The road to acids

Acids - proton donors

Base - proton acceptors

Alkali - a base that dissolves in water forming OH- ions

Qu 1-2  P135

The role of H+ in reactions of acids

HCl(g) + aq g H+(aq) + Cl-(aq)
H2SO4(l) + aq g H+(aq) + HSO4-(aq)

Mono, di and tri - basic acids:

Monobasic: releases 1 proton

HCl(aq) + aq g H+(aq) + Cl-(aq)

Dibasic: releases 2 protons

H2SO4(aq) + aq g H+(aq) + HSO4-(aq)
HSO4-(aq) + aq D H+(aq) + SO42-(aq)

 

Tribasic: releases 3 protons

H3PO4(aq) + aq D H+(aq) + H2PO4-(aq)
H2PO4-(aq) + aq D H+(aq) + HPO42-(aq)
HPO4-(aq) + aq D H+(aq) + PO43-(aq)

 

Acid - base reactions:

        Carbonates

        Bases (metal oxides)

        Alkalis (metal hydroxides)

Carbonates:

 

Full: 2HCl(aq) + CaCO3(s) g CaCl2(aq) + CO2(g) + H2O(l)
Ions: 2H+(aq) + 2Cl-(aq) + CaCO3(s) g Ca2+(aq) + 2Cl-(aq) + CO2(g) + H2O(l)
Ionic: 2H+(aq) + CaCO3(s) g Ca2+(aq) + CO2(g) + H2O(l)
Full: 2HCl(aq) + CaCO3(aq) g CaCl2(aq) + CO2(g) + H2O(l)
Ions: 2H+(aq) + 2Cl-(aq) + Ca2+(aq) + CO32-(aq) g Ca2+(aq) + 2Cl-(aq) + CO2(g) + H2O(l)
Ionic: 2H+(aq) + CO32-(aq) g   + CO2(g) + H2O(l)

 

Bases:

 

Full: 2HNO3(aq) + MgO(s) g Mg(NO3)2(aq) + H2O(l)
Ions: 2H+(aq) + 2NO3-(aq) + MgO(s) g Mg2+(aq) + 2NO3-(aq) + H2O(l)
Ionic: 2H+(aq) + MgO(s) g Mg2+(aq) + H2O(l)

 

Alkalis:

 

Full: H2SO4(aq) + 2KOH(aq) g K2SO4(aq) + 2H2O(l)
Ions: 2H+(aq) + SO42-(aq) + 2K+(aq) + 2OH-(aq) g 2K+(aq) + SO42-(aq) + 2H2O(l)
Ionic: 2H+(aq) + 2OH-(aq) g 2H2O(l)

 

Redox reactions of acids with metals:

Full: 2HCl(aq) + Mg(s) g MgCl2(aq) + H2(g)
Ions: 2H+(aq) + 2Cl-(aq) + Mg(s) g Mg2+(aq) + 2Cl-(aq) + H2(g)
Ionic: 2H+(aq) + Mg(s) g Mg2+(aq) + H2(g)

Qu 1 - 2  P137

Conjugate acid - base pairs

Acids as proton donors:

Hydronium ions:

HCl(aq) + H2O(l) g H3O+(aq) + Cl-(aq)
  hydronium ion    

HCl(aq) g H+(aq) + Cl-(aq)
    hydronium ion    

Acid - base pairs:

Conjugate pairs

    Conjugate pair    
HNO2(aq) + H2O(l) D H3O+(aq) + NO2-(aq)
acid   base   acid   base
Congugate pair
    Conjugate pair    
NH3(aq) + H2O(l) D OH-(aq) + NH4+(aq)
base   acid   base   acid
Congugate pair

Qu 1 - 2  P139

What is pH?

The pH scale

Practical

pH  =  -log10[H+(aq)]

[H+(aq)]  =  10-pH

What does pH mean:

  Low pH High [H+(aq)]
  High pH Low [H+(aq)]

Converting between pH and [H+(aq)]

Calculating the pH of strong acids:

                                  Strong acids donate protons to water completely

                                  Weak acids do no donate protons to water very readily (later)

  Water      
HCl(g) g H+(aq) + Cl-(aq)
1 mole   1 mole   1 mole

Examples:

1)  Calculate the pH of 1 x 10-3 mol dm3 HCl:    

