Chemistry Form 6 Chap 7 New
81
CHEMISTRY FORM 6 PHYSICAL CHEMISTRY CHAPTER 10 : ACID BASE EQUILIBRIA
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Transcript of Chemistry Form 6 Chap 7 New
CHEMISTRY FORM 6 PHYSICAL CHEMISTRY
CHAPTER 10 : ACID BASE EQUILIBRIA
Acid / Base Theory
Arrhenius Theory Bronsted –
Lowry Theory
LewisTheory
7.1 – Theory of Acid and Base7.1.1 – The Arrhenius Theory Acid are substance that dissociate in water to produce
hydrogen ions, H+ (aq)HA (aq) H+ (aq) + A- (aq)
Base are substance that dissociate in water to yield hydroxide ions, OH- (aq)
MOH (aq) M+ (aq) + OH- (aq) Disadvantage – many organic compounds like amine
RNH2 and ammonia contain no OH- (aq) ions but show basic properties in water. So strictly speaking, ammonia is not Arrhenius base.
To solve it, chemist suggested that aqueous solution of ammonia contain contains the compound call ammonium hydroxide, NH4OH. This explanation is not acceptable as there’s no evidence supporting the existence of this compound in solution.
7.1.2 – The Bronsted Lowry Theory A Bronsted-Lowry acid is substance (molecule or ion) that
……………………… to a base (proton donor) A Bronsted-Lowry base is substance that
………………………… from acid (proton acceptor). Thus for a reversible reaction, acid-base reaction involve
proton transferHA + B- HB + A-
Base is usually negative ion or neutral molecule with a lone pair electron that can formed dative bond with proton. When an acid loses proton to a base, itself will formed base. The base formed as a result of donating proton is called as conjugate base. Similarly a base in accepting proton forms an acid, so this acid is known as conjugate acid. HCl (aq) + NH3 (aq) Cl- (aq) + NH4
+ (aq)
Acid base conjugate conjugate base acid
donate protons accept
protons
NH3 (g) + H2O (l) NH4+ (aq) + OH- (aq)
H2CO3 (aq) + H2O (l) HCO3- (aq) + H3O+ (aq)
H2O (l) + HSO4- (aq) H3O+ (aq) + SO4
2- (aq)
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH-
(aq)
Bronsted – Lowry base
Bronsted – Lowry acid
Conjugate base
Conjugate acid
Bronsted – Lowry base
Bronsted – Lowry acid
Conjugate base
Conjugate acid
Bronsted – Lowry base
Bronsted – Lowry acid
Conjugate base
Conjugate acid
Bronsted – Lowry base
Bronsted – Lowry acid
Conjugate base
Conjugate acid
7.1.3 Lewis Theory Lewis acid – a substance which can form a dative
covalent bond by …………………………………………. from Lewis base
Lewis base – a substance which form dative covalent bond by ……………………………………..… to Lewis acid
Neutralisation occur when Lewis base donate a lone pair electron to form dative bond with Lewis acid.
For example the reaction between boron trifluoride and ammonia
Similarly, the reaction between boric acid, [B(OH)3 and H3BO3] and water
BF
F
F
+ N
H
H
H BF
F
F
N
H
H
H
B
OH
OHHO HO H B
OH
OHHO
OH
+ + H+
receiving lone pair electron
donating lone pair electron
In this case, boric acid is classified as acid, not because it produce H+ but …………………………………………………..…………………
One of the most important applications of the Lewis acid – base theory is in describing the formation of complex ions. Ligand act as Lewis base which donate lone pair electron to central metal ion, which act as a Lewis acidCu2+ (aq) + 4 NH3 (aq)
[ Cu (NH3)4 ]2+ (aq)Al3+ (aq) + 4 OH- (aq)
[ Al (OH)4 ]- (aq)
Cu
H3N
NH3
NH3
H3N
2+
Al
HO
OH
OH
HO
-
O on water donate the lone pair electron to B
Lewis acid Lewis base
Positive metal (normally from d-
block)
Molecules with incomplete octet
(central atom)
Negative ion (ligand)
Molecules with extra lone pair
electron
Example
Zinc (II) ion, Zn2+ ; Lead (II) ion, Pb2+
Chromium (III) ion, Cr3+
Example
Water (H2O) ; cyanide ion (CN-)
Oxalate ion, (C2O42-)
Exercise 1 : Identify Bronsted-Lowry acid / base ; conjugate acid / base
a) NH3 + H2O NH4+ + OH–
b) H2O + HCO3– CO3
2- + H3O+
c) H2NCONH2 + H2O H3N+CONH2 + OH–
d) H2NCONH2 + NH3 H2NCONH- + NH4+
No Bronsted – Lowry acid
Bronsted – Lowry base
Conjugate acid Conjugate base
a
b
c
d
H2O NH3 NH4+ OH
H2OHCO3- H3O+ CO3
2-
H2O H2NCONH2 H3N+CONH2
H2NCONH2
OH
NH3H2NCONH-NH4
+
Exercise 2 : Write the formula of the conjugate acid for each of the following bases
a) Ammonia ; NH3 b) Carbonate ion ; CO32-
c) Hydroxylamine ; NH2OHd) Pyridine ; C6H5N
NH4+ HCO3
-
NH3OH+
C6H5NH+
Exercise 3 : Identify the Lewis acid and Lewis base in the following reaction
a) Cl2 + AlCl3 Cl+ + AlCl4-
b) SO3 + CaO CaSO4
c) H3BO3 + H2O B(OH)4- + H+
d) CH3COCl + FeCl3 CH3CO+ + FeCl4-
e) CH3COO- + 2 HF CH3COOH + HF2-
f) Cu2+ + 4 NH3 → [Cu(NH3)4]2+
Lewis
base
Lewis
acid
Lewis
base
Lewis
acid
Lewis
base
Lewis
acid
Lewis
base
Lewis
acid
Lewis
base
Lewis
acid
Lewis
base
Lewis
acid
7.2 The Strengths of Acid and Base
Strong acid : ……………………………………………………………………………….
Equation
Weak acid : ………………………………………………………………………………
Equation
Strong base : ………………………………………………………………………………
Equation
Weak base : .……………………………………………………………………………..
Equation The strength of acid and base can be compared in terms of
degree of dissociation (varies with concentration) the dissociation constant (does not vary with concentration)
Acid which dissociate completely in water
HX + H2O H3O+ + X-
Acid which dissociate partially in water
HA + H2O ↔ H3O+ + A-
Base which dissociate completely in water
MOH M+ + OH-
Base which dissociate partially in water
NH3 + H2O ↔ NH4+ + OH-
7.2.1 The degree of dissociation of acids and dissociation constant (Ka) and p Ka
When a strong acid dissolve in water, it will ……………………… dissociate according to the equation
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
Initial c 0 0
At the end 0 c c As HCl will be completely dissociate in water, the concentration of
HCl will be ……… and this will cause the value of KC to become ………
Consider the dissociation of a weak monoprotic acid, HA in water is
HA (aq) + H2O (l) ↔ H3O+ (aq) + A- (aq)
Initial c 0 0
At equil. c (1 – α) c α c α In the dissociation of weak acid, since the dissociation of acid are
so small, it does not affect much the water. In another word, [H2O] ……….
