__________________________________________________________________________________________________
LECTURE 13 – IONIC EQUILIBRIA: ACID–BASE EQUILIBRIA
LEARNING OUTCOMES
Candidates should be able to:
(a) show understanding of, and apply the Bronsted–Lowry theory of acids and bases, including the concept of conjugate acids and conjugate bases
(b) explain qualitatively the differences in behaviour between strong and weak acids and bases in terms of the extent of dissociation
(c) explain the terms pH; Ka; pKa; Kb; pKb; Kw and apply them in calculations, including the relationship Kw = KaKb
(d) calculate [H+(aq)] and pH values for strong acids, weak monobasic (monoprotic) acids, strong bases, and weak monoacidic bases
(e) explain the choice of suitable indicators for acid–base titrations, given appropriate data (f) describe the changes in pH during acid–base titrations and explain these changes in terms of
the strengths of the acids and bases
(g) (1) explain how buffer solutions control pH
(2) describe and explain their uses, including the role of H2CO3/HCO3– in controlling pH in blood
(h) calculate the pH of buffer solutions, given appropriate data
CONTENT
1 Understanding Acids and Bases
Bronsted–Lowry theory of acids and bases
2 Calculations Involving Acids and Bases
pH, pOH and the ionic product of water, Kw Dissociation constants, Ka and Kb
Salt hydrolysis
3 Buffer Solutions
Buffer action (Independent Learning – Section 3.2 )
pH of buffer solutions
Blood as a buffer system (Independent Learning – Sections 3.5, 3.6)
4 Acid–base Titrations
pH indicators (Independent Learning – Sections 4.1, 4.2)
Titration curves
REFERENCES
1. Peter Cann & Peter Hughes (2002). Chemistry for Advanced Level. John Murray. Chapters 6 & 12. 2. Martin S. Silberberg (2006). Chemistry – The Molecular Nature of Matter & Change,
1 UNDERSTANDING ACIDS AND BASES
1.1 Bronsted–Lowry Theory of Acids & Bases
An acid is defined as any species which donates a proton, H+. An acid must thus contain H in its formula. E.g. HNO3 and H2PO4 A base is defined as any species which accepts a proton, H+.
A base must contain a lone pair of electrons to bind the H+ ion (via dative bond). E.g. NH3, CO32, F and OH
A Bronsted–Lowry acid–base reaction involves the transfer of a proton from an acid to a base.
Acid–base reactions do not only occur in aqueous solutions. They can also occur between gases, in non–aqueous solutions and in heterogeneous mixtures.
EXERCISE 1
Identify the Bronsted acid and base in each of the following (forward) reactions:
(a) HNO3 + NH3 NO3− + NH4+ acid base (b) HNO3 + H2O H3O+ + NO3– acid base (c) NH3 + H2O ⇌ NH4+ + OH– base acid (d) HNO3 + H2SO4 H2NO3+ + HSO4– base acid
What can you say about the acid–base nature of water?
H2O is amphoteric (can act as an acid or a base, depending on the other substance present).
Note:
In aqueous solution, H+ (a ‘bare’ proton) does not exist on its own. Instead, it forms a dative bond with a water molecule to form H3O+, called hydronium, hydroxonium or oxonium ion. Chemists often use H+(aq) and H3O+(aq) interchangeably to refer to the solvated H+ ion.
Other definitions of acids and bases:
Arrhenius acid–base definition
An acid is a substance that dissociates in water to produce H+(aq).
A base is a substance that dissociates in water to produce OH(aq).
Lewis acid–base definition
Acids are species that accept an electron pair, e.g. AlCl3 Bases are species that donate an electron pair e.g. NH3
1.2 Conjugate Acid–Base Pair
Consider the following acid–base reaction:
CH3CO2H + H2O CH3CO2 + H3O+
acid base conjugate base conjugate acid
In the forward reaction:
CH3CO2H acts as a Bronsted acid, donating a proton to the Bronsted base, H2O.
In the reverse reaction:
CH3CO2 acts as a Bronsted base, accepting a proton from the Bronsted acid, H3O+. CH3CO2H and CH3CO2 form a conjugate acid–base pair;
H2O and H3O+ form another conjugate acid–base pair.
In general,
acid 1 + base 2 base 1 + acid 2
An acid after donating a proton forms its conjugate base.
A base after accepting a proton forms its conjugate acid.
A conjugate acid–base pair constitutes two species which differ from each other by a proton.
In any acid–base reaction, there are two conjugate acid–base pairs.
EXERCISE 2
Identify the conjugate acid–base pairs in each of the following reactions:
(a) HNO3 + NH3 NO3− + NH4+ HNO3 / NO3− and NH3 / NH4+
(b) HNO3 + H2O H3O+ + NO3– HNO3 / NO3− and H2O / H3O+
(c) NH3 + H2O ⇌ NH4+ + OH– NH3 / NH4+ and H2O / OH−
(d) HNO3 + H2SO4 H2NO3+ + HSO4– HNO3 / H2NO3+ and H2SO4 / HSO4– conjugate acid–base pair
1.3 Strong and Weak Acids and Bases
A strong acid is one that dissociates completely in water to give H3O+. It gives up a proton more readily compared to the H3O+ ion.
e.g. HCl + H2O Cl + H3O+
strong acid conjugate base
The conjugate base of a strong acid has a very low tendency to accept a proton.
So the reverse reaction is negligible single–headed arrow ().
Other examples of strong acids: HNO3, HBr, HI, HClO3, HClO4, H2SO4
A weak acid is one that dissociates partially in water to give H3O+. It gives up a proton less readily compared to the H3O+ ion.
e.g. CH3CO2H + H2O CH3CO2 + H3O+
weak acid conjugate base
The conjugate base of a weak acid has a tendency to accept a proton.
So the reverse reaction occurs to some extent double–headed arrow (⇌).
A mixture of undissociated CH3CO2H molecules and the ions exists in solution. Other examples of weak acids: H2SO3, HNO2, H2CO3, H2S, HF, HCN, most organic acids
A strong base is one that dissociates completely in water to give OH−.
e.g. NaOH Na+ + OH−
Other examples of strong bases: KOH, Ba(OH)2
A weak base is one that dissociates partially in water to give OH−.
e.g. NH3 + H2O ⇌ NH4+ + OH−
Other examples of weak bases: hydrazine N2H4, amines such as CH3NH2, C6H5NH2
The strength of an acid (or base) is different from the concentration of the acid (or base).
Concentration of an acid (or base) = No. of moles of acid (or base) dissolved in 1 dm3 of solution.
EXERCISE 3
Match each acid on the left with the correct description on the right.