        [HCl(aq)] = 1 x 10-3 mol dm3

        [H+(aq)] = 1 x 10-3 mol dm3     

        pH  =  - log [H+(aq)]

        pH  =  - log [1 x 10-3]

        pH  =  3

2)  Calculate the pH of 1 x 10-6 mol dm-3 HNO3

        [HCl(aq)] = 1 x 10-6 mol dm3

        [H+(aq)] = 1 x 10-6 mol dm3     

        pH  =  - log [H+(aq)]

        pH  =  - log [1 x 10-6]

        pH  =  6

Qu  1 - 3  P141

The ionisation of water

The ionisation of water and Kw

Water: facts and models

        Water conducts electricity.  It has a conductivity of 26.6 x 10-6 Scm-1

H2O(l) D H+(aq) + OH-(aq)
Kc

=

[H+]  x  [OH-]
[H2O]
     
Kc   x   [H2O]

=

[H+]  x  [OH-]
Kc   x   [H2O]

=

[H+]  x  [OH-]
Kw

=

[H+]  x  [OH-]

    Kw has an equilibrium constant of 1 x 10-14

PH and the ionic product of water

Kw

=

[H+]  x  [OH-]
Kw

=

[H+]  x  [OH-] = 1 x 10-14
[H+] =     [OH-]    
[H+]

x

[OH-] = 1 x 10-14  
    [H+]2 = 1 x 10-14  
    [H+] = (1 x 10-14)0.5

where ( )0.5 = square root

    [H+] = 1 x 10-7  

pH  =  -log10[H+(aq)]

The significance of Kw:

Kw

=

[H+]  x  [OH-]
         
  Acidic conditions [H+] > [OH-]. Kw must always equal 1 x 10-14
  Alkaline conditions [H+] < [OH-]. [H+]  x  [OH-]  =  1 x 10-14
         

 

The link between [H+] and [OH-]

Kw

=

[H+]  x  [OH-] = 1 x 10-14
pH 1 3 5 7 9 11 13  
[H+] 10-1 10-3 10-5 10-7 10-9 10-11 10-13  
[OH-] 10-13 10-11 10-9 10-7 10-5 10-3 10-1  

Qu  1 - 3  P147

 

pH of strong bases

 

                                  Strong bases dissociate completely to generate OH- ions in water

                                  Weak bases do not dissociate completely to generate OH- ions in water (later)

  Water      
NaOH(s) g Na+(aq) + OH-(aq)
1 mole   1 mole   1 mole
Kw

=

[H+]  x  [OH-] = 1 x 10-14

Example:  0.1M solution of NaOH

[H+]

x

[OH-] = 1 x 10-14  

[H+]

x

0.1 = 1 x 10-14  
    [H+] = 1 x 10-14  
      0.1  
    [H+] = 1 x 10-13  
           
    pH = -log10[H+(aq)]  
    pH = -log10[1 x 10-13]  
    pH = 13  

There is a quicker way:

[H+]

x

[OH-] = 1 x 10-14  

-log10 [H+]

x

-log10 [OH-] = -log10 (1 x 10-14)  

pH

+

pOH = 14  

Qu  1 - 3  P149

Strong and weak acids, calculating the pH for weak acids

Acid - base equilibria:

Strong acids:

  Water      
HA(aq) g H+(aq) + Cl-(aq)

Weak acids:

  Water      
HA(aq) D H+(aq) + Cl-(aq)
1 mole   0 moles   0 moles
0.99 moles   0.01 moles   0.01 moles
  Water      
CH3COOH(l) D H+(aq) + CH3COO-(aq)
0.99 moles   0.01 moles   0.01 moles

The acid dissociation constant, Ka

CH3COOH(l) D H+(aq) + CH3COO-(aq)
Ka

=

[H+]  x  [CH3COO-]
[CH3COOH]

Units of Ka

  Ka = [H+]  x  [CH3COO-]    
      [CH3COOH]    
           
  Ka = mol dm-3  x  mol dm-3  
      mol dm-3  
           
  Ka = mol dm-3  x  mol dm-3  
      mol dm-3  
           
  Ka = mol dm-3    
           
  Large Ka   Large dissociation - Strong acid
  Small Ka   Small dissociation - Weak acid

 

Ka and pKa - a compressed scale:

Ka

=

[H+]  x  [A-]
[HA]

pKa  =  -log10[Ka]

Calculating the pH for weak acids

HA(aq) D H+(aq) + A-(aq)
1M   0M   0M
0.99M   0.01M   0.01M

    Step 1:  [H+(aq)] must be calculated from the acid equilibrium expression, Ka:

Ka

=

[H+]  x  [A-]
[HA]

        1)    For a weak acid:    [H+(aq)]    =    [A-(aq)]    - as both have been formed from the dissociation of the acid.

        2)    [HA(aq)] remains unchanged - a weak acid dissociates so little that the concentration remains virtually the same.

