note that water slightly dissociates in equilibrium with hydrogen and hydroxide ions: h 2 o h + +...

18 Acids and Bases (AHL)DP Chemistry

Rob Slider

Acid-Base Calculations

Product constant of water Kw

Note that water slightly dissociates in equilibrium with hydrogen and hydroxide ions:

H2O H+ + OH-

So, the equilibrium constant is:

Kw = [H+][OH-] (product constant of water)

Experimentally, it has been determined in a neutral solution at 250C:

[H+] = [OH-] = 10-7 (a very small amount!)

So,Kw = (10-7)(10-7) = 10-

14

This shows how far to the left this equilibrium is

Kw and temperature

In the previous slide we saw:

Kw = (10-7)(10-7) = 10-14

This value applies to 250C

Kw changes with temperature variations and so does the pH!

Temperature effects on Kw

T (°C)Kw (mol2

dm-6)pH

0 0.114 x 10-

14 7.47

10 0.293 x 10-

14 7.27

20 0.681 x 10-

14 7.08

25 1.008 x 10-

14 7.00

30 1.471 x 10-

14 6.92

40 2.916 x 10-

14 6.77

50 5.476 x 10-

14 6.63

100 51.3 x 10-14 6.14

Note: this does not mean that water is more acidic at higher temperatures and more alkaline at lower temperatures. Water is still neutral. It simply means that neutral is a different pH value at different temperatures.

QuestionsTemperature effects on Kw

T (°C)Kw (mol2

dm-6)pH

0 0.114 x 10-

14 7.47

10 0.293 x 10-

14 7.27

20 0.681 x 10-

14 7.08

25 1.008 x 10-

14 7.00

30 1.471 x 10-

14 6.92

40 2.916 x 10-

14 6.77

50 5.476 x 10-

14 6.63

100 51.3 x 10-14 6.14

1. Given the values in the table, what are the [H+] and [OH-] at 00C and 1000C?

2. What do the values of Kw tell you about whether the ionisation of water is endothermic or exothermic? Explain.

Answers:1. At 0C : 3.38 x10-7; at 100C: 7.16x10-7

(square root of Kw)2. An increase of temperature increases

the value of Kw. This suggests the products are favoured. According to LC Principle, this means the forward reaction is endothermic (absorbs heat)

Calculations using Kw

SinceKw = (10-7)(10-7) = 10-14

We can use this constant value to calculate [H+] or [OH-]

Calculating [H+]

An alkali with a hydroxide ion concentration of 0.01M (10-2)

Kw = (10-2)[H+] = 10-14

[H+] = 10-12

Calculating [OH-]

An acid with a hydrogen ion concentration of 0.001M (10-

3)

Kw = (10-3)[OH-] = 10-14

[OH-] = 10-11

pH, pOH and pKwpHWe have seen previously thatpH = -log[H30+]

Calculating pH example 1An alkali with a hydroxide ion concentration of 0.01M (10-2)

Kw = (10-2)[H+] = 10-14

[H+] = 10-12

pH = 12

Calculating pH example 2A substance has been found to have a hydroxide ion concentration of 10-

11

[OH-] = 10-11

pOH = 11pH = 14-11 = 3

pOHThis is similar to pHpOH = -log[OH-]

The ‘p’ is a mathematical operation that means ‘-log’

pKwSince Kw = [H+][OH-] pKw = pH + pOH = 14

Calculating concentration from pH

Recall the inverse of the pH calculation:

pH = -log[H30+]

The inverse:

[H30+] = 10-pH

or[H+] = 10-pH

On your calculator, this may be ‘2nd’ LOG or ‘10x’

Example

The pH of a solution is found to be 4.6

Find[H+]

[H+] = 10-pH

= 10-4.6

= 2.5x10-5 mol dm-3

ExercisesProblem 1Calculate the pH of solutions with the following H3O+ concentrations in mol dm-3:a)10-8 b)6.8 x 10-3 c)0.035

Problem 2Calculate the pH of solutions with the following OH- concentrations in mol dm-3:a)10-2 b)7.6 x 10-3 c)0.055

Problem 3Calculate the H3O+ concentrations in solutions with the following pH values:a)0.00 b)4.3 c)2.35 d)13.7

Problem 4What is the pH of a 3.5M solution of H2SO4? (assume complete ionisation)

Problem 5What is the pH of a 0.001M solution of Ba(OH)2? (Hint: use balanced equation)

Weak solutions (Ka & Kb)

Unlike strong acids, weak acids only partially dissociate:HA + H2O H3O+ + A-

The same is true for weak bases:B + H2O BH+ + OH-

Dissociation ConstantsBecause of these equilibria, we have known values for the ratio between products and reactants.

