \\ \ \\concentration \ (g/dm^3)\ =\ molecular\space mass\ \times\ concentration\ (mol/dm^3)$$​​

Variable units

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Titrations 

The results of a titration can be used to work out the concentration of a solution with a known volume. 

Procedure

Calculating a concentration   

You can calculate the concentration of a solution using the concentration of a known solution and a balanced equation. 

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Example  

In a titration $$30\ cm^3$$ of $$0.2\ mol/dm^3$$ $$NaOH$$ is exactly neutralised by $$25 \, cm^3$$ of $$HCl$$. Calculate the concentration of the hydrochloric acid solution in $$g/dm^3$$​.

Calculate the volume of $$NaOH$$ solution by converting $$cm^3$$ to $$dm^3$$​:

​$$\frac{volume\ in\ cm^3}{1000}\ =\ volume\ (dm^3)$$

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$$\frac{30 \, cm^3}{1000}\ =\ 0.03 \, dm^3$$

Calculate the moles of $$NaOH$$ in solution:

​$$amount \space of \space solute \ =\ concentration\ \times\ volume$$

$$0.2\ mol/dm^3\ \times\ 0.03\ dm^3\ =\ \underline{0.006\ mol}$$ 

Look at the ratio of acid to base required to neutralise the solution. Use this to calculate the amount of $$HCl$$ in moles.

$$NaOH\ +\ HCl\ \rightarrow \ NaCl\ +\ H_2O$$

The mole ratio is $$1:1$$​. The moles of $$NaOH$$​ is equal to the moles of $$HCl$$, which is $$\underline{0.006\ mol}$$.​

Calculate the volume of $$HCl$$ used in $$dm^3$$​:

$$\frac{volume\ in\ cm^3}{1000}\ =\ volume\ (dm^3)$$

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$$\frac{25 \, cm^3}{1000}\ =\ 0.025 \, dm^3$$

Calculate the concentration in $$(mol/dm^3)$$​ of $$HCl$$:

$$concentration\ =\ \frac{amount\space of\space solute}{volume}$$

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$$\frac{0.006\ mol}{0.025\ dm^3}\ =\ \underline{0.24\ mol/dm^3}$$ 

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Calculate the molecular mass of $$HCl$$​:

$$M_r(HCl) = 1\ (A_r \space of \space hydrogen)\ +\ 35.5\ (A_r \space of \space chlorine) \ =\ 36.5$$ 

Calculate the concentration in $$(g/dm^3)$$ of $$HCl$$ in solution: 

$$concentration\ (g/dm^3)\ =\ relative\ formula\ mass\ \times\ concentration\ (mol/dm^3)$$ 

HCl\ =\ 36.5\ \times\ 0.24\ mol\ = 8.76\ g$$36.5\ (M_r)\ \times\ 0.24\ mol/dm^3\ = 8.76\ g/dm^3$$ 

Therefore, the concentration of $$HCl$$ is $$\underline{8.76\ g/dm^3}$$.