20202021 SC Topic 7 Ionic Equilibria

SESSION 2020/2021 TOPIC 7: IONIC EQUILIBRA
CHEMISTRY SK015

TUTORIAL 7
7.1: ACID & BASE
1. (a) Acid is any substance that can donate a proton (H+) to other substances

Base is any substance that can accept a proton from other substances

(b) i. CH3COO-+ HCN CH3COOH + CN-
Base Acid Conjugate Conjugate

Acid Base

ii. HCO3- + HCO3- H2CO3 + CO32-
Base Acid Conjugate Conjugate

Acid Base

iii. HClO + CH3NH2 CH3NH3+ + ClO-
Acid Base Conjugate Conjugate
Base
Acid

(c) i. CO32- ii. C2H5O- iii. ClO2-

2. a) Strong Acid is an acid that dissociates completely in water.
Weak Acid is an acid that dissociates partially in water.

b) Strong acid : ii) HClO4 and iv) HBr
Strong Base: i) K2O and viii) Ca(OH)2
Weak Acid: v) HClO, vi) HCOOH and vii) HF
Weak Base: iii) CH3NH2

3. Given , pH = 13.0
pOH = 1.0

-log [OH-] = 1.0
[OH-] = 0.1M
NaOH → Na+ + OH-
[OH-] = [NaOH] = 0.1 M
nNaOH = 0.1 x 0.250 = 0.0250 mol

➔ mass of NaOH = 0.0250 x 40 = 1.0 g
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4. (a) [OH-] =[NaOH]= 0.20 M
pOH = -log (0.20) = 0.70
pH = 13.3

(b) nHCl = 0.5x 50/1000 = 0.0250 mol

[HCl] = 0.0250 = 0.250 M
100 /1000

[H+] = [HCl] = 0.250 M

pH = -log (0.250) = 0.60

(c) HF (aq) + H2O(l) F-(aq) + H3O+(aq)

[ ]i 0.60 00

[ ]c -x +x +x

[ ]e 0.60-x xx

Ka = [F−][H3O+]
[HF]

6.8 x 10-4 = x2
0.6 − x

x2 + 6.8 x10-4x – 0.6(6.8x10-4) = 0

x = 0.0199M = [H3O+]

pH = 1.69

(d) CH3NH2(aq)+H2O(l) CH3NH3+(aq)+OH-(aq)
--
[ ]i 0.5
+x +x
[ ]c -x xx

[ ]e 0.5-x 122

[CH NH + ][OH − ]
3
Kb = 3

[CH3NH2 ]

4.4 X 10-4 = (x)(x)
(0.5 − x)

x2 + 4.4x10-4 x – 2.2x10-4 = 0

x1 = 0.01461, x2 = -0.0151
x = 0.01461 = [OH-]

SESSION 2020/2021 TOPIC 7: IONIC EQUILIBRA
CHEMISTRY SK015
pOH = 1.84; pH = 12.16

5. HF(aq) + H2O(l) F-(aq) + H3O+(aq)

[ ]I 0.245/0.5 00

=0.49

[ ]c -x +x +x

[ ]f 0.49-x xx

pH = 1.88 = - log x
x = 0.0132 M = [F-] = [H3O+]
[HF] = 0.477 M

Ka = (0.0132)2 = 3.65 x 10-4
(0.477)

6. (a)

N2H4(aq) + H2O(l) N2H5+(aq) + OH-(aq)

[ ]i 0.02 --

[ ]c -x +x +x

[ ]f 0.02 – x x x

%α = x 100 = 0.69
0.02

x = [OH-] = 1.4 x 10-4 M

  (b)
Ka = N 2 H 5 + OH −
N2H4

( )1.410−4 2

Kb = 0.01986 = 9.6 x 10-7

(c) pOH = -log (1.4 x 10-4)

= 3.85

pH = 10.15

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

Compound Kb Conjugate Acid
C6H5NH2 3.810−10 C6H5NH3+
N2H4 1.710−6 N2H5+
NH3 1.810−5 NH4+
NH2OH 1.110−8 +NH3OH

(a) Basicity: C6H5NH2< NH2OH < N2H4< NH3
(b) Acidity: NH4+< N2H5+<+NH3OH < C6H5NH3+

8. (a) Basic* (b) Acidic* (c) Neutral*

*refer to lecture notes for full answer

9. (a) A solution that is able to maintain its pH when small amount of an acid or a base
added to it.

