Chemguide: Core Chemistry 14 - 16

Acid-alkali titration calculations

This page looks at calculations resulting from acid-alkali titrations.

Don't even think about reading this page unless you are happy with all the other calculations in this section. In particular, it follows on from the page about calculations involving solutions.

The page doesn't cover the practical details of doing titrations. You will find these in detail in the second part of the page about making soluble salts. Read the part headed "Making sodium, potassium and ammonium salts" before you carry on with this page.

Example calculations

A titration is used to find an unknown concentration of a solution by reacting it with a solution of known concentration. The examples will make this clear.

Example 1

25.0 cm3 of a solution of sodium hydroxide, NaOH, of concentration 0.100 mol dm-3 was pipetted into a flask and a few drops of methyl orange indicator was added.

Hydrochloric acid of unknown concentration was run in from a burette until the indicator turned from yellow to orange, showing the solution to be neutral. 20.0 cm3 of acid was needed. Calculate the concentration of the acid in mol dm-3.

NaOH + HCl     NaCl + H2O

important:  In this, and the rest of these calculations, I am expressing small numbers in scientific notation (for example, 2.50 x 10-3) because that is almost certainly how it will come off your calculator.

Your calculator will show this as 2.5-03. That is NOT 2.5 to the power of -03. It means 2.5 x 10-3.

Start with what you know most about. You know the volume and concentration of the sodium hydroxide, so start there and calculate the number of moles.

You know that 1000 cm3 (1 dm3) contains 0.100 moles. So 25.0 cm3 contains a lot less: 25.0/1000 of that.

No of moles of NaOH = 25.0/1000 x 0.100 = 2.50 x 10-3

The equation says that 1 mole of NaOH reacts with 1 mole of HCl to make a neutral solution.

Therefore number of moles of HCl = 2.50 x 10-3

This is contained in 20.0 cm-3 of HCl.

Number of moles in 1000 cm-3 (1 dm3) = 1000/20.0 x 2.50 x 10-3 = 0.125

Concentration of HCl = 0.125 mol dm-3

This is all pretty obvious, apart possibly from the last step. If this worries you, insert another step involving 1 cm3.

Number of moles in 1 cm-3 = 2.50 x 10-3/20.0 = 1.25 x 10-4

Number of moles in 1000 cm-3 (1 dm3) = 1000 x 1.25 x 10-4 = 0.125

Concentration of HCl = 0.125 mol dm-3

Note:  Whichever way you do this, you can just leave the intermediate answers on your calculator ready for the next step. If you write down the answers step at a time and they need rounding, you could round them to 1 more significant figures than your answer needs - but still go on using the full number on your calculator.

Example 2

This example adds another step because you are going to have to calculate the concentration of one of the substances in g dm-3 instead of mol dm-3. You also have to be careful because the equation proportions are no longer 1 to 1.

25.0 cm3 of sodium carbonate, Na2CO3, solution was found to need 22.0 cm3 of hydrochloric acid of concentration 0.200 mol dm-3 for neutralisation using methyl orange as indicator.

Na2CO3 + 2HCl     2NaCl + H2O + CO2

Calculate the concentration of the sodium carbonate in
a) mol dm-3; b) g dm-3.

(RAMs: H = 1; C = 12; O = 16; Na = 23)

Part a)

Na2CO3 + 2HCl     2NaCl + H2O + CO2

This time you know everything about the hydrochloric acid, so start there.

No of moles of HCl used = 22.0/1000 x 0.200 = 4.40 x 10-3

The equation shows that you need half as much Na2CO3 as HCl.

No of moles Na2CO3 = 1/2 x 4.40 x 10-3 = 2.20 x 10-3

That's in 25.0 cm3.

Number of moles in 1000 cm-3 = 1000/25.0 x 2.20 x 10-3 = 0.0880

(Or add another step via 1 cm3.)

Concentration of Na2CO3 = 0.0880 mol dm-3

Part b)

1 mol Na2CO3 weighs (2 x 23) + 12 + (3 x 16)
= 106 g

Therefore 0.880 mol weighs 0.088 x 106 g = 9.33 g

Concentration of Na2CO3 = 9.33 g dm-3

Example 3

This time we are going to start with a concentration in g dm-3 and end up with a concentration in g dm-3.

25.0 cm3 of sodium hydrogencarbonate, NaHCO3, solution containing 16.8 g dm-3 was found to need 20.0 cm3 of dilute sulfuric acid for neutralisation using methyl orange as indicator.

