Chemical Reactions: pH Equilibria

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Reversible Reactions and

Chemical Equilibrium

University of Lincoln presentation

http://creativecommons.org/licenses/by-nc-sa/2.0/uk/

http://chemistryfm.blogs.lincoln.ac.uk/


Outline

• Reversible reactions• Chemical Equilibrium• Le Chatelier’s Principle• Equilibrium constants



Reversible Reactions

BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)

CH3CO2H + CH3CH2OH ↔ CH3CO2CH2CH3 + H2O

Cr2O72-(aq) + 2OH-(aq) ↔ 2CrO4

2-(aq) + H2O(l)

CH3CO2H(aq) + H2O(l) ↔ CH3CO2-(aq) + H3O+

(aq)



Chemical Equilibrium

• Reactions not 100% complete– Products and Reactants exist

together• A dynamic equilibrium • Position of equilibrium ???

• Can the position of equilibrium be changed?



Le Chatelier’s Principle

When an external change is made to a system in equilibrium, the system will respond to oppose the change

External Changes• Concentration • Pressure (gases)• Temperature

Link to external videoLink to external video



Concentration

1. BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)

2. Cr2O72-(aq) + 2OH-(aq) ↔ 2CrO4

2-(aq) + H2O(l)

How does reaction 1 respond to addition of hydrochloric acid?

How does reaction 2 respond to addition of alkali?How does reaction 2 respond to addition of acid?



Pressure

N2(g) + 3H2(g) ↔ 2NH3(g)

CO(g) + 2H2(g) ↔ CH3OH(g)

2NO2(g) ↔ 2NO(g) + O2(g)

PCl5(g) ↔ PCl3(g) + Cl2(g)

H2(g) + I2(g) ↔ 2HI(g)

CO(g) + H2O(g) ↔ CO2(g) + H2(g)

How do the above equilibria respond to:An increase in pressureA decrease in pressure



Temperature

N2(g) + 3H2(g) ↔ 2NH3(g) rH = -92.2 kJ mol-1

H2(g) + I2(g) ↔ 2HI(g) rH = -9.4 kJ mol-1

CO(g) + H2O(g) ↔ CO2(g) + H2(g) rH = -41.2 kJ mol-1

PCl5(g) ↔ PCl3(g) + Cl2(g) rH = 87.9 kJ mol-1

How do the above respond to anIncrease in temperatureDecrease in temperature



Equilibrium constantsa measure of equilibrium

positionaA + bB ↔cC + dD

ba

dc

c [B][A][D][C]

K

BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)

]O][H[BiCl

]][HCl[BiOClK

(l)2(s)3

2(aq)(s)

c

Write the expressions for Kc for the reactions given in previous slides



Calculating Equilibrium Constants

HNO2(aq) ↔ H+(aq) + NO2-(aq)

Solution

[HNO2(aq)]

mol litre-1

[H+(aq)]

mol litre-1

[NO2-(aq)]

mol litre-1

A 0.090 6.2 x 10-3 6.2 x 10-3

B 0.20 9.3 x 10-3 9.3 x 10-3

C 0.30 11.4 x 10-

3

11.4 x 10-3

The table shows the equilibrium molar concentrations for three solutions of nitrous acid in water at 25 oC

Calculate the equilibrium constant for this reaction at 25oC



433

2

2c 104.3

0.090106.2106.2

(aq)][HNO(aq)](aq)][NO[H

K

Units of Kc

1

11

c litremol)litre)(mollitre(mol

K

14c litremol104.3K

Solution A

Now try for solutions B and C

1litremol



Acids and Bases



Outline

• Definitions• Weak Acids• Dissociation Constants• Weak Bases• Drugs• pH• Buffers



Acids and Bases• Several definitions available - most common is

Bronsted and Lowry• Acid is a proton donor

– HCl is able to transfer H+

• Base is a proton acceptor

– NH3 is able to accept H+ and become NH4+

• Aqueous solutions

• Proton species is H3O+ (hydroxonium ion)

– HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

– HCl(aq) H+(aq) + Cl-(aq)



Strong Acids• Strong acids are fully dissociated

HCl (aq) + H2O(l) H3O+ (aq) + Cl- (aq)

• all dissolved HCl molecules are ionised

• 1 mol dm-3 HCl(aq) there are:

– Approx 1 mol dm-3 H3O+ (aq)

– Approx 1 mol dm-3 Cl- (aq)

DO NOT confuse ‘strong’ and ‘concentrated’1 x 10-4 mol dm-3 HCl (aq) is a dilute solution

of a strong acid



Other strong acids

• HNO3 (nitric)

• H2SO4 (sulfuric)

• HClO4 (perchloric)

Write equations showing the dissociation of the above acids

Which are monoprotic?Are any diprotic?Chemical equilibrium – K very large



Weak Acids

• Acids that dissociate in a reversible reaction (e.g. CH3COOH; ethanoic (acetic) acid)

CH3COOH (aq) + H2O(l) ↔ H3O+ (aq) + CH3COO-

(aq)

• Solution of CH3COOH (aq) contains:

– CH3COOH (aq)

– H3O+ (aq)

– CH3COO- (aq)

• CH3COOH is partially dissociated



How weak is a weak acid?