These are known as dissociation constants:

Ka – acid dissociation constantKb – base dissociation constant

Formulae

Ka =

Kb =

Notice water is not in these calculations. Its concentration has been incorporated into the dissociation constant.

Calculations using Ka

CH3COOH(aq) ↔ H+(aq) + CH3COO-

(aq)

Ka expression for acetic acid (ethanoic acid) above is: (constant at set temp)

Ka = [H+] [CH3COO-] mol dm3

[CH3COOH]

Ka can be used to find the pKa (like pH)pKa = - log Ka

The larger the value of Ka the stronger the acid (more it dissociates – breaks

apart)What about pKa?

Calculations using Ka


(aq)

Ka = [H+] [CH3COO-] mol dm3

[CH3COOH]

Using the equilibrium constant expression (above) we are able to calculate [H+] and thus the pH of the acid


(aq)

Initial conc x 0 0Final conc x-y y y

𝐾 𝑎=[ 𝑦 ] [𝑦 ][𝑥−𝑦 ]

=[ 𝑦 ]2

[𝑥−𝑦 ] (𝑦 ≈0)≈

[ 𝑦 ]2[𝑥 ]

Assumption made based on relative concentrations of ‘x’ vs. ‘y’ to simplify calculations

Calculations using Kb

NH3(aq) + H2O(l) ↔ NH4+

(aq) + OH-(aq)

Kb = [NH4+] [OH-]

[NH3]

This works the same as in the acid example

NH3(aq) ↔ NH4+

(aq) + OH-(aq)

Initial conc x 0 0Final conc x-y y y

𝐾𝑏=[ 𝑦 ] [𝑦 ][𝑥− 𝑦 ]

≈[ 𝑦 ] 2[𝑥 ]

Ka, Kb and Kw

Consider the following generic acid + water reaction:

HA + H2O A- + H3O+

ACID Conjugate BASE

Now, if the conjugate base reacts with water in this reaction:

A- + H2O HA + OH-

𝐾 𝑎=[ 𝐴− ] ¿¿𝐾𝑏=

[𝐻𝐴 ] [𝑂𝐻−][ 𝐴−]

So, Ka x Kb

= [] = Kw

Ka x Kb = Kw

pKa, pKb and pKw

From the previous slide,

Ka x Kb = Kw

If we take the ‘p’ of each of these values (-log) we find,

pKa + pKb = pKw = 14

Summary

Ka = (acid dissociation constant; larger the number, stronger the acid)

Kb = (base dissociation constant; larger the number stronger the base)

pKa = - log Ka (smaller the value, stronger the acid) (pKb is the

same)

Ka x Kb = Kw= 10-14

pKa + pKb = pKw = 14 = pH + pOH

Exercises

Calculate the H3O+ concentrations, and the pH value for the following weak acids:a) HCO2H conc. = 0.0100 mol dm-3 Ka = 2.04 x 10-4

b) CH3CO2H conc. = 0.1 mol dm-3 Ka = 1.77 x 10-5

c) HCN conc. = 1.00 mol dm-3 Ka = 3.96 x 10-

10

Calculate Ka for the following acids:d) HF conc. = 0.200 mol dm-3 pH = 2.23e) HClO2 conc. = 0.0200 mol dm-3 pH = 1.85

Calculate the OH- and H3O+ concentrations, and the pH value for the following weak bases:f) NH3 conc. = 0.0100 mol dm-3 Kb = 1.80 x 10-5

g) C5H5N conc. = 0.1 mol dm-3 Kb = 1.40 x 10-9

h) CH3NH2 conc. = 1.00 mol dm-3 Kb = 4.38 x 10-4

Buffer Solutions

What is a buffer?