Ka

=

[H+]2
[HA]
[H+]2

=

Ka  x  [HA]
[H+]

=

(Ka  x  [HA])0.5

    Step 2:  [H+(aq)] is put in the pH formula:  pH  =  -log10[H+(aq)]

Example:  Calculate the pH of a 0.100 mol dm-3 ethanoic acid, Ka = 1.7 x 10-5 mol dm-3

CH3COOH(l) D H+(aq) + CH3COO-(aq)

    Step 1:  [H+(aq)] must be calculated from the acid equilibrium expression, Ka:

Ka

=

[H+]  x  [CH3COO-]
[CH3COOH]

        1)    [H+(aq)]    =    [CH3COO-]

        2)    [CH3COOH] remains unchanged

Ka

=

[H+]2
[CH3COOH]
1.7 x 10-5

=

[H+]2
0.100
[H+]2

=

1.7 x 10-5  x  0.100
[H+]

=

1.304  x  10-3 mol dm-3

pH 

=  -log10[H+(aq)]

pH 

=  -log10 [1.3  x  10-3]

pH 

=  2.88

Summary:

    Step 1:  Calculate [H+(aq)]

[H+]

=

(Ka  x  [HA])0.5

Applying assumptions:

        1)    [H+(aq)]    =    [A-(aq)]

        2)    [HA(aq)] remains unchanged

    Step 2:  Calculate pH

                    pH  =  -log10[H+(aq)]

Calculating Ka for weak acids

        1)    pH of the weak acid

        2)    Concentration of the weak acid

    Step 1:  Calculate [H+(aq)]

[H+]

=

10 - pH

    Step 2:  Calculate Ka

Ka

=

[H+]2
[HA]

Example - Calculate Ka for 0.03 mol dm-3 methanoic acid, pH = 2.66

    Step 1:  Calculate [H+(aq)]

[H+]

=

10 - pH
[H+]

=

10 - 2.66
[H+]

=

2.19  x  10-3

    Step 2:  Calculate Ka

Ka

=

[H+]2
[HA]
Ka

=

[2.19  x  10-3]2
[0.03]
Ka

=

1.6  x  10-4 Mol dm-3

 

Qu 1 - 3  P 143  /  Qu 1 - 2  P145

Buffer solutions

Practical

  A buffer solution minimises the change in pH with the small addition of acids or bases

            1)  Weak acid, HA

            2)  Conjugate base of the weak acid, A-

CH3COOH(aq) D H+(aq) + CH3COO-(aq)  
CH3COO-  Na+(aq) g Na+(aq) + CH3COO-(aq)  

Alternativily:

HCOOH(aq) + NaOH(aq) g HCOO-  Na+(aq) + H2O
HCOOH(aq) D H+(aq) + HCOO-(aq)  
HCOO-  Na+(aq) g Na+(aq) + HCOO-(aq)  

How does a buffer act?

HA(aq) D H+(aq) + A-(aq) Reaction 1
H+(aq) + OH-(aq) g H2O(l) Reaction 2

Addition of and acid, H+ ions:

Addition of an alkali, OH- ions:

Overall:

                               

Qu 1,2  P151

pH values of buffer solutions

Calculations involving buffer solutions

            1)    Ka of the weak acid

            2)    Equilibrium concentrations of the conjugate acid - base pair

Method 1:

    1)    Write the reaction:

HA(aq) D H+(aq) + A-(aq)  

    2)    Write the equilibrium expression, Ka

Ka

=

[H+]  x  [A-]
[HA]

    3)    Rearrange the equation to calculate the hydrogen concentration

[H+]

=

Ka    x [HA]
[A-]

            1)  [HA] remains unchanged as it dissociates by such a small amount

            2)  [A-]  =  [NaA]  as is dissociates fully

    4)    calculate the pH

pH 

=  -log10[H+(aq)]

Example:

    1)    Write the reaction:

C2H5COOH(aq) D H+(aq) + C2H5COO-(aq)  

    2)    Write the equilibrium expression, Ka

Ka

=

[H+]  x  [C2H5COO-]
[C2H5COOH]

    3)    Rearrange the equation to get the hydrogen ion concentration:

[H+]

=

Ka    x [C2H5COOH]
[C2H5COO-]

 

[H+]

=

1.3  x  10-5    x 0.600
0.800

 

[H+]

=

9.75  x  10-6  
 

    6)    calculate the pH

pH 

=  -log10[H+(aq)]

pH 

=  -log10 [9.75  x  10-6]

pH 

=  5.01

Method 2:

Proof

pH

=

pKa

+    log

[A-]
[HA]

