SK015

Titration Curves Titration Curves

Strong Acid - Strong Base Titration Strong Acid – Weak Base Titration
HCl and NH3 solution
1. At the beginning of titration
- pH has a low value about 1.0. Conical burette
flask
2. Just before the equivalence point
- as NaOH is added slowly to HCl in conical flask, pH pH ≈ 5
increases slowly until just before the equivalence point.
- when one more drop of NaOH is added, the pH change 133 133
sharply from 3.5 to 10.5.

3. At the equivalence point
- the pH at equivalence point is the pH of the salt solution,
in this case, NaCl, pH 7 (neutral salt).

4. After the equivalence point 132 132
- beyond equivalence point, the pH rises slowly.

Titration Curves Titration Curves

Strong Acid – Weak Base Titration Weak Acid – Strong Base Titration

1. At the beginning of titration
- pH has a low value. (solution in conical flask = strong acid)

2. Just before the equivalence point pH ≈ 9 CH3COOH and NaOH
- as NH3 is added slowly to HCl in conical flask, pH increases
slowly until just before the equivalence point. Conical flask burette
- when one more drop of NH3 is added, the pH change
sharplyfrom about 3 to 7.

3. At the equivalence point

- the pH at equivalence point is the pH of the salt solution,

in this case, NH4Cl, pH 5 (acidic salt)

4. After the equivalence point

- beyond the equivalence point, the pH rises slowly which is
134 134
buffer solution started to form. 135

Titration Curves How to sketch a titration curve

Weak Acid – Strong Base Titration Steps :

1. At the beginning of titration 1. Calculate the initial pH of the solution
- pH is about 2.8 because CH3COOH is a weak acid. - Identify the analyte (solution in the conical
flask and given volume). Whether strong acid,
2. Just before the equivalence point weak acid, strong base or weak base.
- as NaOH is added to CH3COOH in conical flask, pH
increases slowly near equivalence point (formation of buffer 2. Determine the equivalence point :
solution).
- at equivalence point, the pH change sharply from about - Volume ( use formula, Ma Va  a )
6.5 to 10.5. MbBb b

3. At the equivalence point - pH (depends on pH of salt)
- the pH at equivalence point is the pH of the salt solution,
in this case, CH3COONa, pH 9 (basic salt) 137

4. After the equivalence point
- beyond the equivalence point, the pH again rises slowly. 136

248

3. pH jump (steep portion / sharp portion) Example 1 :
- depends on the type of the titration
Sketch the titration curve of 25.0 mL 0.10 M HCl with
Type pH jump 0.10 M NaOH.
Strong acid - Strong base 3 – 11
Strong acid – Weak base 3–7 Step 1 :
Weak acid – strong base 7 - 11
Analyte is a strong acid,
HCl – dissociates completely

4. Identify the final pH HCl(aq) H+(aq) + Cl- (aq)

- Depends on the [ ] of the titrant (solution in the [H+] = [HCl] = 0.10 M
burette). pH = - log [H+]

= 1.0

138 139

Step 2 : At equivalence point Step 3 :

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Type of titration : strong acid-strong base
pH jump : 3 – 11
- pH equivalence:

- the solution is NaCl (aq) Step 4 :
- both Na+ and Cl- do not hydrolyse

- therefore pH = 7.0 Titrant is a strong base, NaOH.

- Volume equivalence: NaOH (aq) Na+ (aq) + OH- (aq)

n HCl = n NaOH [OH-] = [NaOH] = 0.10 M

MHCl VHCl = MNaOH VnaOH pOH = - log 0.10 = 1

VNaOH = 0.10 x 25 = 25 mL pH = 14 – 1
0.10
= 13

Final point approaching pH =13.

140 141

Example 2 :

Sketch a titration curve Sketch the titration curve of 25.0 mL 0.10 M NH3 and
0.10 M HCl.