A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it

There are 2 types:› Acidic› Alkaline

Acidic buffersAn acidic buffer has a pH less than 7

It is often made from equal molar concentrations of a weak acid and it’s conjugate base

Example: ethanoic acid and ethanoate ion

CH3COOH CH3COO- + H+

(weak acid) (conj. base)

Acidic buffersTake a 0.1M solution of ethanoic acid: CH3COOH CH3COO- + H+

At equil: (0.09583M) (0.00417M) (0.00417M)

What’s the pH of this solution?

Now, say you add HCl to this equilibrium. The acetate ion is the only species available to reduce the added H+ and these are very low in concentration, so the pH will drop dramatically.

What can you do to reduce this effect?

-log [H+] = 2.38

Add acetate ions (sodium acetate) in equimolar amounts

Acidic buffers – adding acid


If we buffer the solution by adding 1M sodium acetate, what will happen?

This will have the following effect:1. Increase the amount of acetate ions, shifting the equilibrium to the left2. Increase the pH to 4.76

3. When adding H+ ions, the extra acetate ions will react to reduce the [H+] moving the equilibrium to the left – producing more weak acid.

4. The solution resists changes to pH, meaning it is buffered

Acidic buffers – adding base


Adding base results in more OH- being added to the equilibrium. This

extra amount of ions is removed by:

CH3COOH + OH- CH3COO- + H2O

hydroxide ions are removed by reaction with undissociated ethanoic

acid, shifting equilibrium right making a weak base (CH3COO-)

Add OH-

Acidic buffers – made with a base

An alternative way to make an acidic buffer is to add a strong base to the weak acid to produce high proportions of the weak acid and the conjugate base.

MOH + HA H2O + MA

This drives the equilibrium to the right producing more salt (MA) which dissociates. This leads to:

M+ + OH- + HA H2O + M+ + A-

Adding acid – conjugate base A- reacts with the added H+ to minimise the effect

Adding base – weak acid HA dissociates further and the H+ reacts with the OH- to minimise the effect

A 2:1 molar ratio of a weak acid and a strong base of the same concentration is used to make this buffer solution.

pH of buffers depends on concentration of conjugate pair

pH vol. of 0.1Macetic acid (ml)

vol. of 0.1Msodium acetate (ml)

3 982.3 17.7

4 847.0 153.0

5 357.0 653.0

6 52.2 947.8

Note: you can also get various pH buffers by changing the acid/base pair.

Alkaline buffersAn alkaline buffer has a pH greater than 7

It is often made from a equal molar concentrations of a weak base and it’s conjugate acid

Example: ammonia and ammonium ion

NH3 + H2O NH4+ + OH-

(weak base) (conj. acid)

Alkaline buffers NH3 + H2O NH4

+ + OH-

(weak base) (conj. acid)

Due to the weak nature of ammonia, this equilibrium will be well to the left. We can create a buffered solution by adding ammonium chloride or a strong acid (like the acid buffer).

What will this do to the equilibrium?

Addition of ammonium ions (adding ammonium chloride) will shift the equilibrium even further to the left. The pH of this solution would be 9.25

You can also add ~half the moles of a strong acid (e.g. HCl) which will react with ammonia to form more ammonium ions (NH3 + H+ NH4

+)

Alkaline buffers – adding acid


What will happen if you add acid to this solution?

Reaction of the acid with the weak base

NH3 + H+ NH4+

removal by reaction with ammonia to produce more ammonium ion – a weak acid)

Alkaline buffers – adding base


What will happen to the equilibrium if you add base?

Adding base effectively adds OH-. This means:

The ammonium ion reacts with the OH- to shift the equilibrium to the left, consuming most of the OH- ions.

NH4+ + OH- NH3 + H2O

Summary

Buffer solutions resist changes in pH when acids and alkalis are added

Buffers generally contain:› Sufficient concentrations of a weak acid and

it’s conjugate base OR weak base and it’s conjugate acid

The pH of buffer solutions depend on the concentrations and type of conjugate acid/base pairs that are used.

