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Concentration of Solutions
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Concentration of Solutions
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multiply by 1000
divide by 1000
Measuring concentrations
It is not enough to say that one concentration is higher
or lower than another. Concentrations usually need to be
measured accurately.
There are two ways of measuring concentration:
mass per unit volume, e.g. grams per decimetre
cubed (g/dm3)
moles per unit volume, e.g. moles per decimetre
cubed (mol/dm3).
A cubic decimetre (dm3) is a unit of volume. 1dm3 is
equivalent to 1000cm3.
1cm3 1dm3
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Volume unit conversions
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volume of solution (dm3)
Concentrations in g/dm3
The following equation gives concentration in g/dm3:
If 1.0g of solid sodium hydroxide is dissolved in 250cm3
of water, what is the concentration in g/dm3?
= 4g/dm3
concentration =mass dissolved (g)
concentration =250cm3
1g
0.25dm3
1 g
0.25
1
dm3
g= =
convert the units
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Calculating concentrations in g/dm3
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Concentrations in mol/dm3
To calculate concentration in mol/dm3:
This equation can be added to a formula
triangle to rearrange the formula:
volume of solution (dm3)concentration =
number of moles (mol)
c =nv
v =nc
n = c v×
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c =nv
Concentrations in mol/dm3 – example
The information in the question provides
the volume but not the number of moles.
The following formula is required:
molar massnumber of moles =
mass
40
1= = 0.025mol
c =0.025 mol
0.25 dm3
Now substitute 0.025mol into the original formula:
= 0.1mol/dm3
If 1.0g of solid sodium hydroxide (NaOH) is dissolved in
250 cm3 of water, what is the concentration in mol/dm3?
?
0.25dm3
=
Remember, molar mass = relative formula mass (RFM)
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Concentrations in mol/dm3: practise
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Rearranging formulae – example 1
What volume of 0.80mol/dm3 potassium bromide solution
contains 1.6moles of potassium bromide (KBr)?
v = n / c
This calculation is simply a matter of substituting the values
into the rearranged formula:
v = n / c = 1.6 / 0.8 = 2dm3
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Rearranging formulae – example 2
How many moles of copper sulfate are there in 250cm3 of
0.2mol/dm3 copper sulfate solution (CuSO4)?
n = c × v
Step 1: convert the units
250cm3 = 0.25dm3
Step 2: substitute into the formula
n = c x v = 0.2 × 0.25
= 0.05mol
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Calculating mass of solute
What mass of copper sulfate (CuSO4) was used to make
250cm3 of 0.2mol/dm3 copper sulfate solution?
This is simply an extension of the previous calculation, from
which the moles of copper sulfate was found to be 0.05mol.
Step 1: Calculate the number of moles of copper sulfate
Step 2: Rearrange the moles formula to give the mass value
= 0.05 × 159.6
= 8.0g
mass of copper sulfate = number of moles × molar mass
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Concentration of Solutions
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Titrations
A titration is a technique used to accurately determine the
concentration of a substance in solution.
During a titration, a solution of known
concentration, called a standard
solution, is added to a solution of
unknown concentration.
The purpose of a titration is to determine
the volume of solution required to reach an
endpoint. An endpoint is an observable
physical change, such as a colour change.
Why is measuring the volume useful?
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Finding an unknown concentration
Volume can be used to determine the concentration by
using the following relationship:
volume of solution (dm3)concentration =
number of moles (mol)
An unknown concentration value can be
found if the following values are known:
These values can be found through
a titration.
number of moles in solution
volume of solution.
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Titration equipment
The following laboratory equipment is required to perform
a titration:
stand
beaker
burette
pipette
conical
flask
safety
filler
Use to measure a set volume of
alkali that is then added to the
conical flask.
Indicator then added – either
Phenolphthalein or Methyl Orange
Filled with the standard solution
– likely to be acid with a known
concentration
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Titration process
Stages:
1. Small amount of acid added (a drop at a time) from the burette.
2. Swirl conical flask
3. Stop adding acid when a colour change is seen:
1. Phenolphthalein – Pink (alkali) Colourless (acid)
2. Methyl Orange – Yellow (alkali) Red (acid)
4. The colour change signifies the end point
5. Record the added volume of acid (standard solution) to
neutralise.
6. Repeat a minimum of three times so that a mean can be
calculated.
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Indicators
The endpoint of a titration is often marked by a colour change.
This is provided by an indicator solution.
Indicators are substances which
change colour according to the
pH of a solution.
A small amount of indicator
solution is added to one of the
solutions during a titration.
When the indicator changes
colour, the endpoint of the
titration has been reached.
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Performing a titration
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Titration apparatus
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Concentration of Solutions
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Using titration results
How are the results of a titration used to calculate
the unknown molar concentration of a solution?
Step 1: Write a balanced equation for the reaction.
Step 2: Calculate the number of moles of standard solution
by rearranging the concentration formula:
Step 3: Use the balanced equation to determine the number
of moles for the solution under investigation.
Step 4: Use the concentration formula to determine the
unknown molar concentration:
moles = concentration × volume
molar concentration = moles ÷ volume
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Titration calculations: worked example (1)
1. Write a balanced equation for the reaction:
2. Calculate the number of moles of sodium hydroxide:
The endpoint of a titration was reached when 20cm3 of
0.1mol/dm3 sodium hydroxide was added to 25cm3 of
hydrochloric acid. What is the concentration of the acid?
moles = concentration (mol/dm3) × volume (dm3)
= 0.1 × (20.0 ÷ 1000)
= 0.002mol NaOH
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volume (dm3)
Titration calculations: worked example (2)
3. The balanced equation from step 1 shows that one mole
of HCl reacts with one mole of NaOH:
4. Use the number of moles to calculate the molar
concentration of the hydrochloric acid:
= 0.08mol/dm3
moles 0.002
(25 ÷ 1000)=
Therefore 0.002moles of NaOH will react with:
0.002moles of HCl.
molar
concentration=
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Titration calculations: questions
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Titration calculations
Calculating volume
25 cm3 of 0.2 mol/dm3 sodium hydroxide solution was
neutralised by 0.15 mol/dm3 sulphuric acid. Calculate the
volume of sulphuric acid needed for the neutralisation.
2NaoH + H2SO4 Na2SO4 + 2H2O
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Concentration of Solutions
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Glossary
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Multiple-choice quiz