(b) CH3COO-(aq) + H2O (aq) → CH3COO-(aq) + H3O+(l)

CH3COONa(aq) → CH3COO − + Na +
( aq ) (aq)

The presence of large amount of CH3COO- ions from CH3COONa suppresses the
dissociation of CH3COOH. Hence, in the mixture contains large amount of CH3COOH
(acid) and CH3COO- ions (base).

When a small amount of acid (H3O+) is added to the mixture, the H3O+ ions will be
neutralized by CH3COO- ions (base)to form CH3COOH.

CH3COO−(aq) + H3O+ (aq) → CH3COOH(aq) + H2O(l)

When small amount of base is added, the OH- will be neutralized by CH3COOH (base)
to form CH3COO- ions.

CH3COOH(aq) + OH − (aq) → CH3COO−(aq) + H2O(l)

In both cases, the added H3O+ and OH- are effectively removed. Hence, the pH of the
solution is not much affected.

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10. (a) [NH3] = 0.24 M [NH4Cl] = 0.18 M

( )pOH = − log 1.810−5 + log 0.18 = 4.6
0.24

pH = 9.4

(b) i. NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)
ii.
ni 0.09 1x10-4 0.12

nc -1x10-4 -1x10-4 +1x10-4

nf 0.0899 0 0.1201

[ ] 0.179 0 0.2397

( )pOH = − log 1.810−5 + log 0.179
0.2397

= 4.6
pH= 9.4

NH3(aq) + H3O+(aq) → NH4+(aq) + H2O(l)

ni 0.12 1x10-4 0.09 -

nc -1x10-4 -1x10-4 +1x10-4 -

nf 0.1199 0 0.091

[ ] 0.239 0 0.182

( )pOH = − log 1.810−5 + log 0.182
0.239

= 4.6
pH= 9.4

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11.
ni CH3COOH + NaOH → CH3COONa + H2O
△n 0.10 1 x 10-4 0.15 -
nf -1 x 10-4 -
-1 x 10-4 0 + 1 x 10-4 -
0.0999 0.1501

[CH3COOH] = 0.0999/1.001 = 0.0998
[CH3COONa] = 0.1501/1.001 = 0.14995

pH = pKa + log [CH3COONa]/ [CH3COOH]
= -log 1.75x10-5 + log (0.14995/ 0.0998)
= 4.93

12. a) HNO3(aq) + NaOH(aq) → H2O(l) + NaNO3(s)

ninitial (150.0x0.20)/1000 (75.0x0.20)/1000 - -
= 0.030 mol = 0.015 mol

△ n -0.015 mol -0.015 mol +0.015 mol

nfinal 0.015 mol 0 0.015 mol

The total volume is 225.0 L
Since NaNO3 does not undergoes hydrolysis, pH depends on dissociation of strong acid (HNO3)

[H+] = [HNO3]
[H+] = 0.015 mol / 0.225 L = 0.067 M

pH = -log [H+]
= -log 0.067
= 1.2

b) CH3COOH (aq) + NaOH (aq) → H2O (l) + CH3COONa(aq)

ninitial (25.0x0.9)/1000 (25.0x0.45)/1000 - 0
= 0.022 mol = 0.011 mol

△ n -0.011 mol -0.011 mol +0.011 mol

nfinal 0.011 mol 0 0.011 mol

The solution contain weak acid and salt of weak acid (buffer solution)
pH = pKa + log [CH3COONa]/[ CH3COOH]
= -log 1.8x10-5 + log (0.011/(50 x 10-3) / (0.011/(50x10-3)
= 4.7