Go here http://www.pearsonhotlinks.co.uk/9780435994402.aspx for good animations of this (chapter 8)

http://www.pearsonhotlinks.co.uk/9780435994402.aspx



pH of a buffer solution

Consider the dissociation of a weak acid

HA(aq) + H2O(l) A-(aq) + H3O+

The acid dissociation constant expression is

This can be rearranged to find

Commercially available buffer solutions are used to calibrate pH meters

pH of a buffer solutionHA(aq) + H2O(l) A-

(aq) + H3O+

Assumptions:1. As a large quantity of conjugate base (A-) has been added, the equilibrium

shifts far left, so that equilibrium concentration of the acid is approximately equal to the initial concentration of the weak acid :

[HA]eq ≈ [HA]i = [acid]

2. The equilibrium concentration of the conjugate base ion (A- ) is approximately equal to the concentration of the salt that was added to the equilibrium.

[A-]eq ≈ [A-]i = [salt]

pH of a buffer solution

So, considering our 2 assumptions:

[HA]eq ≈ [HA]I = [acid]

[A-]eq ≈ [A-]i = [salt]

Recall from a previous slide that

¿𝑝𝐻=𝑝 𝐾 𝑎− 𝑙𝑜𝑔

[𝑎𝑐𝑖𝑑][𝑠𝑎𝑙𝑡 ]

Alternatively, you may solve for [H+] first and then solve for pH usingpH=-log[H+]

When [acid] = [salt]pH= pKa

pOH of a buffer solution

Similarly for a base equilibriumB(aq) + H2O(l) HB+

(aq) + OH-(aq)

𝑝𝑂𝐻=𝑝 𝐾𝑏−𝑙𝑜𝑔[𝑏𝑎𝑠𝑒 ]

[𝑠𝑎𝑙𝑡 ]

Alternatively, you may solve for [OH-] first and then solve for pOH usingpOH=-log[OH-]

When [base] = [salt]

pOH= pKb

ExerciseAn aqueous solution of 0.1M ammonia and 0.1M ammonium chloride has a pH of 9.3. The reaction is: NH3 + H2O NH4

+ + OH-

a) Calculate the Kb for ammoniab) If a pH of 9.0 is needed, what should be added? Explainc) Calculate the new concentration of the substance added in b)

Answers:a) = 10-4.7 = 2.00 x 10-5

b) Ammonium chloride should be added as this will shift the equil to the left(towards ammonia), reducing the [OH-] and decreasing the pH

c) pH = 9.0 means pOH = 5, so [OH-] = 10-5

= = 0.2 mol dm-3

Notice the additional volume has a negligible effect on [NH3] and has been ignored

ExerciseIf you wanted to make a buffer solution of pH=4.46 using ethanoic acid and sodium ethanoate, what concentrations could you use? pKa (ethanoic acid) = 4.76. The reaction is: CH3COOH+ H2O CH3COO- + H3O+

Answer:pH = pKa – log 4.5 = 4.76 – log log = 4.76 – 4.46 = 0.30

= 100.30

= 2.0

So, we need a solution with twice as concentrated acid as ethanoate salt

Note: it should be expected that the acid concentration will be higher than the salt.

At equal concentrations, the pH = pKa = 4.76At pH 4.46 (a lower pH), we need to add acid to shift equil right, increasing [H3O+], decreasing pH

ExerciseAn aqueous solution of 0.025M ethanoic acid and 0.050M sodium ethanoate is prepared. a) Calculate the pH of this solution given Ka = 1.74x10-5 mol dm-3

b) If 1.0cm3 of 1.0M NaOH is added to 250cm3 of buffer, what will happen to the pH?

Answers:a) = -log (1.74x10-5) – = 4.76+0.30=5.06b)

Notice the additional volume has again been ignored as it is assumed to have little effect on the overall pH

CH3COOH + OH- CH3COO- + H2O

ni (mol) 0.00625

0.001 0.0125 -

nc(mol) -0.001 -0.001 +0.001 -

ne(mol) 0.00525

0 0.0135 -

[ ] (mol dm-

3)

0.021 0 0.054 -

So,pH = pKa – log pH = 4.76 – log pH = 4.76 + 0.41

pH = 5.17

Acid-Base Titration

TitrationTitration is an analytical technique that is used to determine the end point of a reaction (often Acid + Base) using an indicator or pH meter.

Acid-Base Titration (basic steps):

1. Unknown: Accurately measure a volume of base of unknown concentration using a pipette and place into a flask

2. Indicator: Add an appropriate indicator which will change colour at the end point of the reaction

3. Titrate: Carefully add an acid of known concentration until the end point is reached as indicated by the colour change

PipetteIndicator addition

Titration End point

Titration – end/equivalence

Equivalence Point

The equivalence point in an acid-base reaction is the point where neutralisation occurs.