Equilibrium constant is re-expressed as
completely
0 ∞
HA
]A][OH[K 3
a
CH3COOH + H2O ↔ CH3COO- + H3O+
HOBr + H2O ↔ OBr- + H3O+
C6H5COOH + H2O ↔ C6H5COO- + H3O+
HSO4- + H2O ↔ SO4
2- + H3O+
As for polyprotic acid (…………………………………………………………………………....) ; it goes through stages of dissociation. Each stage of dissociation it has its own Ka value.Consider carbonic acid, H2CO3 ; dissociate in 2 distinct stages in waterStage 1 : H2CO3 + H2O ↔ HCO3
- + H3O+ Stage 2 : HCO3- + H2O ↔ CO3
2- + H3O+
Overall : H2CO3 + 2 H2O ↔ CO32- + 2 H3O+
Ka = Ka1 x Ka2 =
]COOHCH[
]COOCH][OH[K
3
33a
]HOBr[
]BrO][OH[K 3
a
]COOHHC[
]COOHC][OH[K
56
563a
]HSO[
]SO][OH[K
4
243
a
Acid which dissociate more than 1 mole of H+
]COH[
]HCO][OH[K
32
331a
]HCO[
]CO][OH[K
3
233
2a
]HCO[
]CO][OH[
]COH[
]HCO][OH[
3
233
32
33
]COH[
]CO[]OH[
32
23
23
Since the value of Ka is usually small, pKa is often used to compare the strength of acid where pKa = - log Ka. So,
smaller the pKa ; stronger the acid that is.
Example : The table below shows the Ka values of some acids. Calculate the pKa of each acid and arrange in ascending order of their acid strength
Increasing order of acid strength :
Acids Ka (mol dm-3) pKa
Ethanoic acid, CH3COOH 1.80 x 10-5
Chloric (I) acid, HOCl 3.71 x 10-8
Benzoic acid, C6H5COOH 6.31 x 10-5
Methanoic acid, HCOOH 2.09 x 10-4
4.74
7.43
4.20
3.68
CH3COOH < C6H5COOH < HCOOH
7.2.2 The degree of dissociation of base and dissociation constant (Kb) and pKb
In the dissociation of base, it is similar in the way of dissociation of acid
Consider a weak base, B, which undergoes partial dissociation in water
B- (aq) + H2O (l) HB (aq) + OH- (aq) The equilibrium constant is called the base dissociation
constant, Kb, and is given by
The value of Kb measures the strength of base. Higher the value of Kb, stronger the base ; greater the extent of its dissociation in water. Since the value of Kb is usually small, pKb is often used to compare the strength of base where pKb = - log Kb. So, the ……………… the value of pKb ; ………………….. the base that is.
Example : The table below shows the Kb values of some bases. Calculate the pKb of each base and arrange in ascending order of their basic strength.
]B[
]HB][OH[Kb
smaller stronger
Base Kb (mol dm-3) pKb
Ammonia, NH3 1.78 x 10-5
Methylamine, CH3NH2 4.37 x 10-4
Trimethylamine, (CH3)3N 6.31 x 10-5
Phenylamine, C6H5NH2 4.17 x 10-10
Increasing order of basic strength :
4.75
3.36
4.20
9.38
C6H5NH2 < NH3 < (CH3)3N < CH3NH2
7.3 Ostwald dilution Law, Ka and Kb
Consider a weak acid, HA, with concentration, C mol dm-3 and the degree of dissociation, .
In aqueous solution, HA undergoes partial ionisation ;
HA (aq) + H2O (l) ↔ H3O+ (aq) + A- (aq)
Initial c 0 0
At eq c (1 – ) -- c c
Since so replacing all the value into equation
and since is usually small ; (1 – ) ≈ 1
]HA[
]A][OH[K 3
a
)1(c
)c)(c(Ka
Ka = c2 and pKa = - log Ka
Since c = [H3O+] so = [H3O+] / c ]OH[lgpHandcK]OH[so;c]OH[and 3a33
Same thing occur for calculation involving Kb and the concentration of [OH-]. Consider the reaction below, where
B– (aq) + H2O (l) ↔ HB (aq) + OH– (aq)
Initial c 1 0 0
At equilibrium c(1 – ) -- c c
Since so replacing all the value into equation
and since is usually small ; (1 – ) ≈ 1
]B[
]HB][OH[Kb
)1(c
)c)(c(Kb
Kb = c2 and pKb = - log Kb
Since c = [OH-] so = [OH-] / c ]OH[lgpOHandcK]OH[so;c]OH[and b
Example 11 : Calculate (a) the concentration of H+ ions in 0.120 mol dm-3 solution of ethanoic acid (b) the degree of dissociation of ethanoic acid [Ka = 1.80 x 10-5 mol dm-3]
Example 12 : Calculate the concentration of H+ ions in a 0.20 mol dm-3 HCN solution. the value of pKa of HCN is 9.3.
Example 13 : The concentration of OH- in an aqueous solution of ammonia of concentration 0.150 mol dm-3 is 1.64 x 10-3. Calculate (a) degree of dissociation (b) the base dissociation constant of aqueous NH3.
a) [H+] = √Ka x c = √(1.80 x 10-5 x 0.120 = 1.47 x 10-3 mol dm-3
b) [H+] = c = 1.47 x 10-3 / 0.120
= 0.0122
a) Ka = 10-pKa
= 10-9.3
= 5.01 x 10-10 mol dm-3
[H+] = √Ka x c = √(5.01 x 10-10 x 0.20 = 1.0 x 10-5 mol dm-3
a) [H+] = c = 1.64 x 10-3 / 0.150
= 0.0109
b) Ka = c 2
= 0.150 x 0.01092
= 1.79 x 10-5 mol dm-3
Example 14 : A solution of nitric (III) acid, HNO2, is 1.5% ionised. Calculate (a) its concentration in mol dm-3. (b) hydrogen ion concentration, [H+] of the solution. [Given Ka of HNO2 = 5.01 x 10-4 mol dm-3]
Example 15 : An aqueous solution of 1.00 x 10-2 mol dm-3 dimethylamine produce OH- ions of concentration 4.37 x 10-7 mol dm-3. Calculate (a) the dissociation constant of base and (b) degree of dissociation of base
a) Since it is 1.5% ionised, = 0.015
Ka = c 2
c = 5.01 x 10-4 / 0.0152
= 2.23 mol dm-3
b) Since it is 1.5% ionised, = 0.015
[H3O+] = c = 2.23 x 0.015= 0.0334 mol dm-3
a) Kb = [OH-]2 / c = (4.37 x 10-7)2 / 0.0100
= 1.91 x 10-11 mol dm-
3
b) = [OH-] / c = (4.37 x 10-7) / 0.0100
= 4.37 x 10-5
10.4.4 Relative acidity of Oxoacids An oxoacid is acid in which the ionisable hydrogen atom is
bonded to O atom. When an oxoacid dissociates to give one or more protons,
it forms the corresponding oxoanion which is a conjugate base.Oxoacid Oxoanion
N O
O
OH
S
O
O
O
OH H
P
O
OO
OH
H
H
N O
O
O
S
O
O
O
O
2-
3-
P
O
OO
O
S
O
O
O
O
-
H
-
P
O
OO
OH H
2-
P
O
OO
OH
The strength of an oxoacid increase ifa) In oxidation number of the central atom increased.
b) with an increase of electronegativity of the central atom.
Oxoacid HClO HClO2 HClO3 HClO4
Oxidation number
+1 +3 +5 +7
Ka (mol dm-3) 2.8 x 10-8 1.0 x 10-2 ~ 103 ~ 107
Oxoacid HIO HBrO HClO
Electronegativity of halogen atom
I = 2.5 Br = 2.8 Cl = 3.0
Ka (mol dm-3) 2.4 x 10-11 2.6 x 10-9 2.8 x 10-8
Increase acid strength
Increase acid strength
These trend in acid strength can be explain in terms of molecular structure of oxoacid
Consider the dissociation of an oxoacid H – O – X in water :
H – O – X (aq) + H2O (l) H3O+ (aq) + OX- (aq)
i) When an oxoacid dissociate, the H+ ion dissociate from O that is bond with X. Thus, higher the oxidation number of X, greater the tendency of X to draw electrons away from the oxygen atom bonded to H and the easier it will be for H to dissociate, thus increasing Ka
ii) The more electronegative X is, the greater will be the tendency of X to draw electron away from O thus weakening the O – H bond.