0.001 mol dm–3 HNO3 concentrated but weak acid 4.0 mol dm–3 CH3CO2H dilute but strong acid 0.001 mol dm–3 CH3CH2CO2H dilute and weak acid
2 CALCULATIONS INVOLVING ACIDS AND BASES
2.1 Concept of pH, pOH and the Ionic Product of Water
A) pH and pOH
The concentration of H+ and OH in an aqueous solution is usually quite small. A convenient indication of the concentration of H+(aq) is to express it in terms of its negative logarithm to base 10.
pH = lg [H+
(aq)]
Hence, [H+(aq)] = 10−pH
E.g. for 0.01 mol dm–3 HCl, [H+(aq)] = 10–2 mol dm–3 and pH = 2.
Note:
pH is a measure of the (total) concentration of H+ in a solution. It is not a measure of the strength of the acid unless the two acids being compared have the same initial concentration (see Section 2.6).
Similarly, a convenient indication of the concentration of OH(aq) is to express it in terms of its negative logarithm to base 10.
pOH = lg [OH(aq)]
and [OH−(aq)] = 10−pOH
E.g. for 0.01 mol dm–3 NaOH, [OH(aq)] = 10–2 mol dm–3 and pOH = 2.
B) Ionic product of water, Kw
Pure water conducts electricity slightly. This indicates the presence of trace concentrations of ions which arise from the very slight self–ionisation or autoionisation of water.
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
A simpler way of representing the equilibrium is:
H2O(l) H+(aq) + OH–(aq)
Kc =
[H+][OH–] [H2O]
Since the extent of ionisation is very small, [H2O] is almost constant (at 55.6 mol dm–3).
Therefore, Kc x [H2O] = [H+][OH–]
or Kw = [H+][OH–] where Kw is a constant known as the ionic product of water
Taking –lg on both sides of equation:
–lg Kw = (–lg [H+]) + (–lg [OH–])
pKw = pH + pOH * remember the symbol ‘p’ means ‘–log10’
The value of Kw depends only on temperature. It is always a constant at constant temperature.
Experimentally, at 25oC, Kw = 1.0 x 10–14 mol2 dm–6 (value is given in the Data Booklet)
pKw = pH + pOH = 14
C) Relationship between [H+(aq)], [OH–(aq)] and pH
In pure water as well as any aqueous solution, the following equilibrium exists:
H2O(l) H+(aq) + OH–(aq)
Thus, an acidic solution contains a few hydroxide ions, and an alkaline solution contains a few hydrogen ions.
The product of [H+] and [OH–] always equals Kw since Kw is an equilibrium constant and its numerical value is only affected by temperature.
When [H+(aq)] = [OH–(aq)] neutral solution [H+(aq)] > [OH–(aq)] acidic solution [H+(aq)] < [OH–(aq)] alkaline solution
At 25o C, since Kw = 10–14 mol2 dm–6
Type of solution [H+] / mol dm–3 [OH–] / mol dm–3 pH pOH Neutral = 1.0 x 10 –7 = 1.0 x 10 –7 = 7.00 = 7.00 Acidic > 1.0 x 10–7 < 1.0 x 10 –7 < 7.00 > 7.00 Basic < 1.0 x 10 –7 > 1.0 x 10 –7 > 7.00 < 7.00
EXERCISE 4 [Ref: N96/4/8, N03/1/1, N95/4/33]
Like other equilibrium constants, the value of Kw depends only on temperature:
Temperature / oC Kw / mol2 dm–6 [H+] = [OH–] =
K
w / mol dm–3 0 10 25 40 60 1.5 10–15 3.0 10–15 1.0 10–14 3.0 10–14 9.5 10–14 3.9 10–8 5.5 10–8 1.0 10–7 1.7 10–7 3.1 10–7State whether the following statements are true or false.
(i) The ionic dissociation of water, H2O(l) ⇌ H+(aq) + OH–(aq), is an endothermic process.
T
(ii) The dissociation of water increases by a factor of 6 between 10 oC and 60 oC. F
(iii) The association of water molecules by hydrogen bonding increases as temperature increases.
F
(iv) When water is heated, the concentration of H+(aq) increases. T
(v) Water remains a neutral liquid at 60 oC. T
(vi) When water is heated, the concentration of OH–(aq) increases. T (vii) The pH of pure water at temperatures greater than 25 oC is greater than 7. F
(viii) In pure water, the concentration of H+(aq) is higher than that of OH–(aq) at 60 oC.
2.2 pH of Strong Acid/Strong Base Solutions
Strong acids and strong bases are fully dissociated, hence
[H+] = [strong monobasic acid] and [H+] = 2 x [strong dibasic acid] Similarly, [OH−] = [strong monoacidic base] and [OH−] = 2 x [strong diacidic base]
Note: If the concentration of the acid or base is 10–7 mol dm–3 or lower, the contribution of H+ or OH– from auto–ionization of H2O becomes significant. We will need to include it in the calculation.
EXERCISE 5
(a) An aqueous solution contains 1.0 10–5 mol dm–3 of hydrogen iodide (a strong acid). Calculate the concentration of H+ in mol dm–3, and hence the pH of this solution at 25 oC.
Ans: [H+] = 1.00 10–5 mol dm–3, pH = 5.00 HI(aq) H+(aq) + I–(aq)
[H+] = [HI] = 1.0 10–5 mol dm–3 pH = –lg [H+] = –lg (1.0 10–5) = 5.00
(b) An aqueous solution contains 0.10 mol dm–3 of calcium hydroxide (a strong base). Calculate the concentration OH– in mol dm–3, and hence the pH of this solution at 25 oC.
Ans: [OH–] = 0.200 mol dm–3, pH = 13.3
Ca(OH)2(aq) Ca2+(aq) + 2OH–(aq) [OH–] = 2 [Ca(OH)2] = 0.200 mol dm–3 pOH = –lg [OH–] = –lg 0.2 = 0.699 pH = 14 – pOH = 14 – 0.699 = 13.3
(c) An aqueous solution contains 0.010 mol dm–3 of Ba(OH)2 (a strong base) and 0.050 mol dm–3 of KOH. Calculate the pH of this solution at 25 oC.
Ans: pH = 12.8
Ba(OH)2(aq) Ba2+(aq) + 2OH–(aq) KOH(aq) K+(aq) + OH–(aq)
Total [OH–] = 20.010 + 0.050 = 0.070 mol dm–3 pH = 14 – pOH = 14 + lg 0.070 = 12.8
(d) Solution X contains HCl, it has pH 2.2. Solution Y has the same volume as solution X, it contains HNO3 and has pH 1.6. The two solutions are mixed. Calculate the pH of the acid mixture.
Ans: 1.80 Total [H+] = 2 10 102.2 1.6 = 0.0157 mol dm–3 pH = – lg 0.0157 = 1.80
(e) An aqueous solution contains 1 x 10–8 mol dm–3 of HNO3. Calculate the pH of this solution. Ans: 6.96 HNO3 (aq) H+ (aq) + NO3– (aq)
H2O (l) ⇌ H+ (aq) + OH– (aq) (auto–ionization of water) Hence, [H+] = 10–8 + 10–7 = 1.1 x 10–7 mol dm–3 pH = – lg (1.1 x 10–7) = 6.96
2.3 Dissociation Constants
A) Acid dissociation constant, Ka
A weak acid HA dissociates partially in water to give H3O+.
