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Chemical analysisTitration calculations

Chemists monitor our environment using a variety of quantitative and qualitative analysis techniques. The results from quantitative analysis are used in calculations that give essential information.

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Titration calculations

In a titration, sulfuric acid of concentration \(0.1 moll^{-1}\) was added from a burette into a conical flask containing \(20cm^{3}\) of sodium hydroxide solution. The average titre was \(11.2cm^{3}\). Calculate the concentration of the sodium hydroxide solution.

Step 1 - Work out the number of moles of sulfuric acid

We know the concentration of the acid was \(0.1 moll^{-1}\) and the volume added was \(11.2cm^{3}\). It is therefore possible to work out the number of moles using the formula below:

An equilateral triangle pointing upwards and divided into three parts. The top part represents the number of moles. The bottom left part represents the concentration in moles per litre. The bottom right part represents the volume in litres.

(Remember to change the volume into litres by dividing by 1000.)

\(number of moles of acid = concentration \times volume\)

\(= 0.1 \times 0.0112\)

\(= 0.00112\)

Step 2 - Find the mole ratio

\(H_{2}SO_{4} +2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O\)

From looking at the large balancing numbers in front of the compounds it can be seen that 1 mole of acid reacts with 2 moles of sodium hydroxide. The mole ratio is therefore:

\(H_{2}SO_{4} : NaOH\)

\(1 : 2\)

This means that the number of moles of acid must be multiplied by 2 to find the number of moles of NaOH that were in the conical flask. Since the number of moles of acid added was calculated to be 0.00112 the number of moles of NaOH is \(2 \times 0.00112 = 0.00224\)

Step 3 - work out the concentration

We know that the volume in the conical flask was \(20cm^{3}\) and we have just worked out that there were 0.00246 moles. Using the equation in the above triangle:

\(concentration = \frac {number\: of\: moles}{volume\: in\: litres} = \frac {0.00224}{0.02} = 0.112\: moll^{-1}\)

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