7.5 The Dissociation of Water : Ionic Product of Water Pure water is a very weak electrolyte since the degree of
dissociation of pure water is very small.H2O (l) H+ (aq) + OH- (aq) ∆H = + 57.3 kJ/mol
The Kc for this reaction ; Kc = [H+] [OH-] / [H2O] or
Kc [H2O] = [H+] [OH-] Since the dissociation of water is so small that makes it
almost constant, so Kw = Kc [water]. Kw here is called as ionic product of water.
At 25oC (298 K), the dissociation constant for water is 1.0 x 10-14 mol2 dm-6. If [H+] = [OH-] for neutral solution, [H+] = [OH-] = ……………………………
Same as Kc, Kw gives different values at different temperatures. Since the reaction is an endothermic process, equilibrium will shift to the right position upon increasing the temperature and make the value of Kw become large.
1.0 x 10-7 mol dm-3
Temp (oC) 0 5 10 20 30 50
Kw(mol2 dm-6) 1.14 x 10-15 1.86 x 10-15 2.93 x 10-15 6.81 x 10-15 1.47 x 10-14 5.48 x 10-14
pKw 14.9 14.7 14.5 14.2 13.8 13.3
Kw can also be expressed in terms of pKw = - log Kw = -log (1.0 x 10-14) = 14
7.5.1 Relationship between Ka , Kb , and Kw.
Consider the chemical equilibrium between ethanoic acid and its conjugate base
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
Hence, Ka x Kb =
= [H3O+][OH-]
= Kw
]COOHCH[
]COOCH][OH[K
3
33a
]COOCH[
]OH][COOHCH[K
3
3b
]COOHCH[
]COOCH][OH[
3
33
]COOCH[
]OH][COOHCH[
3
3
Similarly, for ammonia and its conjugate acid :NH3 (aq) + H2O (l) NH4
+ (aq) + OH- (aq)
NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)
Hence Ka x Kb
= [H3O+][OH-]
= Kw
]NH[
]NH][OH[K
3
4b
]NH[
]NH][OH[K
4
33a
]NH[
]NH][OH[
3
4
]NH[
]NH][OH[
4
33
7.6 The pH Scale The pH of a solution is used to denote hydrogen ion
concentration in the solution. It is the negative logarithm to the base ten of the hydrogen ion concentration in mol dm-3.
pH = – lg [H+] or – lg [H3O+] If concentration of OH- is considered, pOH = – lg [OH-]
At 25oC, Kw = 1.0 x 10-14 mol2 dm-6 , pKw = 14 = pH + pOH
pH value usually lie within the range of 0 – 14. However, it is possible, in theory, for a very strong acid solution to have a pH less than 0 or more than 14.
[H+]mol dm-3 1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Strength Strong acid Weak acid
Neu-
tralWeak base Strong base
a) 0.01 mol dm-3 of hydrochloric acid, HCl b) 0.01 mol dm-3 of sulphuric acid, H2SO4
c) 0.010 mol dm-3 of ethanoic acid, CH3COOH (Ka = 1.8 x 10-5 mol dm-3)
d) 2.0 x 10-8 of hydrochloric acid
e) 0.01 mol dm-3 of sodium hydroxide, NaOH
f) 8 x 10-3 mol dm-3 of magnesium hydroxide, Mg(OH)2
a) Since HCl, conc of H+ = 0.01
pH = - lg [H+] = - lg 0.01 = 2
Since 1 mol H2SO4 = 2 mol H+
[H+] = 2 x 0.01 = 0.02pH = - lg 0.02
= 1.7
[H+] = √Ka x c = √(1.8 x 10-5 x 0.010 = 4.24 x 10-4 mol dm-3
pH = - lg 4.24 x 10-4 = 3.37
since conc of acid is less than1 x 10-7, water diss, conc of H+ = OH- = 1 x 10-7
So, conc of H+ = 1 x 10-7 + 2.0 x 10-8
= 1.2 x 10-7 mol dm-
3
pH = - lg 1.2 x 10-7 mol dm-3
= 6.92Since NaOH, conc of OH- =
0.01pOH = - lg [OH-] = - lg 0.01 = 2
pH = 14 – pOH = 14 – 2 = 12
Since 1 mol Mg(OH)2 = 2 mol OH-
conc of OH- = 1.6 x 10-2
pOH = - lg [OH-] = - lg 0.016 = 1.80
pH = 14 – pOH = 14 – 1.80 = 12.2
g) 0.01 mol dm-3 of ammonia, NH3 (pKb = 4.7)
h) 4.0 x 10-8 mol dm-3 potassium hydroxide, KOH
i) A 0.100 mol dm-3 of an alkaline solution produced by a weak base has a pH of 9.10. Calculate a) the concentration of OH- b) degree of dissociation.
pKb = -lg [Kb] ; Kb = 10-4.7
Kb = 2.0 x 10-5 mol dm-3
[OH-] = √Kb x c = √(2.0 x 10-5 x 0.01 = 4.47 x 10-4 mol dm-3
pOH = - lg 4.47 x 10-4 = 3.35
pH = 14 – 3.35 = 10.65
since conc of base is less than1 x 10-7, water diss, conc of H+ = OH- = 1 x 10-7
So, conc of OH- = 1 x 10-7 + 4.0 x 10-8
= 1.4 x 10-7 mol dm-
3
pOH = - lg 1.4 x 10-7 mol dm-3
= 6.85 pH = 14 – 6.85 = 7.15
pOH = 14 – pH= 14 – 9.10 = 4.90
[OH-] = 10-4.90
= 1.26 x 10-5 mol dm-3
[OH-] = c = [OH-] / c
= 1.26 x 10-5 / 0.100= 1.26 x 10-4
j) At 298 K, 0.200 mol dm-3 of nitrous acid, HNO2 is 1.5% dissociated. Calculate a) the value of pKa (b) the degree of dissociation for 0.500 mol dm-3 and its pH.
k) A solution was prepared by mixing 0.15 mol dm-3 of ethanoic acid and 0.010 mol of hydrochloric acid and making up the solution 1.00 dm3 with distilled water. [Ka of CH3COOH = 1.80 x 10-5 mol dm-3]. Calculate a) [CH3COO-] at equilibrium b) pH of the mixture solution
a) Since it is 1.5% ionised, = 0.015
Ka = c 2
Ka = 0.200 x 0.0152
= 4.50 x 10-5 mol dm-3
pKa = 4.35
b) Ka = c 2
4.50 x 10-5 = 0.500 x 2
= 9.49 x 10-3 [H+] = c = 0.500 x 9.49 x 10-3
= 4.74 x 10-3 mol dm-3
pH = 2.32
a) CH3COOH + H2O CH3COO- + H3O+ Initial : 0.15 0 0
At equilibrium : 0.15 (1-x) 0.15 x 0.15 xsince the conc of H3O+ = 0.010 + 0.15 x ≈ 0.010 (as x is very small)
Ka = [CH3COO-][H3O+] / [CH3COOH] 1.80 x 10-5 = [CH3COO-] (0.010) / 0.15
[CH3COO-] = 0.15 x 1.80 x 10-5 / 0.010= 2.7 x 10-4 mol dm-3
[H3O+] = 2.7 x 10-4 + 0.010 = 0.01027 mol dm-3
pH = 1.99 ≈ 2
7.7 Acid – Base Indicator ~ is a water soluble dye that change colour
according to the concentration of hydrogen ions in the solution to which indicator is added.