HA + H2O A + H3O+
In dilute aqueous solution, [H2O] is almost constant (at 55.6 mol dm–3). The equilibrium constant may be written as: Ka = [H+]eqm[A–]eqm mol dm–3 [HA]eqm and pKa = −lg Ka
For strong acids e.g. HCl, Ka is very large and not used.
Like other equilibrium constants, the value of Ka depends only on temperature.
The value of Ka indicates the extent to which the weak acid dissociates in water at the specified temperature.
To compare the strength of two weak acids, compare their Ka values.
Some Weak Acids and their Ka values
Equilibrium reaction Ka /mol dm–3
CH3CO2H + H2 O ⇌ CH3CO2– + H3O+ 1.8 X 10–5 HCO2H + H2 O ⇌ HCO2– + H3O+ 1.8 x 10–4 C6H5CO2H + H2 O ⇌ C6H5CO2– + H3O+ 6.5 x 10–5 HCN + H2 O ⇌ CN– + H3O+ 4.9 x 10–10
B) Base dissociation constant, Kb
A weak base dissociates partially in water to give OH–.
B + H2O BH+ + OH–
In dilute aqueous solution, [H2O] is almost constant (at 55.6 mol dm–3). The equilibrium constant may be written as:
Kb =
[BH+]
eqm[OH–]eqm
mol dm–3
[B]eqm
and pKb = −lg Kb
For strong bases e.g. NaOH, Kb is very large and not used.
The value of Kb depends only on temperature.
The value of Kb indicates the extent to which the weak base dissociates in water at the specified temperature.
To compare the strength of two weak bases, compare their Kb values.
At the same temperature, larger Kb (smaller pKb) stronger base.
Some Weak Bases and their Kb values
Equilibrium reaction Kb /mol dm–3
NH3 + H2 O ⇌ NH4+ + OH– 1.8 X 10–5 CH3NH2 + H2 O ⇌ CH3NH3+ + OH– 4.4 x 10–4 HS– + H2 O ⇌ H2S + OH– 1.8 x 10–4 CO32– + H2 O ⇌ HCO3– + OH– 1.8 x 10–4
In aqueous solution, the two large classes of weak bases are nitrogen–containing molecules such as NH3 and the amines (e.g. CH3CH2NH2), and the anions or conjugate bases of weak acids (e.g. ethanoate ion, CH3CO2–).
C) Relationship between Ka, Kb and Kw
Consider the weak acid, CH3CO2H, and its conjugate base, CH3CO2. CH3CO2H ⇌ CH3CO2 + H+ Ka = [CH3CO2][H+] –––––––––––––––––––– (1) [CH3CO2H] CH3CO2 + H2O ⇌ CH3CO2H + OH– Kb = [CH3CO2H][OH] –––––––––––––––––––– (2) [CH3CO2] (1) X (2), Ka x Kb = [H+] [OH–] = Kw
Therefore, for any conjugate acid–base pair,
Ka x Kb = Kw
Taking –lg on both sides of equation:
(–lg Ka) + (–lg Kb) = –lg Kw
pKa + pKb = pKw * remember the symbol ‘p’ means ‘–log10’
Hence, we can determine the value of Ka, given the Kb value of its conjugate base and vice versa.
From Ka x Kb = Kw , we see that for a conjugate acid–base pair,
as the strength of the acid decreases (smaller Ka), the strength of its conjugate base increases (larger Kb) and vice versa.
E.g. CH3CO2H is a weak acid (Ka of CH3CO2H = 1.80 x 10–5 mol dm–3), its conjugate base CH3CO2– is a stronger base than water. Hence CH3CO2– has a tendency to accept a proton from water.
However, CH3CO2– is still a weak base (Kb of CH3CO2– = 5.56 x 10–10 mol dm–3).
2.4 Degree of Dissociation
The fraction of molecules which is ionised into ions in water is called the degree of dissociation, .
For an acid,
= amount ionised = [acid]dissociated initial amount [acid]initial
In general, for a strong acid or strong base, is close to 1 for a weak acid or weak base, << 1
2.5 pH of Weak Acid/Weak Base Solutions
Since weak acids and weak bases dissociate partially in water, [H+] or [OH−] in the solution will be
less than the concentration of the weak acid or weak base.
We use Ka or Kb to calculate the [H+] or [OH−] in the solution at equilibrium.
EXERCISE 6
Find (a) the pH and (b) the degree of dissociation of a solution of ethanoic acid, CH3COOH, of concentration 0.1 mol dm–3. [Given: Ka of CH3COOH = 1.8 x 10–5 mol dm–3]
Ans: 2.87; 1.34 10–2 or 1.34 % (a) HA H+ + A Initial conc 0.1 Change x + x + x Eqm conc 0.1 – x x x Ka = [H + ][A] = x 2 x 2 (assume x << 0.1) [HA] 0.1 – x 0.1 x = [H+] =
1
.
8
10
5
0
.
1
= 1.34 x 10–3 mol dm–3 pH = – lg (1.34 x 10–3) = 2.87 (b) α = 1.34 x 10–3/0.1 = 1.34 x 10–2 (1.34% dissociated) Note:The assumption (x << 0.1) is valid since a weak acid only dissociates partially. Check: x = 1.3325 x 10–3 (by solving quadratic equation) << 0.1
Solving quadratic equations is not required at ‘A’ Level.
Generally, for a weak monobasic acid,
Ka = [H+]eqm[A]eqm = [H + ]eqm2 [H + ]eqm2
[HA]eqm [HA]eqm [HA]initial
EXERCISE 7
The pH of a weak acid, HA, of concentration 0.100 mol dm–3 was found to be 3.0. Calculate the pKa of the acid.
Ans: 5.00
HA(aq) ⇌ H+(aq) + A– (aq)
Since pH = 3.0, [H+] = 10–3 mol dm–3 Ka = [H + ]eqm[A]eqm = (10 –3 ) (10–3) [HA]eqm 0.1 – 10–3 = 1.01 x 10–5 mol dm–3 pKa = – lg Ka = 5.00
Note: If [H+]eqm is known, the assumption [H+] << [HA] is not required in the calculations.
EXERCISE 8
Caffeine, C8H10N4O2, a stimulant found in coffee and tea, is a weak base.
C8H10N4O2 + H2O ⇌ (C8H11N4O2)+ + OH–
At 25 oC, a 0.020 mol dm–3 caffeine solution has a pH of 11.5.
(a) Write the Kb expression of caffeine and calculate its value at 25 oC. (b) Calculate the degree of dissociation, for caffeine at 25 oC.