Calculate the concentration of the sulfuric acid in
a) mol dm-3; b) g dm-3.

(RAMs: H = 1; C = 12; O = 16; Na = 23; S = 32)

2NaHCO3 + H2SO4     Na2SO4 + 2H2O + 2CO2

Part a)

This time you know everything about the sodium hydrogencarbonate, so start there.

1 mol NaHCO3 weighs 23 + 1 + 12 + (3 x 16) = 84 g

16.8 g NaHCO3 is 16.8/84 = 0.200 mol

So the concentration of NaHCO3 is 0.200 mol dm-3.

No of moles of NaHCO3 in 25.0 cm3 is
25.0/1000 x 0.200 = 5.00 x 10-3 mol

The equation shows that you need half as much H2SO4 as NaHCO3.

No of moles H2SO4 = 1/2 x 5.00 x 10-3 = 2.50 x 10-3

That's in 20.0 cm3.

Concentration = 1000/20.0 x 2.50 x 10-3 = 0.125 mol dm-3

(Or add another step via 1 cm3.)

Part b)

1 mol H2SO4 weighs (2 x 1) + 32 + (4 x 16) = 98 g

Therefore 0.125 mol weighs 0.125 x 98 g = 12.25 g

Concentration of H2SO4 = 12.25 g dm-3

(Technically, this should be rounded to 12.3 to 3 significant figures.)

Example 4

This isn't a full example - just a slight modification of the last one.

If you were doing this in the lab, you wouldn't make up a whole litre (dm3) of sodium hydrogencarbonate solution. In most titrations you will do a rough titration and then at least 2 accurate ones which agree to within 0.1 cm3. You won't need anything like a litre of solution.

So typically you would make up 250 cm3 of solution.

Since you are only making up 1/4 of a litre, obviously you would only need to use 1/4 as much sodium hydrogencarbonate to make up a solution of the same strength.

If you wanted a solution of concentration 16.8 g dm-3, you would dissolve 16.8/4 = 4.20 g of solid to give 250 cm3 of solution.

So, if you were told that you had a solution of sodium hydrogencarbonate containing 4.20 g in 250 cm3 of solution, you would have to remember to multiply that by 4 as the first thing you did to give a concentration in mol dm-3.

Example 5

There is anther variant on titration calculations which is less common, and also a bit more fiddly to think about.

If you had 50 cm3 of dilute hydrochloric acid of concentration 0.10 mol dm-3, what volume of 0.20 mol dm-3 NaOH would you need to neutralise it?

NaOH + HCl     NaCl + H2O

As always, start with what you know most about - the hydrochloric acid.

No of moles of HCl = 50/1000 x 0.10 = 5.0 x 10-3

The reaction is 1 : 1 and so you also need 5.0 x 10-3 moles of NaOH.

1000 cm3 of NaOH solution contains 0.20 moles - so what volume will contain 5.0 x 10-3 moles?

You can sort this out using a formula if you have learnt one, but on the whole, I try to avoid students having to learn formulae to do calculations. You can end up learning a whole lot of formulae and understanding nothing! So let's use some common sense.

Start by working out what volume contains 1 mole, and then multiply that by 5.0 x 10-3.

1 mole NaOH is contained in 1000/0.20 = 5000 cm3.

5.0 x 10-3 moles NaOH is contained in
5.0 x 10-3 x 5000 cm3 = 25 cm3.

People often hesitate at the first step of this, because it seems odd to be dividing by 0.20. In a case like this mentally replace the unfriendly number by a friendly one and decide what you would do with that.

Suppose 1000 cm3 contained 2 moles. What volume would contain 1 mole? Obviously you would just divide by 2. There is no reason why you should do anything different with a more difficult number.

Note:  In this example, it is obvious that if the reaction is 1 : 1, and the sodium hydroxide is twice as concentrated, it will only need half as much as the volume of HCl. So why go to all this bother?

I have deliberately chosen the numbers so that they are easy to work with. That won't always be the case, and the reaction you are looking at won't necessarily be 1 : 1. You need to know how to tackle this whatever reaction or numbers you are given.


It is really important tha you practise doing the calculations in the form that your examiners will use. Get hold of as many past papers and mark schemes as you can, and work through all the examples you can find.

You will probably find that the way they ask questions is fairly repetitive, and you need to get used to the way they do it.

Where would you like to go now?

To the calculations menu . . .

To the Chemistry 14-16 menu . . .

To Chemguide Main Menu . . .

© Jim Clark 2021