Example, litmus paper is a weak organic acid which is represented using formula HIn.
HIn (aq) + H2O (l) H3O+ (aq) + In- (aq)
red blue If the solution is acidic (high in H3O+) ; equilibrium
will shift to the left to form more red colour molecules
If the solution is basic, (less in H3O+) ; equilibrium will shift to the right and make the solution look more blue.
The equilibrium constant for the equation above is called indicator dissociation constant, KIn
HIn (aq) + H2O (l) H3O+ (aq) + In- (aq)
If > 10, colour of acid indicator will be seen
If < 10-1, colour of conjugate base will be seen
To the reason above, the detectable colour range of indicator is pKHIn ± 1.
If [HIn] = [In-] , the intermediate colour of the indicator is seen.
baseconjugateofcolour
indicatoracidofcolour
]In[
]HIn[lgpKpH,thus
]In[
]HIn[lgKlg]OH[lg,equationthetolgtaking
]In[
]HIn[K]OH[rearrange
]HIn[
]In][OH[K
HIn
HIn3
HIn33
HIn
][
][In
HIn
][
][In
HIn
Indicator Colour changeAcid – base
Approximate
pH rangepKHIn
Methyl orange Red-orange – Yellow 3.1 – 4.4 3.7
Bromophenol blue Yellow-blue – Violet 3.0 – 4.6 3.8
Methyl red Red – Yellow 4.2 – 6.1 5.0
Bromothymol blue Yellow – Blue 6.0 – 7.6 7.1
Phenol red Yellow – Red 6.4 – 8.2 7.8
Phenolphetalein Colourless – Pink 8.3 – 9.8 9.3
7.7.1 The End-Point of a Titration The equivalence point refers to the point at which
there are equal amounts (in moles) of H3O+ (aq) and OH- (aq) in the titration flask. Equivalence point can be tell by noting the colour change of the acid base indicator. The point in a titration at which the indicator changes colour is called end-point.
At end-point of the titration, All the acid has been neutralised by alkaliThe solution in the conical flask contain salt and
water onlyThe colour of the indicator will be midway between
the colour of acid and its conjugate base
Titration of Strong Acid – Strong Base (HCl – NaOH) The change of pH that occur during titration of hydrochloric acid
and sodium hydroxide is followed by using a pH meter. Alkali solution is added slowly to 25 mL HCl 0.100 M in a beaker and the pH changes occur for each mL of alkali solution added is recorded
1.At the beginning, pH value is low (strong acid)
2.Before the equivalence point / end point, each mL of NaOH added will cause the pH value raise until when it reaches around 24 mL where it reaches around pH 7.
3.Equivalence point / end point occur at 25 mL of NaOH 0.100 M. If one drop of NaOH is added to this mixture now, the pH will change sharply from 3.5 to 10.0. The pH at the equivalence point is the pH of the resultant salt. Thus the pH at equivalence point is 7.0, which is the pH of sodium chloride solution.
4.After the equivalence point / end point, the pH raise slowly for each mL of NaOH added.
The choice of indicator for strong acid – strong base titration is somewhere from pH range 3.5 to 10.0. Thus, methyl orange, Bromothymol blue and phenolphthalein can be used as indicator.
Methyl orange
Bromothymol blue
Phenolphthalein
Titration Curve of Strong Acid – Strong Base (HCl – NaOH)
2
4
6
0
8
10
12
14
5 10 15 20 25 30 35
End point7
Equivalence point
Titration of Strong Acid – Weak Base (HCl – NH3) The pH obtained when 0.10 mol dm-3 ammonia solution (weak
base) is added to 25.0 cm3 of 0.10 mol dm-3 hydrochloric acid is shown in graph below
1.At the beginning, pH value is low (strong acid)
2.Before the equivalence point / end point, each cm3 of NH3 added will cause the pH value raise until when it reaches around 24 cm3 where it reaches pH = 5.3
3.At equivalence point (pH = 5..3), it is less than 7 because the salt of ammonium chloride will undergoes hydrolysis. The ammonium ions react with water to produce hydroxonium ion.
NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)
This will result the solution being acidic.
4.After equivalence point, another drop of ammonium solution added to the mixture will cause the pH to raise gradually to 7
5.Therefore, methyl orange (pH range of 3.1 – 4.4) is a suitable indicator because it changes colour within the pH range of 3 to 7, the vertical portion of graph.
Methyl orange
Phenolphthalein
Bromothymol blue
Titration Curve of Strong Acid – Weak Base (HCl – NH3)
2
4
6
0
8
10
12
14
5 10 15 20 25 30 35
End point
7
Equivalence point
Titration of Weak Acid – Strong Base (NaOH – CH3COOH) The pH curve obtained when 25.0 cm3 of 0.100 mol dm-3 of
ethanoic acid, CH3COOH is titrated against a 0.100 mol dm-3 of sodium hydroxide, NaOH is shown in figure below
1. At the beginning, pH value is rather high (weak acid)
2. Before the equivalence point / end point, each cm3 of NaOH added will cause the pH value raise until when it reaches around 24 cm3 where it reaches pH = 6.5
3. The equivalence point (pH = 9), it is more than 7 because the salt of sodium ethanoate is formed. The ethanoate ions react with water to produce hydroxide ion.
CH3COO- (aq) + H2O (l) CH3COOH(aq) + OH- (aq)
This will result the solution being basic.
4.After equivalence point, another drop of sodium hydroxide added to the mixture will cause the pH to raise gradually to 10.5
5.Therefore, phenolphthalein (pH range of 8.0 – 9.8) is a suitable indicator because it changes colour within the pH range of 6.5–10.5, the vertical portion of graph
Bromothymol blue
Phenolphthalein
Methyl orange
Titration Curve of Weak Acid – Strong Base (CH3COOH – NaOH)
2
4
6
0
8
10
12
14
5 10 15 20 25 30 35
End point7
Equivalence point
Bromothymol blue
Phenolphthalein
Methyl orange
2
4
6
0
8
10
12
14
5 10 15 20 25 30 35
7
Titration of weak acid and weak base (CH3COOH – NH3) There is no sharp rise in pH at equivalence point. Instead, the pH
increases gradually upon increasing of base. Therefore, there’s no suitable indicator for this titration.
Titration of polyprotic acids and strong base When a weak polyprotic acid, like phosphoric (V) acid, is titrated
with NaOH, the titration curve obtained shows separate equivalence points for each displaceable proton.
H3PO4 (aq) + NaOH (aq) NaH2PO4 (aq) + H2O ……………… (1)
NaH2PO4 (aq) + NaOH (aq) NaHPO4 (aq) + H2O ……………… (2)
NaHPO4 (aq) + NaOH (aq) Na3PO4 (aq) + H2O ……………… (3)
1.From the equation 1 – 3 above, there should be 3 level of equivalence point. The first equivalence point occurs at pH 3.6, which is correspondent to the equation (1). At this point, methyl orange is the most suitable indicator.
2.As for the second equivalence point, it occurs at pH 9.0, which is correspondent to the (2). Phenolphthalein, at this moment, is the most suitable indicator.