Ans: 5.94 10–4 mol dm–3; 0.158 or 15.8 % (a) Kb = ] O N H [C ] ][OH O N H [C 2 4 10 8 2 4 11 8 pOH = 14 – pH = 14 – 11.5 = 2.5 [OH–] = 10–2.5 = 3.162 10–3 mol dm–3 Kb = 3 2 3 10 3.162 0.020 ) 10 (3.162 = 5.94 10–4 mol dm–3 (b) = 0.020 10 3.162 3 = 0.158
2.6 Strength of Acids and Bases
The strength of an acid (or base):
• is a measure of how readily the acid (or base) donates its proton (or accepts a proton). • depends on its structure [ref: Topics 15 & 17]
• is indicated by its Ka (or Kb) value [ref: p. 9–p.10]
• is different from the concentration of the acid (or base) [ref: p. 4 Ex 3]
Are pH or degree of dissociation, α, reliable indicators of the strength of an acid (or base)?
EXERCISE 9
In Exercise 6 (p.12), the pH and the degree of dissociation of a solution of ethanoic acid of concentration 0.1 mol dm–3 were found to be:
pH: ………. degree of dissociation, α: ……….
Now, find (a) the pH and (b) the degree of dissociation of ethanoic acid of concentration 0.001 mol dm–3. How have the pH and α been affected by the dilution?
(a) pH = 3.87 (using assumption) or pH = 3.90 (solving quadratic equation) (b) α = 0.134 (13.4% dissociated) or α = 0.125 (12.5% dissociated)
pH is higher (lower [H+] present) pH increases with dilution. α is higher α increases with dilution.
Since pH and degree of dissociation of an acid (or base) will change with change in concentration, they are not reliable indicators of acid (or base) strength.
Hence, the values of pH, α and Ka for a weak acid at a constant temperature will vary with dilution (i.e. volume of water increases) as follows:
(pH 7 as dilution ) (α 1 as dilution )
Ka does not vary with concentration. It is a constant at constant temperature. Therefore Ka is the
best indicator of the strength of a weak acid.
Volume of water Volume of water Volume of water
7
pH α Ka
EXERCISE 10 [Ref: J97/1/3(b)]
Each of the following reactions contains two Bronsted acids and two Bronsted bases. Suggest with brief reasoning which is the stronger acid and which is the stronger base.
(a) NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH(aq) Kc = 3.2 107 weaker base weaker acid stronger acid stronger base
Kc is very small, this suggests that the position of equilibrium lies very much to the left.
So NH4+ is a better proton donor than H2O, OH– is a better proton acceptor than NH3.
(b) C6H5O + CH3CO2H ⇌ C6H5OH + CH3CO2 Kc = 1.3 106 stronger base stronger acid weaker acid weaker base
Kc is very large, this suggests that the position of equilibrium lies very much to the right.
So CH3CO2H is a better proton donor than C6H5OH, C6H5O is a better proton acceptor than
CH3CO2.
Note:
In part (a), NH3 is a weak base (Kb = 1.80 x 10–5 mol dm–3) so its conjugate acid NH4+ is a stronger acid than water. Recall from Section 2.3C, Ka x Kb = Kw.
However, NH4+ is still a weak acid (Ka of NH4+ = 5.56 x 10–10 mol dm–3).
From part (b), we see that the stronger acid is able to protonate the conjugate base of the weaker acid.
Strong acids (e.g. HCl) are often used to liberate the weak acids (e.g. CH3CO2H) from their salt (e.g. CH3COONa+).
HCl(aq) + CH3CO2–(aq) ⇌ Cl–(aq) + CH3COOH(aq)
stronger acid salt weaker acid
2.7 Salt Hydrolysis and pH of Salt Solutions
An ionic salt may be prepared from a reaction between an aqueous acid and an aqueous base. E.g. KCl may be prepared from KOH(aq) and HCl(aq).
Salt hydrolysis is a reaction in which ions react with water to produce OH– or H
3O+ ions.
Salts derived from a strong acid and a strong base (e.g. NaCl, KNO3) do not undergo salt hydrolysis. Hence the resulting solution is neutral pH = 7 at 25 oC.
However, ions in salts derived from weak acids/bases (in particular, the anions of weak acids, and cations of weak bases), as well as metal cations with high charge density, could undergo hydrolysis to result in solutions of pH greater/less than 7 (at 25 oC).
A) Salt derived from a strong acid and a weak base E.g. NH4+Cl–
NH4+Cl– is derived from HCl and NH3.
NH3 is a weak base so its conjugate acid NH4+ is a stronger acid than water. NH4+ undergoes hydrolysis forming H3O+ ions.
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
The anion Cl– does not hydrolyse.
pH of a solution of NH4Cl is < 7 since [H3O+] > [OH–].
Note:
NH4+ is a stronger acid than water (see Ex 10(a)) so is able to donate a proton to water.
However, it is still a weak acid (Ka of NH4+ = 5.56 x 10–10 mol dm–3). Hence a reversible arrow ⇌ is used in the salt hydrolysis equation.
The anion Cl– (derived from strong acid HCl) is a weaker base than water so does not hydrolyse.
B) Salt derived from a weak acid and a strong base E.g. CH3CO2–Na+
CH3CO2–Na+ is derived from CH3CO2H and NaOH.
CH3CO2H is a weak acid so its conjugate base CH3CO2– is a stronger base than water. CH3CO2– undergoes hydrolysis forming OH– ions.
CH3CO2 (aq) + H2O(l) ⇌ CH3CO2H(aq) + OH–(aq) The cation Na+ does not hydrolyse.
pH of a solution of CH3CO2–Na+ is > 7 since [H3O+] < [OH–].
C) Salt derived from a weak acid and a weak base E.g. CH3CO2–NH4+
Both cation and anion hydrolyse.
pH of the salt solution depends on Ka of the acid (cation) and Kb of the base (anion)
If Ka(cation) > Kb(anion) salt solution is acidic
Ka(cation) = Kb(anion) salt solution is neutral
D) Salt containing an aqueous metal cation with high charge density E.g. Al3+, Fe3+, Cr3+
Consider the ionic compound, aluminium nitrate Al(NO3)3. When it dissolves in water, the Al3+ and NO3– ions are separated and is hydrated by water molecules.
Al(NO3)3(s) + 6H2O(l) [Al(H2O)6]3+(aq) + 3NO3–(aq)
Note:
The hydrated cations are coordinated to water molecules through dative (co–ordinate) bonds to form complexes. E.g. [Al(H2O)6]3+, [Fe(H2O)6]3+, [Cr(H2O)6]3+
The Al3+ ion is small and highly charged, it has high charge density. It withdraws sufficient electron
density from the O–H bonds of the coordinated H2O molecules, weakening the O–H bonds. When
one O–H bond breaks, a proton is released:
[Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq)
We call this a hydrolysis reaction. Since [H3O+] > [OH–], Al3+(aq) is acidic, pH < 7.