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c) HCl (aq) + NH3(aq) → NH4Cl (aq)

ninitial (25.0x0.3)/1000 (50.0x0.1)/1000 0
= 7.5 x 10-3mol = 5.0x 10-3mol

△ n -5.0x 10-3mol -5.0x 10-3mol +5.0x10-3mol
0 5.0x 10-3mol
nfinal 2.5 x 10-3mol

The solution contain strong acid (determine the ph of solution) and salt.
[H+] = [HCl] = 2.5 x 10-3/75 x 10-3
= 0.0333 M
pH = 1.47

7.2 ACID –BASE TITRATION

1. (a) Titrant: HCl, Analyte: Ammonia

(b) Possible indicator:methylorange/methyl red/ chlorophenol blue/bromophenolblue
Either one of the proposed indicator is suitable in the titration between HCl and NH3
because the endpoint pH range lies on the steep portion of the titration curve. This
choice ensures that the pH at the equivalent point will fall within the range over which
indicator changes color.

(c) V HCl = 25 mL

NH3 (aq) + HCl(aq) → NH4Cl(aq)

[HCl] = (25 x 0.5)/25 = 0.5 M

2. Exactly 100 mL of 0.100 M nitric acid, HNO3 are titrated against 0.100 M NaOH solution.

Determine pH for;

(a) The initial solution.
HNO3(aq) + H2O(l) → NO3-(aq) + H3O+(aq)

[H3O+] = [HNO3] = 0.100 M
pH = 1

(b) The point at which 95.0 mL of the base has been added.

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

ni 0.01 9.5 x 10-3 - -

nc -9.5 x 10-3 -9.5 x 10-3 +9.5 x 10-3

nf 5 x 10-4 0 9.5 x 10-3

[ ] 2.56x10-3 0 0.0487

Solution contain excess HNO3 and neutral salt NaNO3
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pH = -log(2.56 x 10-3) =2.59

(c) The equivalence point.
At this point, all HNO3 has been neutralized by NaOH. The solution only contains
NaNO3
Both Na+ and NO3- will not hydrolyse.
Therefore the salt solution is neutral .
pH = 7

(d) The point at which 105 mL of base has been added.

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

ni 0.01 0.0105 0

nc -0.01 -0.01 +0.01

nf 0 5 x 10-4 0.01

[] 0 2.44 x 10-3 0.0488

NaOH will dissociate completely.
• [OH-] increase.

• Equilibrium position of hydrolysis reaction shift backward.
• [OH-] produced by hydrolysis << [OH-] produce by base.

[OH-] =[NaOH]= 2.44 x 10-3 M

pOH = - log (2.44 x 10-3) =2.6
pH = 11.4

3. (a) Before the addition of 0.20 M NH3,
[HBr] = [H3O+]= 0.10 M

pH = −log(0.1)
= 1.0

The pH then gradually increases.

• Sharp increases
pH 3 – 7
slightly before and after equivalence volume

• at the equivalence point,
Solution contains acidic salt, NH4Br
pH < 7 at 25C
Volume equivalence = (25.0 x 0.1) / 0.2 = 12.50 mL

• After the equivalence point the pH gradually increases and plateaus off at pH ̴ 11
upon continued addition of 0.20 M NH3.

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TITRATION CURVE FOR STRONG ACID AND WEAK BASE

(b) before the addition of 0.20 M NaOH, H2O F- + H3O+
HF + 0.00 0.00

[ ]i 0.20

[ ] -x +x +x

[ ]eq 0.20-x xX
[H3O+]= x = √ (0.2 x 6.8x10-4)

=0.0117 M

pH = 1.93

• Gradual increase before equivalence point (buffer region)
• Sharp increase at pH 7-11
• At equivalence point,

pH> 7
Volume eq = 25.0 mL

• after the equivalence point the pH gradually increases and plateaus off at pH ̴ 13
upon continued addition of 0.10 M NaOH

THE TITRATION CURVE FOR WEAK ACID AND STRONG BASE

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4. (a) NaOH(aq) + HNO3(aq) → NaNO3(aq) + 2H2O(l)
(b) Mole of NaOH = 5.00g/40gmol- = 0.125 mol
[NaOH] = 0.125/200 x 10-3 = 0.625 M
(c) Mole of NaOH neutralized = 0.625 x 20.00 x 10-3 = 0.0125 mol
(d) From equation,
1 mol NaOH Ξ 1mol HNO3
n HNO3 = 0.0125 mol
[HNO3] = 0.0125/25.00 x 10-3 = 0.500 M
(e) For strong acid and strong base titration, pH of salt is expected as 7 at equivalent point.
Strong acid and strong base will formed weak conjugate base and weak conjugate acid
respectively that are not possible to undergo hydrolysis in water. Thus, pH of the salt will
be based on pH of the water.