Step 1 :

13 Analyte is a weak base, NH3.
pH 11
equivalent point NH3 (aq) + H2O(l) NH4+ (aq) + OH- (aq)
7 [OH-] = [x] = ?

3 1.8 x 10-5 = x2

1 0.10-x

[x] = 1.3 x 10-3 M

pOH = -log (1.3 x 10-3 )

25 = 2.9
Volume of NaOH added (mL)
pH = 14 – 2.9

142 = 11.1 143 143

249

Step 2 : At equivalence point NH4Cl (aq) Step 3 :
HCl(aq) + NH3 (aq) Type of titration : strong acid-weak base
pH jump : 3 – 7
- pH equivalence
- the solution is NH4Cl (aq) Step 4 :
- only NH4+ hydrolyses to form H3O+ Titrant is a strong acid, HCl.
- therefore pH < 7.0
 pH   log H 3 O 
- Volume equivalence   log 0.1
1
n NH3 = n HCl
Final point approaching pH = 1
M VNH3 NH3 =1

MHCl VHCl 1
M VNH3 NH3 = MHClVHCl
= 0.10 x 25.0 = 25 mL
VNH3
0.10

144 145 145

Sketch a titration curve Example 3 :

pH Sketch the titration curve of 25.0 mL 0.10 M CH3COOH and
11 0.10 M NaOH.

Step 1 :

Analyte is a weak acid, CH3COOH.

7 CH3COOH (aq) + H2O(l) CH3COO- (aq) + H3O+ (aq)
pH<7
equivalent point [H3O+] = [x] = ?
3
1.8 x 10-5 = x2

1 0.10-x
x = 1.3 x 10-3 M

25 pH = -log (1.3 x 10-3 )
= 2.9
Volume of HCl added
(mL) 146 147

Step 2 : At equivalence point Step 3 :
Type of titration : weak acid-strong base
CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) pH jump : 7 - 11

- pH equivalence Step 4 :
- the solution is CH3COONa (aq) Titrant is a strong base, NaOH.
- only CH3COO- hydrolyses to form OH-
- therefore pH  7.0  pOH   log OH 
  log 0 .1
- Volume equivalence 1

n NH3 = n HCl pH  14  1
M VCH3COOH CH3COOH =1  13

M VNaOH NaOH 1 Final point approaching pH =13

M V M V=CH3COOH CH3COOH
NaOH NaOH

VNaOH = 0.10 x 25.0 = 25 mL

0.10

148 149

250

Sketch a titration curve

13 150 151 151
pH 11

7

3

25
Volume of NaOH added (mL)

• Indicator is a substance that is generally added to Choosing a Suitable Indicator
the solution in the receiving vessel and undergoes
some sort of colour change when reaction is over. pH Range for Indicators

• Choose an indicator which the Indicator pH Range Acid colour Base colour
end point pH range lies on the
steep part of the titration Thymol blue 1.2 – 2.8 Red Yellow
curve.
Bromophenol blue 3.0 – 4.6 Yellow Bluish purple

Methyl orange 3.2 – 4.4 Red Yellow

Methyl red 4.8 – 6.0 Red Yellow

• This choice ensures that the Bromothymol blue 6.0 - 7.6 Yellow Blue
pH at the equivalent point will
fall within the range over which Cresol red 7.2 – 8.8 Yellow Red
indicator changes colour.
Phenolphthalein 8.3 – 10.0 Colourless Reddish pink

152 152 153

Choosing a Suitable Indicator Exercise 1

pH Range for Indicators What is the colour of the solution when 3 drops
of the below indicators are added separately to
Types of End Point pH Example of water (pH = 7) ?
Titrations
Range Suitable Indicators

Strong Acid - 3 – 10 Any Indicator Indicator pH range Colour Change
Strong Base (except thymol blue)

Weak Acid- 7 – 11 Phenolphthalein, Phenolphthalein 8.2 – 10.0 Colourless → reddish pink
Strong Base bromothymol blue
Methyl orange 3.2 – 4.2 Red → Yellow
Strong Acid-weak Methyl orange, → Blue
Base 3–7 Methyl red Bromothymol blue 6.0 – 7.6 Yellow

The pH range is defined as the range over which the Phenol red 6.8 – 8.4 Yellow → Red
indicator changes from the acid colour to the base
colour. 154 155

251

Exercise 2

The pH range and the colour change for 3
indicator X, Y and Z is shown in the table
below. Determine X, Y and Z.

