10.03.08 Titrations

Syllabus

- The volumes of acid and alkali solutions that react with each other can be measured by titration using a suitable indicator.

Students should be able to

1. describe how to carry out titrations using strong acids and strong alkalis only (Sulphuric, Hydrochloric and Nitric acids only) to find the reacting volumes accurately
2. (HT Only) calculate quantities in titrations involving concentrations in mol/dm³ and in g/dm³ .

Required practical 2: determination of the reacting volumes of solutions of a strong acid and a strong alkali by titration.

(HT only) determination of the concentration of one of the solutions in mol/dm³ and g/dm³ from the reacting volumes and the known concentration of the other solution.

What does this mean?

How a titration is carried out.

Titrations are usually carried out to find out the unknown concentration of a base using an acid of known concentration.

Or, to find out the unknown concentration of a acid using an base of known concentration.

Either way, the volume of the base is carefully measured using a pipette, then placed in a conical flask.

Volumetric pipettes usually have scored mark near the top, filling them to this level is more accurate than using a measuring cylinder.

Now that we know exactly the volume of base we're using we need to know what volume of acid is needed to neutralise it.

So the acid is placed in a burette.

Burette scales are upside down.

This means that, if we fill the burette to 0.00 cm³ and then neutralise the base we can simply read the volume of acid used (reading from the bottom of the meniscus).

But we'll never know if the base has been neutralised unless we use an indicator.

Students usually assume Universal Indicator is best because it has so many colours, but this is why we don't use it.

Instead, we use an indicator with only two colours like Litmus or Phenolphthalein.

The reason being that a gradual colour change is hard to observe accurately but a change from Pink to Colourless is hard to miss.

And it doesn't matter if the indicator doesn't change exactly at pH=7 because the pH can change from around 12 to around 3 in one drop!

Even then we won't know if we've done a good job until we repeat it.

This results table shows the results for three repeat readings.

Trial 1 and Trial 3 are quite close.

We should assume that we did something wrong with trial 2 and ignore it.

So, when asked what volume is needed we take an average of our reliable results.

This is called the Mean Titre -- in this case (4.7 + 4.5)/2 = 4.6 cm³.

Calculating an unknown concentration.

Example 1

A student accurately measured 25cm³ of NaOH and places in a conical flask.

This is then neutralised by titration with 0.5 mol/dm³ Nitric Acid several times.

NaOH + HNO₃ --> NaNO₃ + H₂O

Use the results below to calculate the concentration of the NaOH

We should ignore the Trial Run - they're always wrong.

We should discount Run 2 because it's very different to the other two accurate runs.

Usually, we want two results with 0.20 cm³

So we'll calculate our Mean Titre from Run 1 and Run 3

Mean Titre = (23.00 + 23.10) /2 = 23.05 cm³

Moles of Acid used = Concentration x (volume/1000)

= 0.5 x (23.05/1000) = 0.01175 mol

Ratio NaOH : HNO₃ 1:1

So, moles of NaOH neutralised is also 0.01175 mol

Concentration (NaOH) = Moles of NaOH/ (Volume/1000)

= 0.01175/(25/1000) = 0.47 mol/dm³

Example 2

A student accurately measured 20cm³ of NaOH and places in a conical flask.

This is then neutralised by titration with 0.4 mol/dm³ Sulphuric Acid several times.

2 NaOH + H₂SO₄ --> Na₂SO₄ + 2 H₂O

Use the results below to calculate the concentration of the NaOH

Again, ignore the Trial Run.

We should discount Run 1 this time.

So we'll calculate our Mean Titre from Run 2 and Run 3

Mean Titre = (13.40 + 13.30) /2 = 13.35 cm³

Moles of Acid used = Concentration x (volume/1000)

= 0.4 x (13.35/1000) = 0.00534 mol

Ratio NaOH : H₂SO4 2:1

So, moles of NaOH neutralised is also 2 x 0.00534 = 0.01068 mol

Concentration (NaOH) = Moles of NaOH/ (Volume/1000)

= 0.01068 /(20/1000) = 0.534 mol/dm³

Past Paper Questions

2019

2021

This question was about a titration between citric acid and an alkali

2020

2020 #2

2018

2017

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