a pH value tells us how much H is in a solution It can be defined by pH logH log is to base 10 Also note square brackets are used to denote concentration values For example pH log10 ID: 550913
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Slide1
pH and indicators; the pH scale
a pH value tells us how much H
+
is in a solution. It can be defined by: pH = - log[H
+
] (log is to base 10)
.
(Also note, square brackets are used to denote concentration values); For example: pH = -log[10
-2
] = 2
The pH scale gives a measure of how acidic or alkaline a solution is. It normally runs from 0 (most acidic) to 14 (most alkaline). A neutral substance is pH 7 at 25°C.
Indicators
An indicator changes
colour
depending on the properties of the substance it is added to. Three indicators are commonly used to show whether a solution is acidic or alkaline: 1) Litmus; 2) Phenolphthalein; 3) universal indicator;
Litmus and phenolphthalein are single indicators (they only contain one
colour
-changing substance) whereas universal indicator is a mixed indicator (it contains several different
colour
-changing substances).Slide2
Litmus
Litmus paper can be red or blue.
The table shows its
colours in acidic, neutral and alkaline solutions.Slide3
Phenolphthalein and universal indicator
Phenolphthalein changes
colour
sharply at about pH 8. For most purposes, this means that it is pink in alkaline solutions and
colourless
in acidic solutions.
Universal indicator shows a range of colours depending on the pH of the solution (see below).Slide4Slide5
pH curves: Adding acid to alkali
The pH curve below shows what happens to the pH when a strong acid (such as hydrochloric acid) is added to 25 cm
3
of a strong alkali (such as sodium hydroxide).
The acid and the alkali started off at the same concentration.
Note that the pH falls:
slowly at first as acid is added to the alkali
rapidly at the end-point (the point where the alkali is completely
neutralised
)
slowly again once excess acid is being added
In this example, 25 cm
3
of acid was needed to
neutralise
the alkali. Slide6
Adding alkali to acid
The pH curve below shows what happens to the pH when a strong alkali is added to 25 cm
3
of a strong acid.
As before, they both started off at the same concentration.
Note that the pH rises:
slowly at first as alkali is added to the acid
rapidly at the end-point (the point where the acid is completely
neutralised
)
slowly again once excess alkali is being addedSlide7
Carrying out a titration
The concentration of an acid or alkali can be calculated by carrying out an experiment called a titration.
The apparatus needed includes a:
pipette to accurately measure a certain volume of acid or alkali
pipette filler to use the pipette safely
conical flask to contain the liquid from the pipette
burette to add small, measured volumes of one reactant to the other reactant in the conical flaskSlide8
Method
Use the pipette and pipette filler to add 25 cm
3
of alkali to
a clean conical flask.
Add a few drops of indicator and put the conical flask on a white tile (so you can see the
colour
of the indicator more easily).
Fill the burette with acid and note the starting volume.
Slowly add the acid from the burette to the alkali in the conical flask, swirling to mix.
Stop adding the acid when the end-point is reached (the appropriate
colour
change in the indicator happens).
Note the final volume reading.
Repeat steps 1 to 5 until you get consistent readingsSlide9
The titre
The difference between the reading at the start and the final reading gives the volume of acid (or alkali) added.
This volume is called the
titre
.
For example, if the reading at the start is 1.0 cm
3
and the final reading is 26.5 cm
3
, then the
titre
is 25.5 cm
3 (26.5 – 1.0). Note that the titre
will depend upon the volume of liquid in the conical flask, and the concentrations of the acid and alkali used.
It is important to repeat the titration several times to check that your
titre
value is consistent so that your calculations are reliable.
A single indicator like litmus or phenolphthalein gives a sharp end-point where the
colour
changes suddenly.Slide10
Titration calculations
You should be able to use titration results to calculate the concentration of an acid or alkali.
If several runs have been carried out, any irregular
titres
should be ignored before calculating the mean
titre
.Example: 27.5 cm
3
of 0.2 mol/dm
3
hydrochloric acid is needed to titrate 25.0 cm
3
of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?You can check your answer using this quick method . N1
*V
1
=N
2
*V
2
;
N
1
- unknown concentration; N
2
- known concentration;
V
1
- volume of unknown; V
2
- volume of known;
unknown concentration = 0.2 × 27.5 / 25.0 = 0.22 mol/dm
3Slide11
Titration calculations
You should be able to use titration results to calculate the concentration of an acid or alkali.