3.The third equivalence point should be around 75.0 cm3, according to eq. (3). However this is realised in this titration as no sharp increase in pH about 13 (pH for Na3PO4) as this is the pH of NaOH used in this titration
Bromothymol blue
Phenolphthalein
Methyl orange
Titration Curve of Polyproctic Acid – Strong Base (H3PO4 – NaOH)
2
4
6
0
8
10
12
14
10 20 30 40 50 60 70
7
Titration involving Sodium Carbonate and Hydrochloric Acid the pH curve obtained when 25.0 cm3 of 0.100 mol dm-3 of sodium
carbonate is titrate with 0.100 mol dm-3 of hydrochloric acid.
1.The first corresponding equivalence point occur at pH = 9. This is due to the formation of acid salt, NaHCO3 as
Na2CO3 (aq) + HCl (aq) NaHCO3 (aq) + NaCl (aq)
Phenolphthalein is the most suitable indicator for this equivalence point.
2.The second equivalence point occur at pH = 4 at 50.0 cm3 HCl added. This corresponds to the formation of carbonic acid.
NaHCO3 (aq) + HCl (aq) NaCl (aq) + CO2 (g) + H2O (l)
Or HCO3- (aq) + H+ (aq) H2O (l) + CO2 (g)
Methyl orange is the most suitable indicator for this equivalence point
Bromothymol blue
Phenolphthalein
Methyl orange
Titration Curve of Polyproctic Base – Strong Acid (HCl – Na2CO3)
2
4
6
0
8
10
12
14
10 20 30 40 50 60 70
7
1. When 25.0 cm3 of 0.100 dm-3 HCl (aq) is titrated with 0.100 mol dm-3 NaOH (aq), the pH of the solution increased. Calculate the pH of the solution after the following volume of NaOH have been addeda) 24.90 cm3 b) 25.10 cm3
a) NaOH + HCl NaCl + H2O
Mol H+ = MV / 1000 = (0.100)(25.0)/1000
= 0.0025 molMol OH- = MV / 1000
= (0.100)(24.90)/1000
= 0.00249 mol (lim)Mol of excess H+ = 0.0025-
0.00249 = 0.00001
molConc. of H+ = mol x 1000 / Vtot
[H+]= 0.00001 x 1000/(24.9+25.0)
= 2.00 x 10-4 mol dm-3
pH = - lg 2.00 x 10-4 mol dm-3
= 3.70
a) NaOH + HCl NaCl + H2O
Mol H+ = MV / 1000 = (0.100)(25.0)/1000
= 0.0025 mol (lim)Mol OH- = MV / 1000
= (0.100)(25.10)/1000
= 0.00251 molMol excess OH- = 0.00251-
0.0025 = 0.00001
molConc. of OH- = mol x 1000 /
Vtot
[OH-]= 0.00001x 1000/(25.1+25.0)
= 1.996 x 10-4 mol dm-3
pOH = - lg 1.996 x 10-4 mol dm-3
= 3.70pH = 14 – 3.70 = 10.3
2. When 15.0 cm3 of 0.200 dm-3 H2SO4 (aq) is titrated with 0.100 mol dm-3 NaOH (aq), the pH of the solution increased. Calculate the pH of the solution after the following volume of NaOH have been added
a) 55.0 cm3 b) 65.0 cm3
2 NaOH + H2SO4 Na2SO4 + 2 H2O
Mol of H2SO4 = MV / 1000 = (0.200)(15.0)/1000
= 0.0030 molSince 1 mol of H2SO4 = 2 mol
H+
Mol of H+ = 0.0030 x 2 = 0.0060 mol
Mol OH- = MV / 1000 = (0.100)(55.0)/1000
= 0.0055 mol (lim)Mol of excess H+ = 0.006 -
0.0055 = 0.0005
molConc. of H+ = mol x 1000 / Vtot
[H+]= 0.0005 x 1000/(15.0 + 55.0)
= 7.14 x 10-3 mol dm-3
pH = - lg 7.14 x 10-3 mol dm-3
= 2.15
2 NaOH + H2SO4 Na2SO4 + 2 H2O
Mol of H2SO4 = MV / 1000 = (0.200)(15.0)/1000
= 0.0030 molMol of H+ = 0.0030 x 2
= 0.0060 mol (lim)Mol OH- = MV / 1000
= (0.100)(65.0)/1000= 0.0065 mol
Mol of excess OH- = 0.0065 - 0.0060
= 0.0005 mol
Conc. of OH- = mol x 1000 / Vtot
[OH-]= 0.0005 x 1000/(15.0 + 65.0)
= 6.25 x 10-3 mol dm-3
pOH = - lg 6.25 x 10-3 mol dm-3
= 2.20pH = 14 – 2.20
= 11.80
1. 25.0 cm3 of a solution containing sodium carbonate, Na2CO3 and sodium bicarbonate, NaHCO3 required 17.5 cm3 of 0.100 mol dm-3 hydrochloric acid for neutralisation with phenolphthalein as indicator and 41.3 cm3 of the same acid with methyl orange indicator. Calculate the (a) concentration of Na2CO3 in mol dm-3 (b) concentration of sodium bicarbonate in
mol dm-3
With phenolphthalein as indicator :
End point for the reactions are for
Na2CO3 + HCl NaHCO3 + NaCl
MNa2CO3 = 0.0700 mol dm-3
1
1
VM
VM
HClHCl
CONaCONa 3232
1
1
)50.17)(100.0(
)0.25(M32CONa
With methyl orange as indicator :
End point for the reactions are for
NaHCO3 + HCl NaCl + H2O + CO2
Since NaHCO3 come from 2 sources Na2CO3 NaHCO3 NaClor directly NaHCO3 (from mixture) NaClVHCl for Na2CO3 = 17.5 + 17.5 =
35.0VHCl for NaHCO3 mixture = 41.3
– 35= 6.3
cm3
MNaHCO3 = 0.0252 mol dm-3
1
1
VM
VM
HClHCl
NaHCONaHCO 33
1
1
)30.6)(100.0(
)0.25(M3NaHCO
10.8 Buffer solution A buffer solution is a solution
……………………………………………………………….…… There are 2 types of buffer solutions.
An acidic buffer that keeps the pH below 7 An alkaline buffer that keeps the pH above 7
Acidic buffer solutions consist of a weak acid and salt of the weak acid. The salt contributes the conjugate base of the weak acid. Example:
An aqueous solution which contain equal amounts in moles of ethanoic acid (CH3COOH) and salt, sodium ethanoate (CH3COONa). The salt supplies the conjugate base, CH3COO- (aq) of weak ethanoic acid. This buffer solution pH 5
An aqueous solution which contains a mixture of sodium dihydrogen phosphate, NaH2PO4, which act as the weak acid and disodium hydrogen phosphate, Na2HPO4 act as the salt. The salt supplies conjugate base, HPO4
- (aq) of weak acid. Alkali buffer solutions consist of weak base and salt of the weak
base. The salt contributes the conjugate acid of the weak base. Example :
An aqueous solution which contain equal amounts in moles of ammonia solution (NH3) and salt, ammonium chloride (NH4Cl). The salt supplies the conjugate acid, NH4
+ (aq) of weak base
where pH do not change much when a little acid or base is added
10.8.1 The Action of Acidic Buffer solution An acidic buffer solution has a pH around 4 – 7. To see how
an acidic buffer solution works, consider a mixture of sodium ethanoate and ethanoic acid.