 

pH

=

- logKa

+    log

[C2H5COO-]
[C2H5COOH]

 

pH

=

- log  1.3  x  10-5

+    log

[0.800]
[0.600]

 

pH 

=

5.01

The carbonic acid - hydrogen carbonate buffer system

        Carbonic acid:    H2CO3                   and its conjugate base:    HCO3-                       

H2CO3(aq) D H+(aq) + HCO3-(aq) Reaction 1
H+(aq) + OH-(aq) g H2O(l) Reaction 2

Addition of and acid, H+ ions:

Addition of an alkali, OH- ions:

Overall:

                                               

CO2(aq) D CO2(g)

Stretch

Qu  1,2  P153

Neutralisation - titration curves

Practical

Titrations for AS chemistry

Key features of titration curves - for a base added to an acid:

  • When the base is first added, the pH increases very slightly due to the large excess of acid, the gradually increases.

  • Within a drop, the pH increases sharply from below 7 to above 7, the middle of this section is the equivalence point.

  • With continual addition of base, the increase in pH tails of until the increases is again very slight.

 

Choosing the indicator

HIn(aq) D H+(aq) + In-(aq)

Indicator and titration curves

  • The equivalence point of any neutralisation reaction is pH7

  • The end point (colour change) in indicators do not necessarily correspond to the equivalence point

  • Look at methyl orange - it changes colour in the pH range of 2.8 - 4.0.

  • This means that the end point does not necessarily correspond to the equivalence point for some neutralisation reactions

 

Strong acid Strong base Strong acid Weak base Weak acid Strong base Weak acid Weak base
Phenylphthalein - Suitable Phenylphthalein - Unsuitable Phenylphthalein - Suitable Phenylphthalein - Unsuitable
Methyl orange - Suitable Methyl orange - Suitable Methyl orange - Unsuitable Methyl orange - Unsuitable

Stretch

Qu 1-2  P155

Neutralisation - enthalpy changes

Standard enthalpy change of neutralisation

Full: HCl(aq) + NaOH(aq) g NaCl(aq) + H2O(l)
Ions: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) g Na+(aq) + Cl-(aq) + H2O(l)
Ionic: H+(aq) + OH-(aq) g H2O(l)
 

Is the energy change when aq acid is neutralised by aq base forming 1 mole of water under standard conditions

Determination of enthalpy change of neutralisation:

Example:

25cm3 of 2M nitric acid is added to 25cm3 of 2M potassium hydroxide.  The temperature increases from 22.0 oC to 35.5 oC. 

Calculate the enthalpy change of neutralisation:

1)  The energy change:

Q = m x c x DT  
Q = 50 x 4.18 x 13.5  
Q = 50 x 4.18 x 13.5  
Q = 2821.5 Joules Exothermic so will be negative:
Q = - 2821.5 Joules  

2)  Calculate the number of moles used:

Moles (HNO3) = C x V
   
    1000
Moles (HNO3) = 2 x 25
   
    1000
Moles (HNO3)   0.0500
Moles (KOH) = C x V
   
    1000
Moles (KOH) = 2 x 25
   
    1000
Moles (KOH)   0.0500

3)  Calculate the amount of energy exchanged per mole of water, enthalpy change of neutralisation:

                                                              Enthalpy = Energy

                                                                                    Moles

 

                                                              Enthalpy =  - 2821.5

                                                                                      0.0500

 

                                                              Enthalpy = - 56430 j Mol-1

 

                                                              Enthalpy = - 56.4 Kj Mol-1

Practical

Comparisons of enthalpy changes of neutralisation of different acids:

Acid

Reaction

DHqneut  /  Kj Mol-1

Hydrochloric acid
Full: HCl(aq) + NaOH(aq) g NaCl(aq) + H2O(l)
Ionic: H+(aq) + OH-(aq) g H2O(l)
- 57.9
Nitric acid
Full: HNO3(aq) + NaOH(aq) g NaNO3(aq) + H2O(l)
Ionic: H+(aq) + OH-(aq) g H2O(l)
- 57.6
Hydrobromic acid
Full: HBr(aq) + NaOH(aq) g NaBr(aq) + H2O(l)
Ionic: H+(aq) + OH-(aq) g H2O(l)
- 57.6
Ethanoic acid
Full: CH3COOH(aq) + NaOH(aq) g CH3COONa(aq) + H2O(l)
Ionic: H+(aq) + OH-(aq) g H2O(l)
- 57.1

Stretch

Qu 1 - 2  P157  /  Qu 6 - 9  P159  /  5 - 9  P163