7.3: SOLUBILITY EQUILIBRIA

1. (a) Amount of solid that dissolved in a known value of saturated solution.

(b) Ag2SO4(s) 2Ag+(aq) + SO42-(aq)

[ ]eq 2x x

x=1.5  10−2mol L−1

Ksp = (2x)2(x) = 4x3 = 4 (1.5  10−2)3 =1.4 x 10-5

(c) CaSO4(s) Ca2+(aq) + SO42-(aq)

[ ]eq x x

Ksp = [Ca2+][SO42-]

2.4  10−4 = x2

x = 1.5 x 10-2 M = Molar solubility

2. (a) Ca(IO3)2(s) Ca2+(aq) + 2IO3-(aq)

[ ]eq x 2x

6.3 x 10-7 = 4x3

x = 5.4 x 10-3 M = solubility in water

(b) Ca(IO3)2(s) Ca2+(aq) + 2IO3-(aq)
[ ]eq x+0.1 2x

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6.3 x 10-7 = (x + 0.1) 4x2 ; assume x<< 0.1, x+ 0.1 = 0.1
= 0.1(4x2)

x = 1.3 x 10-3 M = solubility in 0.1 M Ca(NO3)2

(c) Solubility of Ca(IO3)2 in pure water > in 0.10 M Ca(NO3)2.
• In Ca(NO3)2 solution, Ca2+(common ion) is present.

• The equilibrium position shifts backward.
• Solubility of Ca(IO3)2 decrease.
• The present of common ion, Ca2+ reduces the solubility of Ca(IO3)2.

3. (a) i. PbF2 (s) Pb2+ (aq) + 2F-(aq)

[ ]eq s 2s

Ksp= [Pb2+] [2F-]2

3.6 x 10-8= (s) (2s)2

= 4s3

s= 2.1 x 10-3

The solubility of lead(II) fluoride, PbF2 at 25oC in pure water is 2.1 x 10-3 M

ii. PbF2 (s) Pb2+ (aq) + 2F-(aq)

[ ]eq s 2s + 0.10

Ksp= [Pb2+] [2F-]2

3.6 x 10-8= (s) (0.10)2

s= 3.6 x 10-6

The solubility of lead(II) fluoride, PbF2 at 25oC in NaF solution is 3.6 x 10-6 M.

(b) The solubility of Lead(II) fluoride, PbF2 at 25oC in NaF solution has decreased when

compared to the solubility of Lead(II) fluoride, PbF2 at 25oC in pure water due to

common ion effect.

4. BaCl2 (aq) + K2SO4 (aq) → BaSO4 (aq) + 2KCl(aq)

[BaCl2] = 1 x 10-3 M [K2SO4] = 6 x 10-3 M

BaSO4 (s) Ba2+ (aq) + SO42- (aq)
[ ]eq 1 x 10-3 M 6 x 10-3 M

Q = 6 x 10-6

Q > K ; Solution is supersaturated.

• Precipitate will form.

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5. (a) C6H5COO-, H3O+ C6H5COO- (aq) + H3O+ (aq)
00
(b) C6H5COOH (aq) + H2O (l)

2g

[ ]i 122 g mol−1 = 0.0164

1L

[ ]c -x +x +x

[ ]eq 0.0164 – x xx

Assume x <<<< 0.0164, 0.0164 – x ≈ 0.0164

6.5 x 10-5 = x2
0.0164

x = 1.03 x 10-3 M

pH = 3.0

(c) % α= 1.0310−3 100 = 6.3 %
0.0164

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MEKA 7
1.a) Ba(OH)2 is a strong base.