Figure shows [Al(H2O)6]3+ acting as an acid, donating a proton to another H2O molecule
The Ka values of hydrated metal cations are given in the table:
Free ion Hydrated ion Ka /mol dm–3 (at 25 oC) Fe3+ [Fe(H2O)6]3+(aq) 6 10–3
Cr3+ [Cr(H2O)6]3+(aq) 1 10–4 Al3+ [Al(H2O)6]3+(aq) 1 10–5
Note:
In the presence of a base stronger than water e.g. OH–, further abstraction of protons can occur. [Al(H2O)5(OH)]2+(aq) + H2O(l) ⇌ [Al(H2O)4(OH)2]+(aq) + H3O+(aq)
O H H – + – + + +
EXERCISE 11
State whether the following solutions are acidic, alkaline or neutral. Write an equation, including state symbols, for any hydrolysis reaction.
(a) KBr(aq)
neutral, no hydrolysis reaction
(b) NaCN(aq)
alkaline
CN–(aq) + H2O(l) ⇌ HCN(aq) + OH–(aq)
(c) CH3NH3+Cl– (aq)
acidic
CH3NH3+(aq) + H2O(l) ⇌ CH3NH2(aq) + H3O+(aq)
(d) Na2CO3(aq), given: Ka of HCO3– = 4.7 10–11 mol dm–3
alkaline
CO32–(aq) + H2O(l) ⇌ HCO3–(aq) + OH–(aq)
(e) NH4HS(aq), given: Kb of NH3 = 1.8 10–5 mol dm–3 Ka1 of H2S = 9 10–8 mol dm–3
Both ions undergo hydrolysis:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) ––––(1)
HS–(aq) + H2O(l) ⇌ H2S(aq) + OH–(aq) ––––(2)
Ka of NH4+ = 5.6 10–10 mol dm–3
Kb of HS– = 1.1 10–7 mol dm–3
Since Kb of HS– > Ka of NH4+, (2) occurs to a larger extent than (1).
Thus, resultant [OH–] > [H3O+]. The salt solution is alkaline.
(f) FeCl3(aq)
acidic
Fe(H2O)63+(aq) + H2O(l) ⇌ Fe(OH)(H2O)52+(aq) + H3O+(aq)
Ka very small HCO3– is a
weak acid, so CO32– is a
stronger base than water, CO32–
hydrolyses in water
H2S is a weak diprotic acid
Ka1 : H2S(aq) ⇌ HS–(aq) + H+(aq)
EXERCISE 12 (Calculating pH of salt solution)
Calculate the pH of a 0.10 mol dm–3 CH3CO2– Na+(aq) at 25oC. Ka value for ethanoic acid is 1.8 x 10–5 mol dm–3.
Ans: 8.87 CH3CO2– CH3CO2H + OH Initial conc 0.10 Eqm conc 0.10 – x x x Kb (CH3CO2–) = Kw = 5.56 x 10–10 Ka (CH3CO2H) = x2/(0.10 – x) ≈ x2/0.10 (assuming 0.10 >> x) [OH–] = x = 7.45 x 10–6 mol dm–3 pOH = 5.13
pH = 14−pOH = 8.87
3 BUFFER SOLUTIONS
3.1 What are Buffer Solutions?
A buffer solution is one that is able to resist pH changes upon addition of a small amount of acid or
base.
Types of buffer solution:
acidic buffer: an aqueous solution of a weak acid and its salt (conjugate base) alkaline buffer: an aqueous solution of a weak base and its salt (conjugate acid)
So a buffer (usually) consists of two solutes – one provides a weak Bronsted acid and the other provides a weak Bronsted base. Together, they form a conjugate acid–base pair.
EXERCISE 13 [Ref: N2003/I/33]
Which of the following could act as buffer solutions?
1. NaHCO3 and Na2CO3 2. CH3CO2H and NaCl 3. HNO3 and NaNO3
3.2 Principles of Buffer Action (independent learning subtopic)
A) Acidic buffer
E.g. CH3CO2H and CH3CO2Na+
Ethanoic acid is a weak acid and dissociates partially.
CH3CO2H ⇌ CH3CO2 + H+ –––––––––––––––––––––(1)
Adding sodium ethanoate (completely soluble in water) to this solution adds a lot of extra ethanoate ions.
CH3CO2Na+ CH3CO2 + Na+
According to Le Chatelier's Principle, the position of equilibrium (1) shifts further to the left (i.e. the dissociation of ethanoic acid is suppressed).
Concept Check: What happens to the pH of the solution when sodium ethanoate is added to ethanoic acid?
The solution now contains the following: large reservoir of un–ionised ethanoic acid;
large reservoir of ethanoate ions from the sodium ethanoate; and can act as an acidic buffer solution.
Action of Acidic Buffer
(i) When a small amount of H+ is added to this buffer, the H+ ions react with the large reservoir of the CH3CO2 (conjugate base):
CH3CO2 + H+ CH3CO2H The added H+ ions are removed and pH remains almost constant.
(ii) When a small amount of OH is added to this buffer, the OH ions react with the large reservoir of the un–ionized CH3CO2H molecules:
CH3CO2H + OH CH3CO2 + H2O The added OH ions are removed and pH remains almost constant. Note:
Buffer action equations are written with single (non–reversible) arrows.
In (i), after the reaction, [CH3CO2H] increases slightly and [CH3CO2 ] decreases slightly. In (ii), after the reaction, [CH3CO2H] decreases slightly and [CH3CO2 ] increases slightly.
As the original amounts of CH3CO2H and CH3CO2 are large compared to the amount of H+ or OH ions added, the ratio [CH3CO2H] / [CH3CO2 ] remains almost constant,
B) Alkaline buffer
E.g. aqueous NH3 and NH4+Cl
Ammonia is a weak base and dissociates partially.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH– (aq) –––––––––––––––––––(2)
Adding ammonium chloride (completely soluble) to this solution adds a lot of extra ammonium ions.
NH4+Cl NH4+ + Cl
According to Le Chatelier's Principle, the position of equilibrium (2) shifts further to the left (i.e. the dissociation of NH3 is suppressed).
Concept Check: What happens to the pH of the solution when ammonium chloride is added to ammonia?
The solution now contains the following: large reservoir of un–ionised ammonia;
large reservoir of ammonium ions from the ammonium chloride; and can act as an alkaline buffer solution.
Action of Alkaline Buffer
(i) When a small amount of H+ is added to this buffer, the H+ ions react with the large reservoir of the NH3 (weak base):
NH3 + H+ NH4+
The added H+ ions are removed and pH remains almost constant.
(ii) When a small amount of OH is added to this buffer, the OH ions react with the large reservoir of the NH4+ (conjugate acid) from the salt:
NH4+ + OH NH3 + H2O
The added OH ions are removed and pH remains almost constant.