The molar ratios have been reached

End point

The end point is where an indicator solution just changes colour permanently.

Indicators change colours at specific pH ranges and are chosen to be as close to the equivalence point as possible. More later…

Titration Graphs

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Strong Acid - Strong Base

Investigating the titration between:

1M HCl and 1M NaOH

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HCl

10 ml NaOH Start with 10ml of alkali and slowly add acid measuring the pH

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HCl

10 ml NaOH+ almost 10 ml HCl

pH During a Titration

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 2 4 6 8 10 12 14 16 18 20

Volume of acid added / ml

pH

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HCl

10 ml NaOH + 10 ml HCl


0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 2 4 6 8 10 12 14 16 18 20


pH

Now we have equivalent amounts of strong acid and strong base – notice the pH changes dramatically

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HCl

10 ml NaOH


0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 2 4 6 8 10 12 14 16 18 20


pH

Now as we continue to add acid past pH = 7, the pH drops quickly at first and then more slowly

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0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 5 10 15 20 25


pH

Equivalence point at pH7

NaOH + HCl = NaCl + H2O 1M 1M1NaOH + 1HCl NaCl + H2O

10ml 10ml

Solutions mixed in the right proportions according to the equation.

48


Running acid into alkali Running alkali into acid

This nearly horizontal section of the graph is called the buffer region. Here the pH stays relatively constant due to a buffering between the acid and the salt

This point is called half-equivalence where half of the acid has been neutralised. Here, there are equivalent amounts of acid and salt, so

pH = pKa

pH = pKa

pOH = pKb

The same applies to a base being neutralised by an acid

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Strong Acid - Weak Base


1M HCl and 1M NH3

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1M 1M1NH3 + 1HCl NH4Cl

25ml 25mlpH starts to fall quickly as acid is added

pH falls less quickly as buffer soln formed (excess NH3 and NH4Cl present)

Soln at equivalence point is slightly acidic because ammonium ion is slightly acidic

NH4+ + H2O NH3 + H3O+

Weak base so initial pH value is less than 14

acid

alkali

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alkali

acid

Very low pH indicative of a strong acid solution

After equivalence point the soln contains NH3 and NH4Cl – a buffer soln and so resists large increase in pH so graph flattens out.

< pH 7

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Weak Acid - Strong Base


1M CH3COOH and 1M NaOH

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1M 1M1CH3COOH + 1NaOH CH3COONa

25ml 25ml

pH falls less quickly as buffer soln formed (excess CH3COOH and CH3COO- present)

Soln at equivalence point is slightly alkaline because ethanoate ion is slightly alkaline

CH3COO-+ H2O CH3COOH + OH-

acid

alkali

Very high pH – indicative of a strong alkali solution

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alkali

acid

pH starts to rise quickly as alkali is added

pH rises less quickly as buffer soln formed (excess CH3COOH and CH3COO- present)

Excess of alkali present – graph same as when adding strong alkali to strong acid

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Weak Acid - Weak Base


1M CH3COOH and 1M NH3

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acid

alkali

1CH3COOH + 1NH3 = CH3COO- + 1NH4+

About as weak as each other -

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acid

alkali

No steep section – small addition of acid causes a large change in pH, so…

Very difficult to do a titration between a weak acid and a weak base.

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Summary

Salt Hydrolysis(pH of salts)

All salts are not neutral

The pH of a salt depends upon the relative strength of the ions that make up the salt.