pOH = -log [OH-]
= -log (0.0051)
= 2.29

pH + pOH = 14
pH = 14 – pOH
= 14 – 2.29
= 11.71

.b) C5H5N(aq) + H2O(l) → C5H5NH+ (aq) + OH-(aq)

[ ]i 1.23 x 10-3 --

[ ]c -x +x +x

[ ]f 1.23 x 10-3 - x xx

Kb = [C5H5NH + ][OH − ]
[C5H5N ]

(x)(x)

1.23x10−3 − x
( )1.7 X 10-9 =

(assume, x <<1.23 x 10-3, thus 1.23 x 10-3–x = 1.23 x 10-3)
x = 1.45 x 10-6 M= [OH-]
pOH = 5.84
pH =8.16

c) HI is a strong acid.
[H+] =[HI] = 5.04x10-3 M
pH = -log (5.04x10-3)
= 2.30

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d) HCN is a weak acid.

HCN(aq) + H2O(l) ƒ CN-(aq) + H3O+(aq)

[]I 0.1 00

[]c -x +x +x

[]e 0.1-x xx

K = [CN-][H3O+]
[HCN]

4.9 10−10 = 2 (x ≪ 0.1, 0.1-x ≈ 0.1)
0.1 −

4.9x10-10= x2
0.1

x = 7 x 10-6M = [H3O+]

pH = -log(7 x 10-6)

=5.15

2. NH3(aq) + H2O(l) ƒ NH4+(aq) + OH-(aq)

[]I 0.010 00

[]c -x +x +x

[]e 0.010-x xx

x
%∝=4.2= 0.010 x 100

x = 4.2x10-4

[NH4+] = [ OH-] = 4.2x10-4M

[NH3] = 9.58 x 10-3 M

Kb= [NH4+][OH-] = (4.2x10-4)2 = 1.84 x 10−5
[NH3] 9.58x10-3

pOH = -log(4.2 x 10−4) = 3.38

pH =10.62

3. (b), (c), and (d)

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4. n NH3 = 0.12 mol

[NH3] = 0.24 M

n NH4Cl = 0.09 mol

[NH4Cl] = 0.18 M

 NH4Cl (0.18)
NH3  (0.24)
( )pOH
= − log Kb + log = − log 1.8 10−5 + log = 4.62

pH = 9.38

5. pOH = -log [OH-] = -log (6.7×10−2 M) = 1.17
pH = 14-1.17 =12.8
The solution is basic.

6. NH3 (aq) + HCl(aq) → NH4Cl(aq)

In water,

NH4Cl(aq) → NH4+ (aq) + Cl− (aq)

Hydrolysis of NH + produce H3O+ and NH3.
4

NH4+ (aq) + H2O(l) ƒ NH3(aq) + H3O+ (aq)

Acidic solution form.

7. (a) i. pH = -log[H+] = -log (0.10) =1

ii. HNO3 (aq) + NaOH(aq) ƒ NaNO3(aq) + H2O(l)

n HNO3 = 2.5×10-3mol

nNaOH = 2.5×10-3mol

V NaOH = 2.5 10−3 = 0.0125 L = 12.5 mL
0.20

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(b)

pH
13.3

7.0

1.0 12.5 mL V NaOH

Phenolphthalein

8. (a) i. Ag2CrO4(s) → 2Ag+(aq) + CrO42-(aq)
2x x

9.0 x 10-12 = 4x3
x = 1.3 x 10-4 M (solubility in pure water)

ii. Ag2CrO4(s) → 2Ag+(aq) + CrO42-(aq)
2x x+0.005

9.0 x 10-12 = 4x2 (x + 0.005)
x = 2.1 x 10-5 M (solubility in 0.005M K2CrO4)

(b) • Solubility of Ag2CrO4 in pure water > in 0.005 M K2CrO4.
• In K2CrO4 solution, [CrO42-] is increased.
• The equilibrium position will shift backward.
• Solubility of Ag2CrO4 will decrease.
• The addition of common ion will reduce the solubility of Ag2CrO4.