0.1 mol dm-3 HCl is dissociated 91.4% [H3O+] = 0.091 mol dm-3 pH=1.04

0.1 mol dm-3 CH3COOH is dissociated 1.34%

[H3O+] = 0.0013 mol dm-3 pH=2.87

• Extent given by K



Weak Acids

HA(aq) + H2O(l) ↔ H3O+(aq) + A-

(aq)

HA Bronsted acidH2O Bronsted base

H3O+ Bronsted acid

A- Bronsted base

O][HA][H]][AO[H

K2

3



Acid dissociation constant (Ka)

• The higher the Ka value:

– greater degree of ionisation– stronger the acid– Data tables

O][HA][H]][AO[H

K2

3

[HA]]][AO[H

O]K[H 32

[HA]]][AO[H

KO]K[H 3a2



Ka Values

• HCO2H 1.8 x 10-4 mol dm-3

• CH3CO2H 1.7 x 10-5 mol dm-3

• Are these weak or strong acids?• Which is the stronger acid?

aa logKpK

HCO2H 3.75CH3CO2H 4.77



pKa values (data tables)

Acid pKa Conjugate base

H3PO4 2.12 H2PO4-

HNO2 3.34 NO2-

H2CO3 6.37 HCO3-

HCN 9.31 CN-

HCO3- 10.25 CO3

2-



pKa Values

• Controlling the ionisation of weak acids• pH = pKa then [HA] = [A-]• pH > pKa then [A-] > [HA]• pH < pKa then [HA] > [A-]CH3COOH (aq) + H2O(l) ↔ H3O+

(aq) + CH3COO- (aq)

• CH3COOH: CH3COO- at pH = 4.77 ?

• CH3COOH: CH3COO- at pH = 3 ?

• CH3COOH: CH3COO- at pH = 7 ?



Henderson-Hasselbach

ionised][un[ionised]

logpKpH a

For weak acids

Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?



Weak Bases

B(aq) + H2O(l) BH+(aq) + OH-

(aq)

CH3NH2(aq) +H2O(l) ↔ CH3NH3+

(aq) + OH-(aq)

pKa = 10.66 (of conjugate acid) [B]=[BH+]

pH = 10.66pH =8 what happens to CH3NH3

+(aq): CH3NH2(aq)

pH =13 ?



Henderson-Hasselbach

[ionised]ionised][un

logpKpH a

For weak bases

Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?



Acidic drugs

2-[4-(2-methylpropyl)phenyl]propanoic acid

How does this molecule ionise?

pKa=4.5

pH =3 (stomach pH)?

pH=6 (intestine)?

ibuprofen

CH3

CH3OH

O

CH3



Basic drugs

amphetamine (C6H5CH2CH(NH2)CH3)Write an equation for the reaction of

amphetamine with water.The pKa of the conjugate acid is 9.8. What

will happen to the ratio of ionised to unionised amphetamine at:pH 7pH 12

Why might this be important?

CH3

NH2



Water

• Can dissociate:H2O(l) ↔ H+

(aq) + OH-(aq)

2H2O(l) ↔ H3O+(aq) + OH-

(aq)

H2O is amphoteric

22

3

O][H]][OHO[H

K



Water

Kw = [H3O+][OH-]= 1 x 10-14 mol2 dm-6

• Kw the ionic product of water

• In pure water what is [H3O+] and [OH-] ?

• Kw is a very small constant– water is only very partially ionised



pH

• pH is defined as:pH = -log10[H3O+]

• pH is a measure of the H3O+ concentration in solution and can vary from 1 to 14

• pH=7 – neutral [H3O+] = [OH-]

= 1 x 10-7 mol dm-3 at 25 oC

• pH<7 – acidic [H3O+] >[OH-]

• pH>7 - alkaline/basic [H3O+] <[OH-]



pH-examples

• 0.1M HNO3

• 0.1M CH3COOH

• What is the pH?

• pH is dependent on the ionisation of the acid



pH-examples

• What about alkaline solutions?• E.g. 0.1M NaOH solution• Will also depend on degree of

ionisation• use equation: [H+] x [OH-] = 10-14



Buffers


Buffers

• A buffer solution resists pH changes on addition of small amounts of acid or base (alkali) to a system.

• Very important– e.g. blood has a pH of 7.4. If it varies by

± 0.4, death can occur

• Buffer solutions rely upon the effects of a weak acid or base and the salt of that acid or base


Buffers

• Ethanoic acid (a weak acid) and sodium ethanoate (salt) CH3COOH CH3COO- + H+ (1)

CH3COONa CH3COO- + Na+ (2)

• (1)-partially ionised• (2)-fully ionised


Buffers

[acid][salt]

logpKpH a

Henderson-Hasselbach equation

Acidic buffers


Making a buffer solution

• Choose a weak acid with a pKa close to the required pH of the buffer.

• Choose an appropriate salt of the weak acid

• Determine [salt]/[acid] ratio needed to give correct pH


An acidic buffer: Ethanoic acid and sodium ethanoate

[acid][salt]

logpKpH a 04.77pH

What is the [salt] if the acid is 0.1 mol dm-3 to give buffer solutions of

pH = 5

pH = 4

What would be the pH of an ethanoate buffer with equal acid and sodium ethanoate concentrations?


An alkaline buffer:ammonia solution and ammonium chloride

[salt][base]

logpKpH a

Note the base/salt ratioWhat is the pH of a buffer with base:salt ratio = 1?

9.24log19.24pH

Calculate the base:salt ratios for pH 8.5 and pH 10.5


Acknowledgements

• JISC• HEA• Centre for Educational Research and Developmen

t• School of natural and applied sciences• School of Journalism• SirenFM• http://tango.freedesktop.org

http://www.jisc.ac.uk/

http://www.heacademy.ac.uk/

http://www.lincoln.ac.uk/cerd/

http://www.lincoln.ac.uk/cerd/

http://www.lincoln.ac.uk/fabs/

http://www.lincoln.ac.uk/journalism/

http://www.sirenonline.co.uk/

http://tango.freedesktop.org/

http://tango.freedesktop.org/


http://www.lincoln.ac.uk/

http://www.sirenonline.co.uk/

http://www.heacademy.ac.uk/

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Chemical Reactions: pH Equilibria

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