CH3COONa (aq) CH3COO- (aq) + Na+ (aq) …. fully dissociates
CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) ……… partially dissociates
The mixtures contain a high concentration of conjugate base, CH3COO- (aq) yet the concentration of H+ (aq) is low If small amount of acid is added to the buffer solution, the
large number of ethanoate ion will react with the H+
contribute from the acid added. Therefore, adding H+ will not effect much the system thus the pH remain virtually constants
If a small amount of alkali is added to the buffer solution, the hydrogen ion from weak acid will react with OH- from the base. Removing hydrogen ion will cause the equilibrium of ethanoic acid shift to the right. So, the H+ lost will be compensated and the pH of the buffer solution will remain constant.
10.8.2 The Action of Alkali Buffer Solution An alkali buffer solution has a pH around 7 – 10. To see how
an alkali buffer solution works, consider a mixture of ammonia solution and ammonium chloride
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) …… partially
dissociates
NH4Cl (aq) NH4+ (aq) + Cl- (aq) …... completely
dissociates The mixtures contain a high concentration of conjugate
acid, NH4+ (aq) yet the concentration of OH- is low.
If a small amount of acid is added to the buffer solution, it will neutralised the OH- from the ammonia solution. Here, the equilibrium will shift to the right to produce more OH- so the hydroxide ion lost will be compensated and the pH remains approximately constant.
If a small amount of base is added, the NH4+ from the salt
will react with the OH- supplied, to form ammonia solution. Again, this will keep the pH of alkaline solution remain constant.
The pH value of an acidic solution can be calculated from the acid dissociation constant, Ka and also the concentration of acid and salt.
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
and salt : MA M+ (aq) + A- (aq)
When –lg the equation :
Since [A-] contributed by salt while[HA] contributed by weak acid
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
and salt : NH4Cl NH4+ + Cl-
When –lg the equation :
Since [NH4+] contributed by salt while
[NH3] contributed by weak base
]HA[
]A][OH[K 3
a
]NH[
]OH][NH[K
3
4b
]A[
]HA[K]OH[ a3
]A[
]HA[lgKlg]OH[lg a3
]salt[
]acid[lgpKpH a
]NH[
]NH[K]OH[
4
3b
]salt[
]base[lgpKpOH b
]NH[
]NH[lgKlg]OH[lg
4
3b
Example : Calculate the pH of a buffer solution prepared by dissolving 32.8 g of sodium ethanoate, NaCH3COO in 1.00 dm3 of 0.100 mol dm-3 ethanoic acid, CH3COOH. [Ka = 1.75 x 10-5 mol dm-3]
Example : Calculate the pH of a buffer solution prepared by adding 10.7 g of ammonium chloride to 500 cm3 of 0.500 mol dm-3 ammonia solution (Kb = 1.80 x 10-5 mol dm-3)
pH = 5.36][
][lgsalt
acidpKpH a
)(
)100.0(lg)1075.1lg(
828.32
5 pH
)(
)500.0(lg)1080.1lg(
35.0
5.537.10
5
dm
pOH
][
][lgsalt
basepKpOH b
pOH = 4.65pH = 14 – pOH = 9.35
10.8.4 Buffer Capacity ~ amount of acid / base that can be absorbed by the buffer
solution without a significant change of pH. Larger the capacity, greater the amount of acid / base must be added in order to produce a significant change.
Buffer capacity depends on the concentration of the weak acid (HA) and its conjugate base (A-) present.
The pH range in which a buffer effectively neutralise added acids and bases and
If the ratio = 1.0, then pH = pKa ;
if the ratio = 0.1, then pH = pKa + 1 ;
if the ratio = 10, then pH = pKa – 1.
So the significant concentration of both acid and salt
present should be in between 10-1 < < 10
][
][
acid
salt
][
][
acid
salt
][
][
acid
salt
][
][
acid
salt
Where maximum buffer capacity is achieved
Titration Curve of Weak Acid – Strong Base (CH3COOH – NaOH)
2
4
6
0
8
10
12
14
5 10 15 20 25 30 35
End point7
Equivalence point
½ V
pH = pKa
buffer
zone
Titration Curve of Strong Acid – Weak Base (HCl – NH3)
2
4
6
0
8
10
12
14
5 10 15 20 25 30 35
End point
7
Equivalence point
½ V
14-pHpOH = pKb
buffer zone
Example : 25.0 cm3 of 0.100 mol dm-3 ethanoic acid was titrated with 0.100 mol dm-3 sodium hydroxide solution
a) Write an equation for the reaction above.
b) Sketch a pH curve for the reaction and indicate in your graph the region where the mixture can function as a buffer.
c) What is the pH of the solution when it has maximum buffer capacity.
a) CH3COOH + NaOH CH3COONa + H2O
10.8.5 Preparing a buffer solution of specified pH Acidic buffer solution can be prepared by adding
………………… to ……………………………… while basic buffer solution can be prepared by adding ……………..to …………………
The following steps are used to prepare buffer solutions of a specified pH
1 Choose a weak acid / base whose pKa / pKb value is close to the desire pH of buffer solution
2 Choose a suitable salt of the weak acid / base3 Calculate the required amount of mixture by adjusting
[acid] / [salt] or [base] / [salt] ratio
Weak acid salt of the weak acid Weak base salt of weak
base
Example : Explain how a buffer solution with a pH 5.40 can be prepared from ethanoic acid and sodium ethanoate [ pKa of CH3COOH = 4.76]
Example : Explain how a buffer solution with a pH 10.2 can be prepared from ammonia aqueous solution and ammonium chloride, NH4Cl [ Kb of NH3 = 1.78 x 10-5 mol dm-3]
Mol of salt = 4.365Mass of salt use = mol x RMM
= 4.365 x [12(2) +1(3) + 16(2) + 23= 358 g
Buffer Solution with pH 5.40 is prepared by dissolving 358 g of CH3COONa into 1.0 dm3 of CH3COOH 1.00 mol dm-3
]acid[
]salt[lgpKpH a
365.4]acid[
]salt[64.0
]acid[
]salt[lg
]acid[
]salt[lg76.440.5
112.0]base[
]salt[95.0
]base[
]salt[lg
]base[
]salt[lg)1078.1lg(8.3 5
]base[
]salt[lgpKpOH
8.32.1014
pH14pOH
b
Mol of salt = 0.112 molMass of salt use = mol x RMM = 0.112 x [14(1) + 1(4) +
35.5(1) = 5.99 g
Buffer solution with pH 10.2 can be prepared by dissolving 5.99 g of NH4Cl into 1.0 dm3 of NH3 1.00 mol dm-3
Example : calculate the mass of sodium ethanoate solution that has to be added to 1.00 dm3 of 0.100 mol dm-3 ethanoic acid to prepare a buffer solution of pH 4.50 [ Ka of CH3COOH = 1.74 x 10-5 mol dm-3]
Since conc of acid = 0.100[salt] = 0.5495 x 0.100Mol = MV = 0.05495 x 1 = 0.05495
molMass of salt use = mol x RMM= 0.05495 x [12(2) +1(3) + 16(2)
+ 23= 4.51 g
]acid[
]salt[lgpKpH a
5495.0]acid[
]salt[26.0
]acid[
]salt[lg
]acid[
]salt[lg1074.1lg50.4 5
10.8.6 Application of Buffer Solution Chemical Analysis – Some titration, for example EDTA
titrations, have to be carried out in the solutions in which the pH is maintained at a constant value
Research in bacteriology – buffer solution is needed to maintain the pH of culture media so that the bacteria can propagate under the right pH condition.
Fermentation processes – buffer system are essential in biological reactions that are catalysed by enzymes. Enzymes have very narrow and specific pH limit at which it work the best. Hence fermentation processes that utilised yeast as the fermenting organism must take place in buffer solution. Otherwise, small changes in pH will cause the death of yeast and stop the fermentation.