Indicator pH range Colour Change
X 1.2 – 2.8
Y 6.0 – 7.6 Red → Yellow
Z 8.3 – 10.5 Yellow → Blue
Colourless → Reddish pink

Refer to the table on slide 153

156

252

At the end of this topic , students should be able to:

a)Define solubility, molar solubility and solubility
product, Ksp.
7.3 b)Calculate Ksp from concentration of ions and vice
SOLUBILITY EQUILIBRIA
versa.
157 c) Predict the possibility of precipitation of slightly
soluble ionic compounds by comparing the values
of ion-product, Q to Ksp.
d)Define and explain the common ion effect.
e)Perform calculations related to common ion
effect.
158

SOLUBILITY EQUILIBRIUM  The solubility of a salt is the amount of
• Some salts are soluble but most are solid that dissolved in a known value of
saturated solution.
insoluble or slightly soluble in water.
 The unit of solubility used may be gL-1 or
• A saturated solution is a solution that molL-1
contains the maximum amount of solute
that can dissolve in a solvent.  Molar solubility is the maximum number of
moles of solute that dissolves in a certain
159 quantity of solvent at a specific temperature.

160

THE SOLUBILITY PRODUCT CONSTANT, Ksp Consider the equilibrium system below :

Ksp is the product of the molar concentrations of MX (s) M+ (aq) + X- (aq)
the ions involved in the equilibrium, each raised
to the power of its stoichiometric coefficient in Kc = [M+] [X-]
the equilibrium equation. [MX]

 Ksp is called the solubility product constant. Kc [MX] = [M+] [X-]
*since [MX] is a constant ;
 The degree of solubility of a salt is shown by
the value of Ksp. Ksp = [M+] [X-]

161 162

253

 Soluble salt such NaCl and KNO3 has Determining Ksp from Solubility
an extremely high value of Ksp .
At 25 C , solubility of CaF2 in water is
 The smaller the value of Ksp the less 1.69  10-2 g/L. What is the value of Ksp for
soluble the compound in water. CaF2 at this temperature?
Solution:
 Temperature ↑ , solubility ↑, Ksp ↑ Solubility of CaF2 = is 1.69  10-2 g/L

163 In mol/l, the solubility is  1.69 102 gL
78 g mol

= 2.17  10-4 M 164

CaF2 (s) Ca2+ (aq) + 2F-(aq) Determining Solubility from Ksp
change -x
x 2x The Ksp for SrCO3 is 1.1  10-10. What is the
molar solubility (in moles per liter) for this
x = solubility of CaF2 = 2.17  10-4 M substance in water?

Solution:

In saturated solution; Let molar solubility of SrCO3 is x
[Ca 2+] = x = 2.17  10-4 M
[F -] = 2x = 2  2.17  10-4 M = 4.34  10-4 M SrCO3 (s) Sr2 + (aq) + 2 -(aq)
change -x
CO3

xx

Ksp = [Ca2+ ] [F-]2 165 Ksp = [Sr2+ ] [CO32-] 166
= (2.17  10-4)(4.34  10-4)2 1.1  10-10 = x2

= 4.09  10-11 M3 x = 1.05  10-5 M

Example

Write the solubility product expression for each
of the following ionic compounds.

a) Ca3(PO4)2 b) Ag2CO3

16716.6 168

254

Solution: Solution:

a) Ca3(PO4)2 b) Ag2CO3
Ag2CO3(s)
Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = 2Ag+(aq) + CO32-(aq)
[ Ag+]2 [ CO32-]
Ksp = [ Ca2+]3 [ PO43-]2

169 170

Example Solution:
The solubility product of silver chromate(VI), Ag2CrO4 is
2.4 x10-12 mol3dm-9. Calculate the concentration of Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq)
Ag+(aq) and CrO42-(aq) in the saturated solution. 2x x
Ans: [Ag+] = 1.68 x 10-5 M
Ksp = [ Ag+]2 [ CrO42-]
[CrO42-] = 8.43 x 10-5 M 2.4 X 10-12 = (2x)2 (x)

171 X= 8.4 x 10-3 = [CrO42-]
[Ag+] = 2(8.4 x 10-3)
1.7 x 10-3 M
=

172

Example Solution:
The solubility of silver sulphide, Ag2S is
5.0x10-17. Calculate the solubility product Ag2S(s) 2Ag+(aq) + S2-(aq)
of Ag2S. Ksp 2x x
Ans: 5.0 x 10-49
= [ Ag+]2 [ S2-]
173 = (2x)2 (x)
= 4X3
= 5.0 x 10-49