Note:
As the original amounts of NH3and NH4+ are large compared to the amount of H+ or OH ions added, the ratio [NH3] / [NH4+ ] remains almost constant,
3.3 pH of Buffer Solutions
A) Acidic buffer
In a buffer consisting of HA and its salt A, the following equilibrium exists: HA(aq) + H2O(l) ⇌ H3O+(aq) + A– (aq)
and Ka =
[H+]eqm[A]eqm [HA]eqm
As the presence of excess A(aq) suppresses the dissociation of HA, [HA]eqm [HA]initial (since α is very small)
[A]eqm [Na+A ]initial (since [A–] from HA is very small)
Hence pH can be found easily. Note:
When the ratio [acid]/[salt] remain almost constant, [H+] and therefore pH will remain almost constant (see Section 3.2A).
In a buffer, [H+] [A] [H+] = (Ka x [acid]) cannot be used. Recall [H+] = [A ] if the solution contains only the weak acid HA.
Alternatively, to find pH of a buffer, take –lg on both sides of boxed equation,
[salt]
[acid]
lg
lgK
lg[H
a
]
[acid]
[s alt]
lg
pK
pH
a
Henderson–Hasselbalch EquationIt is important to use the correct concentrations in the numerator and denominator of the last term in the equation.
B) Alkaline buffer
In a similar manner as above, the pOH and pH of an alkaline buffer can be found using: therefore Ka
[H+]eqm[A]initial
= [H
+
]eqm[salt]
[HA]initial [acid]
and [H+] = Ka [acid] [salt] and [OH] = Kb [base] [salt]
EXERCISE 14 (Calculating pH of buffer solution)
What is the pH of a buffer solution made by adding 3.28 g of sodium ethanoate (Mr = 82) to 2 dm3 of 0.010 mol dm–3 ethanoic acid? (Ka of ethanoic acid = 1.8 x 10–5 mol dm–3)
Ans: 5.35
[H+] = 9.0 x 10–6 mol dm–3
pH = – lg [H+] = 5.05
EXERCISE 15
Calculate the change in pH when 1 cm3 of a 1.00 mol dm–3 NaOH is added to 1 dm3 of
(a) water; (b) the buffer in Exercise 14
Ans: +4 units; +0.05 units (a) [OH–] = 001 . 1 1 1000 1 = 9.99 x 10–4 mol dm–3 pOH = 3 pH = 11 Change in pH = 11 – 7 = +4 units
(b) CH3CO2H + OH CH3CO2 + H2O
No. of moles of NaOH added = 1/1000 x 1 = 0.001 After the reaction, [CH3CO2] = (0.04 + 0.001) / V
[CH3CO2H] = (0.010 – 0.001) / V
[H+] = 3.95 x 10–6 mol dm–3 pH = – lg [H+] = 5.40 Change in pH = 5.40 – 5.35 = +0.05 units
Concentration of ethanoate ions, [CH3CO2– ] = 3.28/82 = 0.02 mol dm–3 2 Ka = [H + ][CH3CO2– ] [CH3CO2H] 1.8 x 10–5 = [H + ] (0.02) 0.010 Ka = 1.8 x 10–5 = [H + ] (0.04 + 0.001) 0.010 – 0.001
3.4 Buffer Capacity and Buffer Range
The capacity of a buffer is the quantity of H3O+ or OH– it can remove before its pH changes drastically.
A buffer should have high [HA] and [A–], or high [B] and [BH+], to provide a large reservoir of the relevant components.
The more concentrated the components of a buffer, the greater the buffer capacity.
The more similar the concentrations of the buffer components, the greater the buffer capacity.
A) Maximum Buffer Capacity
Using the boxed equations in Section 3.3, it can be shown that
pH = pKa (for acidic buffer) or pOH = pKb (for alkaline buffer)
This buffer is said to have the maximum buffer capacity and it can most effectively resist a change in pH in either direction (i.e. when either acid or base is added).
B) Effective Buffer Range
The further the buffer–component concentration ratio is from 1, the less effective the buffering action (i.e., lower buffer capacity).
In practice, a buffer is no longer effective when [salt]/[acid] or [salt]/[base] > 10/1 or < 1/10.
Since log10 (10/1) = +1 and log10 (1/10) = –1,
for an acidic buffer, [H+] = Ka [s a l t]
[a ci d]
effective buffer pH range = pKa 1
for an alkaline buffer, [OH] = Kb
[s a l t]
[ba s e]
effective buffer pOH range = pKb 1
It can be shown that the most effective buffer has [salt] = 1 or [salt] = 1
EXERCISE 16
Which of the following acids would be the best choice to combine with its sodium salt to make a solution buffered at pH 4.25?
For your choice, calculate the ratio of the salt (conjugate base) to the acid concentrations required to attain the desired pH.
Acid pKa
chlorous acid (HClO2) 1.95
nitrous acid (HNO2) 3.34
methanoic acid (HCO2H) 3.74 hypochlorous acid (HClO) 7.54
The best choice is methanoic acid because its pKa lies closest and within 1 unit of the desired pH.
3.5 Buffer System in Human Blood (independent learning subtopic)
The pH of our blood must be kept constant at about 7.4 since enzymes operate only within a narrow pH range.
The most important buffer system in blood plasma involves the weak acid H2CO3 (an aqueous
solution of CO2) and its conjugate base, HCO3 ions.
The following equilibria operate in the H2CO3/HCO3– buffer system:
H+(aq) + HCO3 (aq) ⇌ H2CO3(aq) ⇌ H2O(l) + CO2(g)
Read the article on the following page for an elaboration on this buffer system.
If blood pH rises above about 7.45, you can suffer from a condition called alkalosis. This can arise from hyperventilation or oxygen deficiency at high altitude. It can lead to over excitability of the central nervous system, muscle spasms and death. One way to treat alkalosis is to breath into a paper bag. The CO2 exhaled is recycled into the body. Can you explain how that helps?
Ka = [H + ][salt] [acid] 10–3.74 = 10 –4.25 [salt] [acid]
[salt] = 3.24 Check: Ratio is between 0.1 and 10 (effective range) [acid]
3.6 Importance of Buffer Solutions in Everyday Life (independent learning subtopic)
Injections and drips into a patient’s body must be carefully buffered so that the pH of body fluids does not change much.
In bacteriological research, pH of culture medium is maintained to control the growth of bacteria.
In agriculture, the pH of soil is maintained to optimize plant growth.
The pH of many industrial processes must also be carefully controlled, e.g. electroplating and manufacture of dyes and leather.
Baking soda, which is sodium hydrogencarbonate, NaHCO3, is sometimes added to swimming pools to control the pH of the water.
Although most buffer systems consist of two separate species that react with H+ or OH–, the HCO3– ion is an example of a single ion that is able to serve both functions:
On addition of small amount of acid, HCO3–(aq) + H+(aq) H2CO3(aq)
4 ACID–BASE TITRATIONS
4.1 Acid–Base Indicators (independent learning subtopic)
An indicator is a weak acid, whose acid form, HIn, is a different colour from its ionised form, In–.