Industrial processes – buffer solution play an important role in many industrial processes. They are used in the manufacture of dyes and photographic material
The importance of buffers in biological systems. Buffer solutions are very important in biological systems. Blood, for example, is a buffer solution that can soak up the acids and bases produced in biological reactions. The pH of human blood is carefully controlled at a value very close to 7.4 by conjugate acid–base pairs, primarily H2CO3 and its conjugate base HCO3
-. The oxygen-carrying ability of blood depends on control of the pH to within 0.1 pH unit.
11.0 Solubility Equilibrium ~ equilibrium between .............. undissociated
solute (in solid) and its .................................... ions in the solution.
When increasing quantities of a sparingly soluble ionic solid are added to water, a saturated solution is eventually formed.
The dissolving and precipitating of ionic compounds are phenomena that occur both within us and around us. Examples:
1. The dissolving of tooth enamel in acidic solutions causes tooth decay.
2. The precipitation of certain salts in our kidney produces kidney stones.
3. The precipitation of CaCO3 from groundwater is responsible for the formation of stalactites and stalagmites within limestone caves.
Salt ofOppositely charged
11.1 Solubility product When a sparingly soluble salt can no more dissolve in
solvent, a ..................... solution is formed For example, when dissolving silver chloride, AgCl (s) ;
............................................................. The product of the concentrations of Ag+ (aq) + Cl-
(aq) ions in a saturated solution of silver chloride is called as ..............................of silver chloride. Based on the equation
Since the concentration of a pure solid is constant, the above simplifies to become
Another example is lead (II) hydroxide, which is a
base. It is sparingly soluble in water. In aqueous solution, lead (II) hydroxide dissociate according to the equation ,
Pb(OH)2 ↔
saturated
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
solubility product K x [AgCl] = [Ag+][ Cl- ]
Ksp = [Ag+][ Cl- ]K x [constant] = [Ag+][ Cl- ]
Pb2+ (aq) + 2 OH- (aq)
Compound Dissociation equation Solubility product, Ksp Unit
HgCl
PbBr2
Ag2CrO4
BaSO4
Fe2(CO3)3
Ca3(PO4)2
So, the Ksp of lead (II) hydroxide is written as
In general, AxBy ↔ x Ay+ (aq) + y Bx- (aq)Ksp = [Pb2+][OH-]2
Ksp =[Hg+][Cl-]HgCl ↔ Hg+ + Cl-
Ksp = [Ay+]x [Bx-]y
mol2 dm-6
Ksp =[Pb2+][Br-]2PbBr2 ↔ Pb2+ + 2 Br- mol3 dm-9
Ksp =[Ag+]2[CrO4
2-]Ag2CrO4 ↔ 2 Ag+ + CrO4
2- mol3 dm-9
Ksp =[Ba2+][SO42-]BaSO4 ↔ Ba2+ + SO4
2- mol2 dm-6
Ksp =[Fe3+]2[CO3
2-]3
Fe2(CO3)3 ↔ 2Fe3+ + 3CO32- mol5 dm-15
Ksp=[Ca2+]3[PO43-]2Ca3(PO4)2 ↔ 3 Ca2+ + 2PO4
3- mol5 dm-15
11.2 Definition : Solubility & Solubility Product11.2.1 Solubility Solubility is the quantity of solute that dissolved in unit
volume of saturated solution at a given temperature. Thus solubility is often measured in g dm-3 or mol dm-3.
11.2.2 Solubility Product, Ksp
The solubility product, Ksp, of a sparingly soluble electrolyte is the product of the concentrations of the ions in a saturated solution at a given temperature. The concentration of ions are raised to appropriate powers depending on the stoichiometry of the equation. Therefore, there is no fixed units for Ksp.
Ksp measures the extent a sparingly soluble salt can dissolve in water. The higher the Ksp value, the .................. the salt.
g dm-3 mol dm-3 RMM
RMM x
more soluble
Example 1 : Calculation of Ksp from Solubility values. Given the solubility of AgI is 6.24 x 10-8 mol dm-3, calculate its Ksp
Example 2 : Calculation of Solubility given Ksp. Calculate the solubility of CaCO3 given that its solubility product, Ksp, is 4.72 x 10-15 mol2 dm-6.
Given the solubility of Pbl2 is 1.21 x 10-3 mol dm-3, calculate its Ksp.
Calculate the solubility of Ag2CrO4, given that its solubility product, Ksp, is 2.4 x 10-12 mol3 dm-9.
Agl (s) ↔ Ag+ (aq) + I- (aq) x x xKsp = [Ag+][I-] = (x)(x) = (6.24 x 10-8)2
= 3.89 x 10-15 mol2 dm-6
CaCO3 (s) ↔ Ca2+(aq)+ CO32-(aq)
x x xKsp = [Ca2+][CO3
2-]4.72 x 10-15 = (x)(x) = 6.87 x 10-8 mol dm-3
PbI2 ↔ Pb2+ + 2 I-
x x 2xKsp = [Pb2+][I-]2
= (x)(2x)2 = 4x3
= 4 (1.21 x 10-3)3
= 7.09 x 10-9 mol3 dm-9
Ag2CrO4 ↔ 2 Ag+ + CrO42-
x 2x xKsp = [Ag+]2 [CrO4
2-] = (2x)2 (x) = 4x3
2.4 x 10-12 = 4 (x)3
x = 8.4 x 10-5 mol dm-3
Given the solubility of Fe(OH)3 is 3.49 x 10-5 mol dm-3 calculate its Ksp
Calculate the solubility of Al(OH)3, given that its solubility product, Ksp, is 1.9 x 10-33 mol4 dm-12.
Fe(OH)3 ↔ Fe3+ + 3 OH-
x x 3xKsp = [Fe3+][OH-]3
= (x)(3x)3 = 27x4
= 27 (3.49 x 10-5)4
= 4.01 x 10-17 mol4 dm-12
Al(OH)3 ↔ Al3+ + 3 OH-
x x 3xKsp = [Al3+] [OH-]3
= (x) (3x)3 = 27x4
1.9 x 10-33 = 27 (x)4
x = 2.9 x 10-9 mol dm-3
Be sure not to confuse solubility with solubility product (Ksp).
The solubility of a salt is the amount present in a unit amount of a saturated solution expressed in mol dm-
3 or g dm-3. The solubility product is an equilibrium constant. Ksp
values are always small, usually < 10-4. Generally, a low vaiue of Ksp means that:
i) The concentration of the ions is low at equilibrium.ii) The substance is not very soluble There is a connection between them - if either is
known, the other can be calculated.
Example 3: The Ksp values for the silver haiides are given below.
Arrange these salts in increasing order of solubility.
Note : Direct comparisons of solubility on the basis of Ksp values can only be made for salts having the same ion ratio. For example, the ion ratio for silver haiides is 1:1, the ion ratio for lead halides is 1:2.
Salt AgCl AgBr AgI
Ksp / mol2 dm-6 1.8 x 10-10 3.3 x 10-13 1.5 x 10-16
AgI < AgBr < AgCl
11.3 Solubility and the Common Ion Effect The solubility of a substance is affected not only by
temperature but also by the presence of other solutes. Example: The value for Ksp for manganese(ll)
hydroxide, Mn(OH)2, is 1.6 x 10-13 mol3 dm-9. Calculate the solubility of Mn(OH)2 in
a) In waterIn water, Mn(OH)2 dissociate according to the equationMn(OH)2 ↔ Mn2+ + 2 OH-
Ksp = [Mn2+][OH-]2
= (x)(2x)2 = 4x3
1.6 x 10-13 = 4x3
x = 3.4 x 10-5 mol dm-3
b) In 0.020 mol dm-3 NaOHIn base, OH- is contributed mainly by NaOH, so [OH-] = 0.020Mn(OH)2 ↔ Mn2+ + 2 OH-
Ksp = [Mn2+][OH-]2
1.6 x 10-13 = (x)(0.020)2 x = 4.0 x 10-10 mol dm-3
Conclusion: In the presence of a common ion, OH- in this case, the solubility of the Mn(OH)2 ........................... in NaOH solution.