174

255

Example Solution:
Calculate the solubility of copper (II) hydroxide,
Cu(OH)2, in g L-1. Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)
Ksp Cu(OH)2 = 2 x 10-20 M3, x 2x
Mr Cu(OH)2 = 97.57g mol-1
Ksp = [Cu2+][OH-]2
175
2 x 10-20 = (x) (2x2) = 4x3

x = 1.709 x 10-7mol/L

Solubility of = 1.709 x 10-7mol/LX 97.57g/mol
Cu(OH)2 = 1.667 x 10-5 g/L

176

Predicting the formation of a precipitate ✔ If a solution containing M+ ions is mixed with a
solution contains A- ions, then the ion product
✔ Precipitate is an insoluble solid formed (in the expression, Q is given as :
solution) and is separated from the solution. MA (s) ⇋ M+ (aq) + A- (aq)
Q = [M+] [A-]
✔ A mixture of two solutions may produce a
precipitate depends on the value of Q. ✔ At equilibrium, the solubility product constant
expressioncan be written as
✔ Q has the same expression as Ksp but the Ksp = [M+] [A-]
concentrations of ions involved are not in
equilibrium (at any given time). 178

177

Three possible situations: Example

Q < Ksp; Will precipitate form when 50 mL 0.001 M
Solution is not saturated. NaOH is added to 150 mL of 0.01 M MgCl2.
Solid will dissolve and no precipitate formed yet. (Ksp Mg(OH)2 = 2 x 10-11)

Q = Ksp; 180
Solution is saturated. Little precipitate formed.

System is in equilibrium.

Q > Ksp;
Solution is supersaturated;

Ions will form precipitate until the ionic concentration

product of the system equals the Ksp (until the
system reaches equilibrium).
179

256

Solution: 50 mL Solution:
0.001 M NaOH
MgCl2 + 2NaOH  Mg(OH)2 + 2NaCl
150 mL
0.01 M MgCl2 MgOH(s) ⇋ Mg2+(aq) + 2OH-(aq)
Q = [ Mg2+] [ OH-]2
Mg(OH)2 (s) Mg2 + (aq) + 2OH- (aq)

No. of mole of NaOH = MV/1000 = (0.001x50)/1000 = (7.5 x 10-3) (2.5 x 10-4)2
= 5 x 10-5 mol
= 4.69x10-10
[NaOH] = 5x10-5mol/(0.05 + 0.15)L = 2.5 x 10-4M

No. of mole of MgCl2 = MV/1000 = (0.01x150)/1000 Q > Ksp. Solution is supersaturated. Ions
= 1.5 x 10-3 mol will form precipitate until the system reach

[MgCl2] = 1.5x10-3mol/(0.05 + 0.15)L = 7.5 x 10-3M equilibrium. 182
181

Example Solution:

Exactly 200 mL of 0.0040 M AgNO3 are No. of mole of AgNO3 = MV/1000 = (0.004x2000)/1000
added to 800 mL of 0.008 M K2SO4. Will a = 8.0 x 10-4 mol
precipitate form?
(Ksp Ag2SO4 = 1.4x10-5) [AgNO3] = 5x10-5mol/(0.2 + 0.8)L = 8.0 x 10-4 M

183 No. of mole of K2SO4 = MV/1000 = (0.08x800)/1000
= 6.4 x 10-3 mol

[K2SO4] = 6.4x10-3mol /(0.2 + 0.8)L = 6.4 x 10-3M

184

Solution: COMMON ION EFFECT

Ag2SO4(s) 2Ag+(aq) + SO42-(aq) Common ion is an ion that is common to
Q = [ Ag+]2 [ SO42-] two or more components in a mixture of
= (8.0x10-4)2(6.4x10-3) an ionic solution.
= 4.096 x 10-9
Common ion effect is the shift in
Q < Ksp. Solution unsaturated solid will equilibrium caused by the addition of a
dissolve and not precipitate formed. compound having an ion in common
with the dissolved substances.
185
186

257

The Common Ion Effect Consider a beaker contains silver chloride, AgCl with
its solid and ions in equilibrium.
The effect of a common ion on solubility
(Ksp AgCl = 1.6 x 10-10)
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
The equilibrium system is:
PbCrO4(s) Pb2+(aq) + CrO42-(aq) CrO42- AgCl (s) ⇋ Ag+ (aq) + Cl- (aq)
ad1d8e7d
If AgNO3 is added to the saturated AgCl solution :
AgNO3 (aq) → Ag+ (aq) + NO3- (aq)
(common ion)

