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pH and  indicators;  the pH scale pH and  indicators;  the pH scale

pH and indicators; the pH scale - PowerPoint Presentation

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pH and indicators; the pH scale - PPT Presentation

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

concentration acid mol alkali acid concentration alkali mol volume solution hydroxide titration indicator sodium unknown added titre conical molarity

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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).Slide4
Slide5

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