# AQA C3 2.3 Titration Calculations

A powerpoint to accompany my lesson on Titration Calculations.

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

## Other slides in this set

### Slide 2

Extracted text:

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

Extracted text:

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

Extracted text:

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

Extracted text:

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

Extracted text:

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

### Slide 10

Wed 8th May, 2013 @ 20:46

Thank you it was quite useful :)

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

Sat 18th May, 2013 @ 16:16