188

 [Ag+] will increase. Example

 Equilibrium position will shift backward and more Calculate the solubility of AgCl (mol L-1) in :
precipitate of AgCl (s) formed.
a. liquid water
 The solubility of AgCl in the solution decreases b. 0.05 M of silver nitrate solution.
(compare the solubility in water).
(Ksp AgCl = 1.6 x 10-10)
 So, the addition of a common ion will reduce the
solubility of a slightly soluble salt. 190

189

Solution: b. 0.05M of silver nitrate solution.
(Ksp AgCl = 1.6 x 10-10)

a. liquid water AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp (0.05 + x) (x)
AgCl (s) Ag+ (aq) + Cl- (aq)
= [Ag+][Cl-]
(x) (x)

Ksp = [Ag+][Cl-] 1.6 x 10-10 = (0.05 + x) (x)

1.6 x 10-10 = x2 K ↓↓, x<<0.05 until 0.05 + x ≈ 0.05
sp

x = 1.26 x 10-5M 1.6 x 10-10 = 0.05x
Solubility of AgCl in water is 1.26 x 10-5M
x = 3.2 x 10-9M
Solubility of AgCl in AgNO3 is 3.2 x 10-9M.

191 192

258

The Common Ion Effect The Common Ion Effect AgBr (s)

Example: Solution: 1 x 10-3 M
What is the molar solubility of AgBr in; ii)
NaBr (aq)
i) pure water NaBr  Na+ + Br-
ii) 0.0010 M NaBr
110-3 M 110-3 M
(Ksp AgBr = 7.7  10-13)

Solution: Ag + (aq) + Br- (aq) AgBr (s) Ag + (aq) + Br-(aq)
i) AgBr (s) -y
0 1  10-3
Ksp = x2 = 7.7  10-13 [ ] initial
Change + y +y
x = 8.77  10 -7

x is molar solubility of AgBr in water 193 [ ] At equilibrium y y + 1  101-394

The Common Ion Effect Exercises:

Continue.. 1. Write the balance equation for the solubility equilibrium
and the expression for the solubility product of:
Ksp = (y) (y + 110-3) = 7.7  10-13 (i) Lead(II) chloride,
Assume y <<< 1  10-3, (ii) Zinc sulfide,
Thus, y + 1  10-3 = 1  10-3 (iii) Magnesium phosphate,
(iv) Silver chromate
(y) (1  10-3) = 7.7  10-13
y = 7.7  10-10 M 2. Given the value of Ksp calculate solubility for each ionic salts

Ionic salt Ksp Solubility

PbSO4 1.6 x 10-8

BaF2 1.5 x 10-6

Al(OH)3 1.3 x 10-33

Mg3(PO4)2 1 x10-25

Ag2CrO4 2.6  10-12

195 196

Exercises:

3. Calculate molar solubility for Ag2SO4 at 25C in
i) Water
(1.6  10-2 M)

ii) Solution of Na2SO4 0.100 M (6.1  10-3 M)
iii) Solution of AgNO3 0.100 M. (1.5  10-3 M)
(Ksp Ag2SO4 = 1.5  10-5)

197

259

CHAPTER 7: IONIC EQUILIBRIA

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

260

CHEMISTRY
TUTORIAL TOPIC 7: IONIC EQUILIBRIA

7.1 : Acids and Bases

1. Define acid and base according to the Arrhenius, Bronsted-Lowry and Lewis theories.
Write a suitable example by using a different equation for each theory.

2. For each of the following, identify the conjugate acid-base pairs:

(a) NH3 + H2PO4- NH4+ + HPO42-

(b) HClO + CH3NH2 CH3NH3+ + ClO-

3. (a) Define pH.
(b) Calculate the pH of 0.05 M of hydrochloric acid solution.
(c) The concentration of hydroxide ion in a blood sample is 2.5 x 10-7 M. What is
the pH of the blood?

4. Calculate the mass of NaOH needed to prepare 500.0 mL of solution with a pH of
10.00.

5. (a) Write an expression for the dissociation constant, Ka of propanoic acid,
CH3CH2COOH.