HIn(aq) ⇌ H+(aq) + In(aq)
E.g. for litmus red blue
for methyl orange red yellow
for phenolpthalein colourless pink
At different pH values, the proportion of HIn to In– is different, giving rise to different colours.
For litmus:
In acidic solutions, [H+] is high. The position of equilibrium lies on the left. Predominant form of litmus is HIn so the solution will appear red.
In alkaline solutions, [OH–] is high so [H+] is low. The position of equilibrium lies on the right. Predominant form of litmus is In– hence the solution will appear blue.
4.2 Working pH Range of Indicators (independent learning subtopic)
For most indicators, the predominant form must be at least 10 times more concentrated than the other form for its colour to be distinguished from the other coloured form.
Given that the dissociation constant of an indicator, termed KIn, is
Taking –lg on both sides and rearranging,
For litmus:
If the ratio of [In−]/[HIn] ≥ 10/1, i.e. pH ≥ pKIn + 1, the solution appears blue.
If the ratio of [In−]/[HIn] ≤ 1/10, i.e. pH ≤ pKIn – 1, the solution appears red.
Therefore, the litmus changes from red to blue over a pH range of
pH = pKIn – 1 to pH = pKIn + 1
That is, the working range of an indicator (i.e. pH range where a marked colour change occurs) is approximately pKIn ± 1 . KIn = [H+][ In] [HIn] pH = pKIn + lg [ In] [HIn]
Common indicators and colour changes
EXERCISE 17
At low pH, the colour of bromocresol green is yellow, and at high pH, the colour is blue. Calculate the pH at which the colour of bromocresol green (KIn = 1.58 x 10–5) is green.
HIn(aq) ⇌ H+(aq) + In(aq)
yellow blue
When [HIn] = [In–],
bromocresol green appears green (yellow + blue) pH = pKIn = 4.80
4.3 Choice of Indicators
During an acid–base titration, one solution is run into the other in the presence of an indicator.
End–point: the point in a titration at which the indicator changes colour when 1 drop of excess
titrant is added.
Equivalence point: the point in a titration at which the exact number of moles of acid and base
react completely.
The end–point coincides with the equivalence point only when a suitable indicator is used.
The indicator must have a distinct colour change.
The working range of the indicator must coincide with the rapid pH change for the titration.
The rapid pH change for the titration depends on the strengths of acid and alkali used. Either the acid or the alkali, or both, may be strong or weak, so that there are four possible combinations.
Type suitable indicator
strong acid – strong alkali methyl orange or phenolphthalein strong acid – weak alkali methyl orange
weak acid – strong alkali phenolphthalein weak acid – weak alkali no suitable indicator
4.4 Acid–Base Titration Curves
A titration curve (a graph of pH against the volume of the titrant added) shows how the pH changes during a titration. It can be obtained by measuring the pH of the solution using a pH meter.
Four types of acid–base titration curves are shown on the following pages. Note that in all of them, the acid is in the conical flask while the base is added from the burette.
(A) Strong Acid – Strong Base Titration
E.g. addition of 0.10 mol dm3 NaOH to 25.0 cm3 of 0.10 mol dm3 HCl
Volume of NaOH added / cm3
Note for teachers:
Graph is not given in the students’ notes. Students will sketch the titration curve during the lecture.
Features of the curve:
At VNaOH = 0 cm3
Initial pH = pH of HCl = –lg 0.10 = 1
Low as HCl is a strong acid (fully dissociated in water).
At 0 < VNaOH < 25 cm3
pH gradually rises as an increasing amount of H+ is n eutralized by the NaOH added.
At VNaOH = 25 cm3
Equivalence point is reached.
pH = 7 as the salt formed does not undergo hydrolysis.
Suitable indicator:
methyl orange or phenolphthalein
Note the sharp pH change near equivalence point (see Homework 1, parts (c) and (e)). At VNaOH > 25 cm3
There is excess NaOH present in the solution. As NaOH is a strong base, [OH] is high, hence pH is high (exact value depends on [NaOH] that is in excess).
Equivalence point, pH = 7
Homework 1
Determine the pH of the solution in the conical flask at the point when each of the following volumes of NaOH is added.
(a) 0.00 cm3 Ans: 1.00
(b) 12.50 cm3 (This is called the half–neutralisation or half–equivalence point.) Ans: 1.48
(c) 24.90 cm3 Ans: 3.70
(d) 25.00 cm3 Ans: 7.00
(e) 25.10 cm3 Ans: 10.3
B) Weak Acid – Strong Base Titration
E.g. addition of 0.10 mol dm3 NaOH to 25.0 cm3 of 0.10 mol dm3 CH3CO2H
Volume of NaOH added / cm3
Features of the curve:
At VNaOH = 0 cm3
Initial pH = pH of CH3CO2H
pH is not equal to 1.0 as CH3CO2H is a weak acid (only ionises partially in water).
At 0 <VNaOH < 25 cm3
1. Once the NaOH is added and some acid is neutralized, the solution contains a mixture of weak acid and its salt a buffer solution forms.
2. The pH at the initial part of the titration curve and the part immediately before complete neutralisati on changes more sharply as these lie outside the effective buffer range of the buffer.
3. The maximum buffer capacity occurs when half of the number of moles of acid is neutralized at VNaOH = 12.50 cm3 (where [acid] = [salt]). At this point, pH = pKa of the weak acid.
At VNaOH = 25 cm3
Equivalence point is reached. pH > 7 as the anion undergoes hydrolysis to produce OH: CH3CO2–(aq) + H2O(l) ⇌ CH3CO2H(aq) + OH–(aq)
Suitable indicator: phenolphthalein
The sharp change in pH in the curve near equivalence point occurs over a smaller pH range compared to the previous strong acid–strong base titration.
At VNaOH > 25 cm3
There is excess NaOH present in the solution. As NaOH is a strong base, [OH] is high and hence pH is high (exact value depends on [NaOH] in excess).
Homework 2
Determine the pH of the solution in the conical flask at the point when each of the following volumes of NaOH is added. (Ka of CH3CO2H = 1.8 10–5 mol dm–3)
(a) 0.00 cm3 Ans: 2.87
(b) 12.50 cm3 (half–neutralisation point) Ans: 4.74
(c) 25.00 cm3 (see Section 2.7, Exercise 12) Ans: 8.72
buffer region
Equivalence point, pH > 7
C) Strong Acid – Weak Base Titration
E.g. addition of 0.10 mol dm3 NH3 to 25.0 cm3 of 0.10 mol dm3 HCl
Volume of NH3 added / cm3
Features of the curve:
At Vweak base = 0 cm3
Initial pH = pH of HCl = –lg 0.10 = 1
Low as HCl is a strong acid (fully dissociated in water). At 0 < Vweak base < 25 cm3
pH gradually rises as increasing amount of H+ is neutralized by added weak base. However, no buffer is formed as there is no weak base present in excess in the conical flask.
