# AQA C3 2.3 Titration Calculations

- Subject: Chemistry
- Level: GCSE
- Exam board: AQA
- Author: Mrs Boyson
- Year created: 2011

A powerpoint to accompany my lesson on Titration Calculations.

Shows the worked examples and summary questions given in the student book.

## Related resources:

## Other slides in this set

### Slide 2

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A Disclaimer...

This part of C3 appears to be difficult.

It isn't, it's actually VERY straightforward.

However, you will need to concentrate,

and attempt all the exercises.

Make sure your notes are clear.

You WILL need a calculator

### Slide 3

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Introduction & a definition

Titrations are used to work out the

concentration of an unknown solution.

You can also work out how much of a

solution will be needed for neutralisation

of a known concentration.

Concentration: the number of moles of

a solute in 1dm3 solution.

= mol/dm3

(1 dm3 = 1000cm3)

### Slide 4

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Example 1 (follow carefully...)

Calculating concentration of a solution from a known

mass of solute.

If 40g NaOH is used to make 500cm3 of solution,

what is the concentration in mol/dm3?

40 x1000 = 80g of NaOH in 1000cm3

500

1 mole of NaOH = 40g (RAM in g)

So, 80g = 2 moles

So, the concentration of the NaOH solution is 2

mol/dm3

### Slide 5

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Example 2 (still following?)

Calculating the mass of a substance in a known

concentration.

What is the mass of H2SO4 in 25ocm3 of 1mol/dm3?

1 mole H2SO4 = (1x2) + 32 + (16x4) = 98g

1mol/dm3 concentration = 98g H2SO4 in 1000cm3 of

solution.

98x 250 = 24.5g H2SO4 in 250cm3 of 1mol/dm3

1000 solution

### Slide 6

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Titration Calculations

In a titration, we have 2 solutions

One (in burette) is a known concentration

The other (in the conical flask, measured to a

known volume using the pipette & filler) is a

known substance, but unknown

concentration.

The result of the titration can help us

calculate the number of moles of a

substance in the solution in the conical

flask.

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## Comments

Wed 15th May, 2013 @ 17:49

This was really useful! thank you as I actually understand Titration Calculations now! :)

Sat 18th May, 2013 @ 13:26

THANK YOU SO MUCH i've been trying to understand this for weeks aahha

Wed 8th May, 2013 @ 20:46