(b) Calculate the pH of a 0.35 mol L−1 solution of propanoic acid at 25oC.
[Ka = 1.3510−5 mol L-1 ]

6. The pH of a 0.036 M solution of nitrous acid, HNO2, is 2.40. What is the Ka of the
acid?

7. Calculate the percentage dissociation and degree of dissociation,α of 0.10 M

hydrocyanic acid, HCN solution. [Ka HCN = 5.0 10−10 mol L−1] .

8. The percentage ionisation of 0.010 M NH3 solution was 4.2 % ionisation. Calculate
Kb.

9. What is the pH of a 0.10 M solution of phenylamine, C6H5NH2?
[Kb C6H5NH2 = 4.0 10−10 mol L−1].

10. The table shows the base ionisation constant, Kb, for several selected compounds.

Compound Kb
C6H5NH2 3.810−10
N2H4 1.710−6
NH3 1.810−5
NH2OH 1.110−8

(a) Arrange the compounds in order of increasing strength of base.

261

CHEMISTRY
TUTORIAL TOPIC 7: IONIC EQUILIBRIA
(b) Give the structure of conjugate acid for each compound and arrange them in
order of increasing strength of acid.

11. Write the salt hydrolysis equation for the following salts and classify them as acidic,
basic or neutral.
(a) NaCN
(b) N2H5Cl

12. (a) Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 M
CH3COONa. [Ka CH3COOH = 1.8 x 10-5]

(b) Calculate the pH of the 0.20 M CH3COOH solution if there is no salt present.
[Ka CH3COOH = 1.810−5]

(c) Explain the change in pH the solution in (a) when
i. a small amount of strong acid is added.
ii. a small amount of strong base is added.

13. A student is asked to prepare a buffer solution at pH 4.6 using 50.0 mL of 0.5 M
benzoic acid, C6H5COOH and sodium benzoate, C6H5COONa. Calculate the mass of
sodium benzoate required to prepare the buffer solution.
[Ka C6H5COOH = 6.5 x 10-5]

14. A buffer solution was prepared by dissolving 0.15 mol of sodium ethanoate in 1 dm3
of 0.10 dm3 of 0.10 M ethanoic acid.
(a) Calculate the pH of the buffer solution above.
(b) Calculate the change in pH when 1.0 cm3 of 1.0 M hydrochloric acid is added
to 100 cm3 of the buffer solution.

15. Calculate pH of the solution formed when:
(a) 150 mL of 0.200 M HNO3 is mixed with 75.0 mL of 0.200 M NaOH.
(b) 25.0 mL of 0.450 M H2SO4 is mixed with 25.0 mL of 0.900 M NaOH.

262

CHEMISTRY
TUTORIAL TOPIC 7: IONIC EQUILIBRIA

7.2 Acid-base titrations

16. a) A titration experiment at 25oC involving a total of 30 mL of 0.1 M NaOH and
25mL of 0.1 M of HNO3. NaOH was added dropwise along the experiment.
Show the variation of pH of the solution
i) before the addition of NaOH,
ii) at half equivalence point,
iii) at equivalence point and
iv) at final volume of the titration.

b) Sketch a suitable graph for the titration process. Choose the most suitable
indicator based on the following table and give reason for your choice.

Indicator pH range
Phenoplhtalein
Cresol red 8.3 - 10.0
Bromothymol blue 7.0 – 8.8
6.0 – 7.6
[given Kw at 25 oC = 1 x 10-14]

7.3 Solubility equilibria

17. (a) State the differences between solubility and solubility product
(b) Calculate the solubility in g L-1 for calcium fluoride, CaF2.
[Ksp CaF2 = 3.98 x 10-11]
(c) The solubility of calcium fluoride, CaF2 in water is 0.01679 gL-1. Calculate
the value of Ksp

18. Will precipitate form if 200 mL of 0.004 M BaCl2 is added to 600 mL of 0.008 M
K2SO4? [Ksp BaSO4= 1.110−10]

19. Calculate the solubility of silver chromate, Ag2CrO4 at 25C in
(a) liquid water.
(b) solution of 0.005 M K2CrO4 solution.
[Ksp Ag2CrO4 = 9.010−12]
Explain the difference in the values of solubility.

