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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