a “recall the Brønsted-Lowry theory and use it to identify acid base behaviour, and identify acid base conjugate pairs and relate them by means of suitable equations”
Brønsted-Lowry th eory An acid is a proton donor:
E.g. HCl (aq) +H2O (l) ⇌ H3O+ (aq) + Cl- (aq)
In the forward reaction the HCl is acting as an acid.
A base is a proton acceptor:
E.g. NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
In the forward reaction the NH3 is acting as a base.
Acid-base conjugate pairs
E.g. HCl (aq) + H2O (l) ⇌ H3O+ (aq) + Cl- (aq) Acid 1 Base 2 Acid 2 Base 1
Conjugate base 1 (Cl- ) is related to acid 1 (HCl) because it has been formed by the loss of a proton.
Conjugate acid 2 (H3O+) is related to base 2 (H2O) because it has been formed by the gain of a proton.
More examples of acid – base pairs. Identify the acid-base conjugate pairs in these equilibria.
H2SO4 + H2O ⇌ H3O+ + HSO4
-HNO3 + H2SO4 ⇌ H2NO3+ + HSO4
-HClO4 + H2O ⇌ H3O++ ClO4
-c “understand the terms ‘strong’ and ‘weak’ as applied to acids and bases”
Basicity of an acid
The basicity of an acid can be thought of as the number of replaceable hydrogens that it has:
HCl monobasic
H2SO4 dibasic H3PO4 tribasic
CH3COOH tetrabasic monobasic (why?) Strong and weak acids
Strong acids - fully ionised in aqueous solution.
E.g. HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)
The equilibrium lies to the right and the [H3O+] is very high. Strong acids have weak conjugate bases.
Weak acids - only partially ionised in aqueous solution.
E.g. CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)
The equilibrium lies to the left and the [H3O+] is very low. CH3COO- is the strong conjugate base of the weak acid CH3COOH. Weak acids have strong conjugate bases.
Stro ng and weak alkalis
Strong alkalis - fully ionised in aqueous solution.
E.g. NaOH (aq) Na+ (aq) + OH- (aq)
The equilibrium lies to the right and the [OH-] is very high. Strong alkalis have weak conjugate acids.
Weak alkalis - only partially ionised in aqueous solution.
E.g. NH4OH (aq) ⇌ NH4+ (aq)+OH- (aq)
The equilibrium lies to the left and the [OH-] is very low. NH4+ is the
strong conjugate acid of the weak base NH4OH. Weak alkalis have strong conjugate acids.
33691600 Harrow Chemistry 24
b “d efine pH”
d “define Ka and Kw and recall their units”
e “define pKa and pKw”
f “calculate the pH of solutions of strong acids and strong bases”
g “calculate pH of solutions of weak acids given Ka
and vice versa”
Ionic product of water H2O (l) ⇌ H+ (aq) + OH- (aq) Kc = [H + ][OH- ]
[H2O]
[H2O] is so large that it remains effectively constant.
Kw = [H+][OH-], where Kw incorporates Kc and [H2O]
At 298K and 1 atm, Kw = 1 X 10-14 moles2 dm-6
Kw = [H+][OH-] = 1 X 10-14 moles2 dm-6 [pKw = -log10Kw = 14]
For a neutral solution [H+]=[OH-] ([H+]= 1 X 10-7 mole dm-3) For an acid solution [H+]>[OH-] ([H+]> 1 X 10-7 mole dm-3) For an alkali solution [H+]<[OH-] ([H+]< 1 X 10-7 mole dm-3) pH
pH= -log10[H3O +]/moles dm-3 More simply, pH=
-log10[H+] This is the definition of pH!!
Stro ng acids
E.g. calculate the pH of 0.1M HCl.
HCl - strong acid fully ionised.
So [H+] = 0.1 moles dm-3 pH = - log10(0.1) = 1
E.g. calculate the pH of 0.2M H2SO4.
H2SO4 - strong acid ∴ fully ionised* (assume H2SO4 →2H+ + SO42-) So [H+] = 2 X 0.2 moles dm-3
pH = - log10(0.4) = 0.398 Strong alkalis
E.g. calculate the pH of 0.01M NaOH.
NaOH - strong acid ∴ fully ionised.