263

CHEMISTRY
TUTORIAL TOPIC 7: IONIC EQUILIBRIA
20. (a) The solubility of Ag2SO4 in water at temperature 25C is 0.506 g for 100.00
mL solution. Calculate the solubility product at temperature 25C.
(b) If an aqueous solution of Na2SO4 was added progressively into an aqueous
solution of 0.01 M Ag2SO4, determine the minimum concentration of Na2SO4
needed just to begin the precipitation of Ag2SO4.

264

CHEMISTRY
MEKA TOPIC 7: IONIC EQUILIBRIA

1. Classify each of the following compounds as strong acid, strong base, weak acid or

weak base and calculate its pH. b) 1.23×10‒3 M C5H5N
d) 0.1 M HCN
a) 2.55 M Ba(OH)2
c) 5.04×10‒3 M HI

2. The percentage ionization of 0.010 M NH3 solution is 4.2%. Calculate Kb of NH3 and
pH of the solution.

3. Which of the following solutions forms a buffer solution?
a) 0.10 dm3 of 0.25 mol dm−3 NH4Cl + 0.05 dm3 of 0.25 mol dm−3 HCl.
b) 0.10 dm3 of 0.25 mol dm−3 NH3 + 0.05 dm3 of 0.25 mol dm−3 HCl.
c) Equal volume of 0.20 mol dm−3 HF and 0.10 mol dm−3 NaOH.
d) 0.10 mol dm−3 CH3COOH + 0.125 mol dm−3 CH3COONa.
e) 0.10 mol dm−3 NaCl + 0.10 mol dm−3 NH4Cl.

4. A buffer solution is made up by mixing 300 cm3 of a 0.40 mol dm−3 ammonia
solution with 200 cm3 of a 0.45 mol dm−3 ammonium chloride solution. What is the
pH of this buffer solution?

5. A solution contains 6.7×10−2 M of OH− ions at 25oC. Determine whether the solution
is neutral, acidic or basic.

6. A 20.00 mL of 0.15 M NH3 was titrated with 0.20 M HCl until the end point was
obtained. Explain whether the solution formed at equivalence point is acidic, neutral
or basic.

7. A 25.0 mL of 0.10 M HNO2 is titrated with 0.20 M NaOH solution.
a) Calculate
i. the pH of the solution before NaOH is added.

ii. the volume of NaOH required to reach the equivalence point.

b) Sketch the titration curve and suggest a suitable indicator for the titration.

8. a) Calculate the solubility of silver chromate, Ag2CrO4 at 25C in

i. pure water.
ii. 0.005 M K2CrO4.
[Ksp Ag2CrO4 = 9.0×10−12]
b) Compare the solubility in both conditions and state your reason(s).

265

CHEMISTRY
KUMBE TOPIC 7: IONIC EQUILIBRIA

1. Explain why NH3 gas is classified as a base according to Bronsted-Lowry but not as
Arrhenius theory?

2. The percentage dissociation of 0.0015 M HBrO is 0.137% at 25 oC. Calculate

a) the acid dissociation constant, Ka. [2.8 x 10-9]

b) the concentration of HBrO , BrO- and OH- ions in the solution.

[1.498x10-3, 2.055x10-6, 4.87x10-9]

3. Calculate the mass of hydrogen chloride gas, HCl that must be dissolved into 500 mL

of 0.10 M solution of sodium ethanoate, CH3COONa to produce a buffer solution

with a pH 4.74.

[Ka CH3COOH = 1.7510−5] [0.93 g]

4. A solution is prepared by adding 10.0 cm3 of a 0.100 M NaOH to 50.0 cm3 of a 0.100
M CH3COOH.
a) Calculate the pH of the solution.
[4.14]
b) What is the pH of the solution after the addition of 1 mL of 0.100 M HCl is
added.
[4.09]
c) Comment on your answer in (a) and (b).

5. Sketch the curve of pH against volume for a titration between 50.0 mL of 1.0 M HF
and 1.0 M KOH. Explain the characteristic of the curve at the following points;
a) the initial stage.
b) when 20.0 mL of KOH is added.
c) at the equivalence point.