Mn(OH)2 (s) ↔ Mn2+ (aq) + 2 OH- (aq)
This is in accordance with Le Chatelier's principle. As [OH-] increases, the position of equilibrium shifts to the ............................................... ]
decreased
left, forming more Mn(OH)2
11.4 Applications of Solubility Product1. To Determine the Solubility of Sparingly Soluble Saits2. Predicting Precipitation • At a particular temperature, Ksp value indicates the
maximum product of ion concentrations in solution at equilibrium. MX (s) M+ (aq) + X- (aq)
Ionic product, Q = [M+] [X-] If ionic product Q = Ksp
The system is at equilibrium and the solution is saturated, formation of precipitate is observed.
If ionic product [M+] [X-] < Ksp
The system is not at equilibrium and the solution is not saturated (ie. more solid can dissolve until equilibrium is achieved), so no precipitate formed
If ionic product [M+] [X-] > Ksp The system is not at equilibrium and the solution is
supersaturated. The concentrations of the ions in solution are too high and thus the salt will precipitate.
Example 1: Will a precipitate form when 50 cm3 of 0.050 mol dm-3 AgNO3 is added to 50 cm3 of 0.10 mol dm-3 KBrO3?
[Ksp AgBrO3 = 6.0 x 10-5 mol2 dm-6]
Mol AgNO3 = MV / 1000
= (0.050)(50) / 1000 = 2.5 x 10-3 mol
Mol KBrO3 = MV / 1000
= (0.10)(50) / 1000 = 5.0 x 10-3 mol
[AgNO3] = mol x 1000 / V = (2.5 x
10-3)(1000)/(50+50) = 0.025 mol dm-3
[KBrO3] = mol x 1000 / V
= (5.0 x 10-3)(1000)/(50+50)
= 0.050 mol dm-3Since AgBrO3 ↔ Ag+ + BrO3
-
Q = [Ag+][BrO3-]
= (0.025)(0.050) = 1.25 x 10-3 mol2 dm-6
Since Q > Ksp ; so precipitate of AgBrO3 will form.
Example 2: Will a precipitate of BaF2 form if 150 cm3 of 0.050 mol dm-3 of Ba(NO3)2 is mixed with 50 cm3 of 0.0050 mol dm-3 of KF? [Ksp BaF2 = 1.7 x 10-6 mol3 dm-9]
Mol Ba(NO3)2 = MV / 1000
= (0.050)(150) / 1000 = 7.5 x 10-3 mol
Mol KF = MV / 1000 = (0.0050)(50)/1000 = 2.5 x 10-4 mol
[Ba(NO3)2] = mol x 1000 / V = (7.5 x
10-3)(1000)/(150+50) = 0.0375 mol dm-3
[KF] = mol x 1000 / V = (2.5 x
10-4)(1000)/(150+50) = 0.00125 mol dm-3Since BaF2 ↔ Ba2+ + 2 F-
Q = [Ba2+][F-]2
= (0.0375)(0.00125)2
= 5.86 x 10-8 mol3 dm-9
Since Q < Ksp ; so no precipitate of BaF2 will form.
There are limitations of solubility products. For example, Ksp is not applicable when there is the formation of .........................................................
Example 3: PbCl2 in dilute HCI: PbCI2 (s) ↔ Pb2+ (aq) + 2Cl- (aq) (1)
HCI (aq) H+ (aq) + Cl- (aq) In the presence of dilute HCI, solubility of
PbCI2 ....................... due to the .......................................... which shifts the position of equilibrium (1) to ........
When PbCl2 in concentrated HCI: PbCI2 (s) ↔ Pb2+ (aq) + 2CI- (aq) (2)
Pb2+ (aq) + 4CI- (aq) ↔ PbCl42- (aq) The formation of the complex ion shifts the position of
equilibrium 1 to the ......... Thus PbCI2(s) dissolves readily in concentrated HCI.
Insoluble / soluble complex salt
decreasepresence of chloride ion
left
right
1 When NaOH is added to a 0.10 mol dm-3 magnesium nitrate solution, at what pH will magnesium hydroxide begin to precipitate?(Ksp of Mg(OH)2 =1.2 x 10-11 mol3 dm-
9)
2. The solubility product of lead (II) iodide is 1.4 x 10-8 mol3 dm-9. a. Calculate the mass of lead(II) iodide that will dissolve in
500 cm3 of water.
[Mg2+] before Mg(OH)2 precipitated = 0.10 mol dm-3.Mg(OH)2 ↔ Mg2+ + 2 OH-
Q = [Mg2+][OH-]2 = 1.2 x 10-11 Q = 0.10 (x)2 = 1.2 x 10-11
x = [OH-] = 1.1 x 10-5 mol dm-3
pOH = -lg (1.1 x 10-5) = 4.96 pH = 14 – 4.96 = 9.0
PbI2 (s) ↔ Pb2+ (aq) + 2 I- (aq) Ksp = [Pb2+] [I-]2
Ksp = (x)(2x)2 1.4 x 10-8 = 4x3
x = 1.5 x10-3 mol dm-3
Mol of PbI2 in 500 cm3 water = (1.5 x 10-3) x 500 / 1000
= 0.00075 molMass = mol x RMM ; Mass = 0.0075 x (207) + 2(127)
= 0.35 g.
b. When 500 cm3 0.10 mol dm-3 potassium iodide solution is added to lead(II) iodide, what happens to the solubility of lead(II) iodide?
c. What is this effect called?
d. Calculate the mass of PbI2 that will dissolve in 500 cm3 0.10 mol dm-3 potassium iodide solution.
Solubility of PbI2 decrease.
Common ion effect
PbI2 (s) ↔ Pb2+ (aq) + 2 I- (aq) Ksp = [Pb2+] [I-]2
1.4 x 10-8 = [x] (0.10)2
x = 1.4 x 10-6 mol dm-3
Mol of PbI2 that dissolves in 500 cm3 KI solution mol= 1.4 x 10-6 x 500 / 1000 mol= 7.0 x 10-7 molMass = (7.0 x 10-7) x [207 + 2(127)]
= 3.2 x 10-4 g
11.5 Systematic Qualitative Analysis The principles of solubility equilbria can be applied in
many areas of qualitative analysis, for example, in the precipitation of metal hydroxides.
The .............................. the Ksp values, the more soluble the hydroxide is.
In systematic qualitative analysis, the precipitation of the different hydroxides can be controlled by using either NaOH (aq) or NH3 (aq) or NH4CI / NH3 (aq).
MOH (s) ↔ M+ (aq) + OH- (aq) Ksp = [M+] [OH-]
Hydroxide Ksp value Hydroxide Ksp value
Fe(OH)3 2.0 x 10-39 Co(OH)2 6.3 x 10-15
AI(OH)3 1.0 x 10-33 Mn(OH)2 2.0 x 10-13
Cr(OH)3 1.0 x 10-33 Mg(OH)2 1.1 x 10-11
Zn(OH)2 2.0 x 10-17 Ca(OH)2 5.5 x 10-6
Fe(OH)2 7.9 x 10-16 Ba(OH)2 5.0 x 10-3