Very few salts are neutral Salts completely dissociate into their ions

when sufficiently diluteNaCl(s) Na+

(aq) + Cl-(aq)

It is possible for these ions to interact with water to produce H+ or OH- ions which results in acidic or alkaline solutions. These are known as hydrolysis reactions

Remember: the stronger the acid/base, the weaker it’s conjugate

Strong Acid +Base Neutral ions

Neutral anions are formed from strong acids

Neutral cations are formed from strong bases

NaCl is a neutral salt because the ions that are formed derive from a strong acid and base› HCl + NaOH NaCl + H2O

Na+ and Cl- are weak conjugates, so there is no tendency for these ions to undergo hydrolysis reactions.

pH = 7

Strong Base/Weak Acid Basic anions

Weak acids form conjugate bases that can react with water to form hydroxide ions

› H2CO3 + 2NaOH Na2CO3 + H2O

› Na2CO3 2Na+ + CO32-

› CO32- + H2O HCO3- + OH-

In the above reactions: NaOH is a strong base so the weak

conjugate, Na+ will not react with water to form hydrogen ions

Carbonic acid is weak, so the carbonate ion will react with water to a small extent to form OH- ions

Basic anions are formed by weak acids

pH > 7

Strong Acid/Weak Base Acidic cations

HCl + NH3 NH4+ + Cl-

A donatable hydrogen must be available on the cation.

NH4+ + H2O NH3 + H3O+

The ammonium ion comes from the weak base, ammonia, so a hydrolysis reaction can occur to a small extent producing hydronium ions.

pH < 7

Weak Acid + Weak Base Depends on Ka/Kb

When combining weak acids and bases, the pH of the salt will depend on their relative strengths. › If Ka > Kb – acidic

› If Kb > Ka – basic

› If Ka = Kb - neutral

Example:CH3COOH + NH3 NH4

+ + CH3COO-

Both of the salts formed in this reaction will react with water, but the effects cancel resulting in a nearly neutral solution

Ammonium Acetate

Acidic complex ions

Fe3

+

O

O

OO

OO

H HHH

H

H

H

H

H H

H

H

+ H+Fe3

+

O

O

OO

OO

H HHH

H

H

H

H

H H

HAn O-H bond may be broken releasing a proton

2+

Small ions with multiple charges can hydrolyse water. The high charge density pulls electrons towards the metal ion causing a proton to be released, decreasing the pH

Other metal ions that are able to hydrolyse water include Be2+

and Al3+

Salt Hydrolysis Summary

Type of reaction

Example Reaction

Salt produced

Ion that hydrolyses water

Nature of final solution

Strong acid Strong Base

HCl + NaOH NaCl Neither ion

Neutral pH = 7

Weak acidStrong Base

CH3COOH + NaOH

NaCH3COO Anion BasicpH>7

Strong AcidWeak Base

HCl + NH3 NH4Cl Cation AcidicpH<7

Weak AcidWeak Base

CH3COOH + NH3 NH4CH3COO Anion & Cation

Depends on Ka/Kb

Metal complex

[Al(H20)6]3+ [Al(H20)5OH]2+

+ H+

NA Metal complex with multiple charge

AcidicpH<7

Indicators

Indicators – colour changers

Indicators can change colours depending upon the pH of the solution they are in.

Indicators are themselves weak acids or weak bases and as they lose or gain H+, they form substances that have distinct colours.

Let’s look at some examples…

In general, an indicator molecule (In)

Hin(aq) H+(aq) + In-

(aq)

Litmus (red or blue)

“HLit” is litmus – a weak acid – is red in solution

When hydroxide (base) is added to this weak base, “Lit-” is formed and is blue

When hydrogen ions (acid) is added, we shift back to the red “HLit” .

So,• Acid turns blue litmus

red• Alkali turns red litmus

blue

Base

Acid

Add base (hydroxide)

Add acid (hydrogen ion)

Methyl Orange (yellow or red)

Methyl orange is yellow below pH 3.7

Methyl orange is red above pH 3.7

Notice the transfer of a proton between the different coloured compounds

Phenolphthalein

Here is a common acid-base indicator that changes colour at pH 9.3

What colour is the acid?

What colour is the base?

Phenolphthalein

Here is a common acid-base indicator that changes colour at pH 9.3

What colour is the acid? colourless

What colour is the base? pink

Kind (pH = pKa)Recall the generic equation for an indicator

Hin(aq) H+(aq) + In-

(aq)

We can write an Equilibrium expression for this weak acid

When the equilibrium is balanced, ]= so,

This meansKa = [H+] orpKa = pH

So the pKa of the indicator, tells us where the colour change will occur

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Choosing an Indicator3.7 9.3

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Weak Acid – Weak Base

18 Acids and BasesThe end

note that water slightly dissociates in equilibrium with hydrogen and hydroxide ions: h 2 o h + +...

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