Your discussion may include the type of solution formed at each stage.
[Ka HF=6.8×10−4]

6. A sample of 25.0 mL of 0.100 M HCl is titrated against a 0.100 M NH3 solution.
a) Sketch the titration curve.

b) Determine the pH of the solution at equivalence point and after adding 35.0

mL NH3 solution. [<7, 8.86]

7. a) State the difference between solubility and solubility product.
b) The solubility of silver sulphate, Ag2SO4 in water at 25C is 0.506 g in 100
mL. Calculate the solubility product at the same temperature. [1.71×10-−5]
c) If a solution of sodium sulphate, Na2SO4 is gradually added to 0.010 M of
silver nitrate, determine the minimum concentration of sodium sulphate
needed to precipitate the silver sulphate.
[0.17 M]

266

TABLE OF RELATIVE ATOMIC
MASSES AND LIST OF
CONSTANTS

lecture notes & questions

267

CHEMISTRY UNIT, KOLEJ MATRIKULASI MELAKA
TABLE OF RELATIVE ATOMIC MASSES

Element Symbol Proton number Relative atomic mass
Aluminum Al 13 27.0
Antimony Sb 51
Ar 18 121.8
Argon As 33 40.0
Arsenic Ba 56 74.9
Barium Be 4
Beryllium Bi 83 137.3
Bismuth B 5 9.0
Boron Br 35
Bromine Cd 48 209.0
Cadmium Ca 20 10.8
Calcium C 6 79.9
Carbon Ce 58
Cerium Cs 55 112.4
Cesium Cl 17 40.1
Chlorine Cr 24 12.0
Chromium Co 27
Cobalt Cu 29 140.1
Copper F 9 132.9
Fluorine Au 79
He 2 35.5
Gold H 1 52.0
Helium I 53 58.9
Hydrogen Fe 26 63.5
Iodine Kr 36 19.0
Pb 82 197.0
Iron Li 3
Krypton Mg 12 4.0
Mn 25 1.0
Lead Hg 80 126.9
Lithium Ne 10 55.9
Magnesium Ni 28 83.8
Manganese N 7 207.2
Mercury O 8 6.9
Neon P 15 24.3
Nickel Pt 78 54.9
Nitrogen K 19 200.6
Oxygen Pa 91 20.2
Phosphorus Ra 88 58.7
Platinum Rn 86 14.0
Potassium Rb 37 16.0
Protactinium Sc 21 31.0
Radium Se 34 195.1
Radon Si 14 39.1
Rubidium Ag 47 231.0
Scandium Na 11 226.0
Selenium Sr 38 222.0
Silicon S 16 85.5
Silver Sn 50 45.0
Sodium W 74 79.0
Strontium U 92 28.1
Sulphur Zn 30 107.9
23.0
Tin 87.6
Tungsten 32.1
Uranium 118.7
183.8
Zinc 238.0
65.4

268

CHEMISTRY UNIT, KOLEJ MATRIKULASI MELAKA
LIST OF SELECTED CONSTANT VALUES

Ionization constant for water at 25C Kw = 1.01014 mol2 dm6
= 22.4 dm3 mol1 at STP
Molar volume of gases Vm = 24 dm3 mol1 at room condition (25 C, 1 atm)
= 3.0108 m s1
Speed of light in a vacuum c = 4.18 kJ kg1 K1
Specific heat of water = 4.18 J g1 K1
= 4.18 J g1 C1
Avogadro’s number NA = 6.021023 mol1
Planck constant h = 6.631034 J s
RH = 6.631034 Kg m2 s1
Rydberg constant R = 1.097107 m1
= 2.181018 J
Ideal gas constant = 8.314 m3 Pa mol1 K1
= 8.314 J mol1 K1
Density of water at 25C  = 0.08206 L atm mol1 K1
= 62.36 L mmHg mol1 K1
= 1.00 g cm3

Freezing point of water = 0.00 C
= 273.15 K

Vapour pressure of water at 25C P H2O = 23.76 mmHg

UNIT AND CONVERSION FACTOR

VOLUME 1 L = 1 dm3
1mL = 1 cm3

ENERGY 1J = 1 kg m2 s2
1 calorie = 1Nm
= 1  107 erg
= 4.184 J

PRESSURE 1 atm = 760 mmHg
= 760 torr
= 101 325 Pa
= 101 325 N m–2

OTHERS 1 faraday (F) = 96 500 C
1 newton (N) = 1 kg m s2

269

Success is the sum of small
efforts, repeated. -R. Collier-