So [OH-] = 0.01 moles dm-3
Since Kw = [H+][OH-] = 1 X 10-14 moles2 dm-6
∴[H+] =(1 X 10-14)/0.01 = 1 X 10-12 moles dm-3
⇒ pH = 12 Weak acids (1)
E.g. calculate the pH of 0.1M CH3COOH given Ka = 1.7 X 10-5 moles dm-3. Eqm. CH3COOH (aq) ⇌ CH3COO-(aq) + H+(aq)
∴
Ka=[CH3COO- ][H ] +[CH3COOH]
1.7 X 10-5 =[H + ] 2
(0.1)
[H+] = √ (1.7 X 10-5 X 0.1) = 1.3 X 10-3 moles dm-3
∴ pH = -log10(1.3 X 10-3) = 2.88
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Note the assumptions made:
➢ CH3COOH is a weak acid, so only partially ionised ∴ [CH3COOH] is effectively constant/unaltered
➢ [CH3COO-] = [H+]
*Dr. Beavon, Chief Examiner says not!! Check his website:
www.rod.beavon.clara/net.learning.
htm
Weak acids (2)
E.g. calculate the ionisation constant, Ka, of 0.01M CH3CH2COOH given that the pH of the acid is 3.44. What is the pKa value for propanoic acid?
Eqm. CH3CH2COOH ⇌ CH3CH2COO- + H+ Ka=[CH3CH2COO- ][H + ]
[CH3CH2COOH]
∴Ka = [H+ ] 2
[CH3CH2COOH]
Ka = (3.6 X 10-4 ) 2
0.01
= 1.30 X 10-5 moles dm-3 Since pKa = -log10Ka
∴ pKa = -log10(1.30 X 10-5) = 4.89 Effect of temperature on pH
The ionisation of water is slightly endothermic.
H2O (l) ⇌ H+ (aq) + OH- (aq) Kw = [H+][OH-]
At 25 oC and 1 atm, Kw = 1 X 10-14 moles2 dm-6 At 50 oC and 1 atm, Kw = 1.2 X 10-14 moles2 dm-6
So [H+] and [OH-] increase with increasing temperature. Thus the
equilibrium for the ionization of water moves to the right with increasing temperature. This means that the equilibrium must be endothermic in the forward direction.
At 25 oC [H+] = (1.0 x 10-14) moles dm-3 At 50 oC [H+] = √(1.2 x 10-14) moles dm-3
So, [H+] increases with increasing temperature.
Also pH (=-log10[H+]) decreases as [H+] increases.
But the solution does not become more acidic as the temperature increases because there is a simultaneous increase in [OH-] as the
temperature increases, thus “cancelling out” the effect of the increase in [H+].
33691600 Harrow Chemistry 28
Note the assumptions made:
➢ CH3CH2COOH is a weak acid, so only partially ionised
∴ [CH3CH2COOH] is effectively constant/unaltered
➢ [CH3CH2COO-] = [H+]
(From pH = 3.44)
Since pH = -log10[H+] = 3.44∴ [H+]= 3.6 X 10–4 moles
5 10 15 20 25 30
Volume of alkali added/cm3
h “understand the principles involved in acid base titrations”
i “recall the sketch curves for the variation in pH during the following titrations; strong acid – strong base, weak acid – strong base and strong acid – weak base”
j “use titration curves to determine Ka for a weak acid”
k “explain the choice of a suitable indicator for an acid-base titration given pKInd values”
Acid-base titrations
pH changes during an acid-base ti tration (0.1 mole dm-3 concentrations) Strong acid vs strong alkali (e.g. HCl and NaOH)
25 cm3 of 0.1M HCl in a conical flask and add small portions of 0.1M NaOH from a burette measuring the pH after each addition. Plot a graph of pH against volume alkali added.
pH of 0.1M HCl = 1 pH of NaOH
NaOH in excess
Equivalence point - halfway point of the straightest vertical part of the graph (pH = 7)
5 10 15 20 25 30 Volume of acid added/cm3 5 10 15 20 25 30
Volume of acid added/cm3
Weak acid vs strong alkali (e.g. CH3COOH and NaOH)
Place 25 cm3 of 0.1M HCl in a conical flask and add small portions of 0.1M NaOH from a burette measuring the pH after each addition. Plot graphs of pH against volume alkali added.
Weak alkali vs strong acid (e.g. NH4OH v s HCl)
[Apologies for the “switch” but this was the only set of curves that was available at the time].
Place 25 cm3 of the alkali in a conical flask and add small portions of acid from a burette measuring the pH after each addition. Plot graphs of pH against volume alkali added.
Equivalence points during acid-base titrations
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pH of vertical part of the graph (pH =
You will have noticed that the pH values at equivalence are not always equal to 7.
Explanations:
HCl vs NaOH: resulting salt NaCl (aq) - the salt of a strong acid and a strong alkali.
Ions present in aqueous solution:
Na+ Cl- (NaCl(aq)) OH- H+ (H2O)
Na+ will not remove OH- as NaOH as NaOH is a strong electrolyte ∴ fully ionised in aqueous solution.