If several runs have been carried out, any irregular
titres
should be ignored before calculating the mean
titre
.Example: 27.5 cm
3
of 0.2 mol/dm
3
hydrochloric acid is needed to titrate 25.0 cm
3
of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?You can check your answer using this quick method . N1
*V
1
=N
2
*V
2
;
N
1
- unknown concentration; N
2
- known concentration;
V
1
- volume of unknown; V
2
- volume of known;
unknown concentration = 0.2 × 27.5 / 25.0 = 0.22 mol/dm
3Slide12
Titration calculations
You should be able to use titration results to calculate the concentration of an acid or alkali.
If several runs have been carried out, any irregular
titres
should be ignored before calculating the mean
titre
.Example: 27.5 cm
3
of 0.2 mol/dm
3
hydrochloric acid is needed to titrate 25.0 cm
3
of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?You can check your answer using this quick method . N1
*V
1
=N
2
*V
2
;
N
1
- unknown concentration; N
2
- known concentration;
V
1
- volume of unknown; V
2
- volume of known;
unknown concentration = 0.2 × 27.5 / 25.0 = 0.22 mol/dm
3Slide13
Titration calculations
Example 1:
Given the equation
NaOH
(
aq
) + HCl
(
aq
) ==>
NaCl
(
aq) + H2O;
25.0 cm
3
of a sodium hydroxide solution was
pipetted
into a conical flask and titrated with 0.200 mol dm
-3
(0.2M) hydrochloric acid. Using a suitable indicator it was found that 15.0 cm
3
of the acid was required to
neutralise
the alkali. Calculate the
molarity
of the sodium hydroxide and its concentration in g/dm
3
.
(4.80 g/dm
3
)
Calculate the
molarity
of the sodium hydroxide and its concentration in g/dm
3
.
moles =
molarity
x volume (in dm
3
=
cm
3
/1000)
moles
HCl
= 0.200 x (15.0/1000) = 0.003 mol
moles
HCl
= moles
NaOH
(1 : 1 in equation)
so there is 0.003 mol
NaOH
in 25.0 cm
3
scaling up to 1000 cm
3
(1 dm
3
), there are ...
0.003 x (1000/25.0) = 0.12 mol
NaOH
in 1 dm
3
molarity
of
NaOH
is 0.120 mol dm
-3
(or 0.12M)
since mass = moles x formula mass, and
Mr
(
NaOH
) = 23 + 16 + 1 = 40
concentration in g/dm3 =
molarity
x formula mass
concentration in g/dm3 is 0.12 x 40 = 4.80 g/dm3Slide14
Titration calculations
Example
2:
Given the equation
2KOH(
aq
) + H
2
SO
4
(
aq
) ==> K2
SO
4
+ 2H
2
O (l).
20.0 cm
3
of a
sulphuric
acid solution was titrated with 0,0500 mol dm
-3
potassium hydroxide. If the acid required 36.0 cm
3
of the alkali KOH for
neutralisation
what was the concentration of the acid?
(4.41 g/dm
3
)
moles =
molarity
x volume (in dm
3
= cm
3
/100)
mol KOH = 0.0500 x (36.0/1000) = 0.0018 mol
mol H
2
SO
4
= mol KOH / 2 (because of 2 : 1 ratio in equation above)
mol H2SO4 = 0.0018/2 = 0.0009 (in 20.0 cm
3
)
scaling up to 1000 cm
3
of solution = 0.0009 x (1000/20.0) = 0.0450 mol
mol H
2
SO4 in 1 dm
3
= 0.0450
so
molarity
of H
2
SO
4
= 0.0450 mol dm
-3
(0.045M)
since mass = moles x formula mass, and
Mr
(H
2
SO
4
) = 2 + 32 + (4x16) = 98
concentration in g/dm
3
is 0.045 x 98 = 4.41 g/dm
3