At Vweak base = 25 cm3
Equivalence point is reached. pH < 7 as the cation undergoes hydrolysis to produce H3O+: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Suitable indicator: methyl orange
At Vweak base > 25 cm3
There is excess weak base present. The solution contains a mixture of weak base and its salt a
buffer solution forms.
What is the Vweak base and pH at which the buffer with maximum capacity occurs?
Ans: 50 cm3, 9.26
Homework 3
Determine the pH of the solution in the conical flask at the equivalence point.
(Kb of NH3 = 1.8 10–5 mol dm–3) Ans: 5.28
Equivalence point, pH < 7
D) Weak Acid – Weak Base Titration
E.g. addition of 0.10 mol dm3 NH3 to 25.0 cm3 of 0.10 mol dm3 CH3CO2H
Volume of NH3 added / cm3
Features of the curve:
At Vweak base = 0 cm3
Initial pH = pH of CH3CO2H
pH is not equal to 1.0 as CH3CO2H is a weak acid (only ionises partially in water).
At 0 < Vweak base < 25 cm3
Once the weak base is added and some acid is neutralised, the solution contains a mixture of weak acid and its salt a buffer solution forms.
At Vweak base = 25 cm3
Equivalence point is reached. There is no sharp change in pH, hence no suitable indicator can be
used, a pH meter may be used to help detect the equivalence point.
Both the cation and anion undergo hydrolysis (see Section 2.7C). At Vweak base > 25 cm3
There is excess weak base present in the solution together with its salt another buffer region is formed.
4.5 Titration Curves with More than One Equivalence Point
Polyprotic acids have more than one ionisable proton. The successive Ka values may differ by several orders of magnitude. This means the first H+ is lost more easily than subsequent ones.
E.g. H2SO3(aq) + H2O(l) ⇌ HSO3–(aq) + H3O+(aq) Ka1 = 1.4 x 10–2 mol dm–3 HSO3–(aq) + H2O(l) ⇌ SO32–(aq) + H3O+(aq) Ka2 = 6.5 x 10–8 mol dm–3
In a titration of H2SO3, we can assume that all the H2SO3 molecules will lose one H+ ion before any of the HSO3– ions loses its H+.
OH– OH–
H2SO3 –––––––> HSO3– ––––––> SO32–
Volume of NaOH added /cm3
Note that the titration curve has two steep portions, which corresponds to two equivalence points.
A similar curve is also obtained for a titration involving a mixture of two acids with an appreciable difference in Ka values (or a mixture of 2 bases with an appreciable difference in Kb values).
In practice, we can observe the two separate end–points with the proper choice and use of two different indicators. Consider the titration of sodium carbonate with dilute HCl.
There are 2 stages in the titration:
CO32–(aq) + H+(aq) HCO3–(aq) –––––––––––––––––––(1) HCO3–(aq) + H+(aq) H2O(l) + CO2(g) –––––––––––––––––––(2)
The first rapid change in pH for reaction (1) occurs at pH = 8.5. Therefore a suitable indicator for the first equivalence point is phenolphthalein.
The second rapid change in pH for reaction (2) occurs at pH = 4. Therefore a suitable indicator for the second equivalence point is methyl orange.
To observe the two end–points separately, phenolphthalein is first added. Methyl orange is added only after the equivalence point is reached.
If methyl orange is used at the start of the experiment, only one end–point is observed and the titre obtained is twice that obtained with phenolphthalein.
EXERCISE 18 (Double indicator titration)
A 25.0 cm3 portion of a solution containing sodium carbonate and sodium hydrogencarbonate needed 22.50 cm3 of a solution of hydrochloric acid of concentration 0.100 mol dm–3 to decolourise phenolphthalein. On addition of methyl orange, a further 28.50 cm3 of the acid was needed to turn this indicator orange. Calculate the concentration of sodium carbonate and sodium hydrogencarbonate in the solution.
Ans: 0.0900 mol dm–3; 0.0240 mol dm–3
With phenolphthalein indicator:
Na2CO3(aq) + HCl(aq) NaHCO3(aq) + NaCl(aq) –––––––––––––(1)
No of moles of HCl used = 0.100 x 0.0225 = 2.25 x 10–3
No of moles of Na2CO3 = 2.25 x 10–3
[Na2CO3] = 2.25 x 10–3/0.025 = 0.0900 mol dm–3
With methyl orange indicator:
NaHCO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) (from original NaHCO3 and
the NaHCO3 formed in (1))
No of moles of HCl used = 0.100 x 0.0285 = 2.85 x 10–3
No of moles of HCl used for NaHCO3 formed from Na2CO3 = 2.25 x 10–3
No of moles of HCl used for original NaHCO3 = 2.85 x 10–3 – 2.25 x 10–3 = 0.60 x 10–3
Ionic Equilibria – Summary
Concept Application
An acid is a proton donor. A base is a proton acceptor.
To identify if a molecule or ion is behaving as an acid or base given an equation or vice versa
pX = log10X X = 10 pX To define quantities such as pH, pOH, pKw, pKa and pKb
Kw = [H+][OH] = 1 x 1014 (at 25C)
Kw shows the relationship between [H+] and [OH] in any
aqueous solutions.
To calculate [H+] or [OH] given the value of the other. In pure water, [H+] = [OH] = 1 x 107 (at 25C)
pKw = pH + pOH = 14 (at 25C) To calculate pH from pOH, or vice versa
For any conjugate acid–base pair,
Ka x Kb = Kw
pKa + pKb = pKw = 14 (at 25 oC)
To calculate Ka from Kb, or vice versa.
You use this equation when you need one of these K values but have the other.
HA H+ + A Ka = eqm eqm eqm [HA] ] [A ] [H To calculate the pH of a weak acid, Ka initial eqm 2 [HA] ] [H buffer, Ka initial initial eqm [acid] alt] [ ] [H s
To compare the strength of weak acids. The larger the Ka value, the stronger the acid.
B + H2O BH+ + OH Kb = eqm eqm eqm [B] ] [OH ] [BH To calculate the pH of a weak base, Kb initial eqm 2 [B] ] [OH buffer, Kb initial initial eqm [base] [salt] ] [OH
To compare the strength of weak bases. The larger the Kb value, the stronger the base.
Salt hydrolysis – a reaction in which ions react with
water to form acidic or alkaline solutions e.g. NH4+ + H2O NH3 + H3O+
CH3CO2 + H2O CH3CO2H + OH
To explain why the pH of a salt derived from
strong acid and weak base (e.g. NH4Cl) is < 7 weak acid and strong base (e.g. CH3CO2Na) is > 7
Checklist for pH calculations
Do you know how to calculate the pH of a solution containing a strong acid or strong base
a weak acid or weak base a salt such as CH3CO2Na or NH4Cl
a buffer
Using the above, can you sketch a titration curve, and mark the various key points on the curve (initial pH, equivalence point, buffer with maximum capacity)?