Cl- will not remove H+ as Cl- is a weak base ∴ does not remove H+. So [H+] = [OH-] ∴ the salt is neutral (pH = 7 at 25OC)
HCl vs NH4OH: resulting salt NH4Cl (aq) - the salt of a strong acid and a weak alkali.
Ions present in aqueous solution:
NH4+ Cl- (NH4Cl(aq)) OH- H+ (H2O)
NH4+ will remove OH- as NH4OH as NH4OH is a weak electrolyte ∴ only partially ionised aqueous solution.
Cl- will not remove H+ as Cl- is a weak base does not remove H+. So [H+] > [OH-] ∴ the salt is acidic (pH < 7 at 25OC)
CH3COOH vs NaOH: resulting salt CH3COONa (aq) - the salt of a weak acid and a strong alkali.
Ions present in aqueous solution:
Na+ CH3COO- (CH3COONa (aq)) OH- H+ (H2O)
Na+ will not remove OH- as NaOH as NaOH is a strong electrolyte ∴ fully ionised in aqueous solution.
CH3COO- will remove H+ as CH3COO- is the strong conjugate base of a weak acid ∴ removes H+ as unionised CH3COOH.
So [H+] < [OH-] ∴ the salt is alkaline (pH > 7 at 25OC).
Acid - base indicators
An acid-base indicator is (usually) a weak acid whose conjugate base is a different colour from the acid itself.
Use HIn as a general formula for an indicator:
HIn ⇌ H+ + In -Colour Colour
A B
During a titration:
(1) Acid (e.g. HCl) + indicator in conical flask colour A predominates as H+ ions from HCl cause indicator equilibrium to shift to the left.
Alkali (e.g.NaOH) added from a burette removes H+ ions from HCl (H+ + OH- H2O) until no more H+ ions from acid left. The next drop of alkali must then remove H+ ions from the indicator equilibrium thus causing the equilibrium to shift to the right ∴ colour B
predominates.
(2) Alkali (e.g. NaOH) + indicator in conical flask colour B
predominates as OH- ions from NaOH cause indicator equilibrium to shift to the right because they remove H+ ions as water (H+ + OH- H2O).
Acid (HCl) added from a burette removes OH- ions from NaOH as water until no more OH- ions from alkali left. The next drop of acid must then increase the H+ ions in the indicator equilibrium thus causing the equilibrium to shift to the left
colour A predominates.
Acid - base indicators – quantitative approach HIn ⇌ H+ + In
-Colour -Colour A B
KInd = [H + ] X [In- ] [HIn]
[H+] = KInd X [HIn]
[In-] When [Hin] = [In-] [H+] =KInd
pH = pKInd
So, the solution changes colour when its pH = pKInd. Choice of indicators
3 common indicators :
33691600 Harrow Chemistry 32
5 10 15 20 25
Indicator pH range Colour in acid
(Hin) Colour in alkali (In-)
Methyl orange 3.2 - 4.5 Red Yellow
Phenolphthalein 8.2 - 10.0 Colourless Pink Bromothymol
blue 6.0 - 7.0 Yellow Blue
The indicator should change colour anywhere on the straight vertical portion of the curve.
So for the three different types of titration:
Titration pH range on curve† Indicator(s) Strong acid
-Strong alkali 4 - 10 All three
Strong acid - Weak
alkali 4 - 8 Methyl orange or
bromothymol blue Weak acid - Strong
alkali 6 - 10 Phenolphthalein or
bomothymol blue Weak acid – Weak
alkali For weak acids and weak alkalis - equivalence is not very sharply defined difficult to use an indicator (anyway, indicators are weak acids or bases!)
†On the straightest vertical portion of the curve.
We can obtain the pKa for the weak acid from the weak acid – strong alkali curve:
Find the volume of alkali needed for equivalence (e.g. 25 cm3) and divide by 2 (∴12.5cm3) and then extrapolate to the pH value (e.g. 4.7)
l “explain the action of a buffer solution and calculate its pH from Compare
suitable data”
Buffer solutions - definition/constituents
A buffer solution is one which tends to resist changes in pH when small amounts of acid or alkali are added to it.
Constituents of a buffer
Weak acid/(usually sodium salt of) its conjugate base e.g.
CH3COOH/CH3COO-Na+
Weak base/(salt of) its conjugate base e.g. NH4OH/NH4+Cl -Buffer solutions - how do they work?
Buffer reactions in (aqueous solution):
CH3COO-Na+ → CH3COO- + Na+ (I)