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KUMBE 7

1. NH3(g) + HCl(g) → NH4Cl (aq)

NH3 gas is a bronsted Lowry base because it can accept H+ and is also a Lewis base because
there is a lone pair at N atom therefore it can donate its lone pair to form covalent bond. While it
is not an Arrhenius base because in the reaction there is no water involved, therefore it cannot
have given off OH- ion.

2. a) HBrO (aq) + H2O (l) BrO- (aq) + H3O+ (aq)
b) [ ]i 0.0015 --
[ ]c -x +x +x
[ ]f 0.0015-x xx

x = 0.137
0.0015 100

x =2.055 x10−6 M

   BrO− = H3O+ = 2.055x10 −6 M

HBrO = 1.498 x 103 M

BrO− H3O + 

Ka = HBrO

( )=
2.055x10 −6 2
1.498 x 103

= 2.819 x 10−9

   BrO− = H3O+ = 2.055x10 −6 M

HBrO = 1.498 x 103 M

pH = - log 2.055x 10-6
= 5.69

pOH = 8.31
[ OH-] = 4.898 x 10-9 M

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3. CH3COO-(aq) + H+(aq) → CH3COOH(aq)

ni 0.05 x -

nc -x -x +x

nf 0.05-x 0 x

( )4.74 = − log 1.7510−5 0.05 − x v
x
+ log

v

x =0.0256 mol ; mass = 0.93 g

4. a) CH3COOH (aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

ni 5×10-3 1×10-3 - -

nc -1×10-3 -1×10-3 +1×10-3 -

nf 4×10-3 0 1×10-3 -

[CH3COOH] = 4 10−3 = 0.067 M
0.06

[CH3COONa] = 110−3 = 0.017 M
0.06

pH = - log Ka + log CH3COONa 
CH3COOH

= 4.74 + log (0.017) = 4.14
(0.067)

b) CH3COO- (aq) + HCl(aq) → CH3COOH(aq) + Cl- (aq)

ni 1×10-3 1×10-4 4×10-3 -

nc -1×10-4 -1×10-4 +1×10-4 +1×10-4

nf 9×10-4 0 4.1×10-3 1×10-4

[CH3COOH] = 4.110−3 = 0.067 M
0.061

[CH3COONa] = 9 10−4 = 0.015 M
0.061

pH = - log Ka + log CH3COONa 
CH3COOH

= 4.74 + log (0.015) = 4.09
(0.067)

c) The addition of a small amount of strong acid to a buffer solution causes only a very
small change in the pH of the buffer solution.

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5. a) At this point, the solution contain H3O+ that come from dissociation of weak acid, HF.

HF (aq) + H2O (l) F- (aq) + H3O+ (aq)
[ ]i 1.0 --

[ ]c -x +x +x
[ ]f 1-x xx

6.8 x 10-4 = x2
1− x

x = 0.0257 M
pH = 1.6

b) HF (aq) + KOH (aq) → KF (aq) + H2O(l)

ni 0.05 0.02 -

nc -0.02 -0.02 + 0.02
nf 0.03 0 0.02

[ ] 0.429 0.286

At this point, buffer solution is form.

( )pH = − log 0.286
6.8 10−4 + log 0.429 = 3.0

c) All HF has been neutralized by KOH. The solution only contain KF.
F-willhydrolysed water to form OH-.

F- (aq) + H2O (l) HF (aq) + OH- (aq)

∴The salt form is basic salt. pH> 7.

13
11

pH 7

1.6

6. a) At initial : 2Volume of KOH added(mL)

pH= -log (0.1) = 1

At equivalent point :

V of NH3 = (0.1)(25) / (0.1) = 25ml

All NH3 has been neutralized by HCl. The solution only contain NH4Cl.

NH4+ will hydrolysed water to form H3O+.
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SESSION 2020/2021 TOPIC 7: IONIC EQUILIBRA
CHEMISTRY SK015

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)

∴The salt form is acidic salt. pH< 7.

pH jump : pH 3-7 (strong acid – weak base)

At final:

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
[ ]i 0.1 --
[ ]c -x +x +x

[ ]eq 0.1-x xx

1.8 x 10-5 = x2
0.1− x

x = 1.33 x 10-3 M
pOH = 2.88

pH = 11.12
Final pH approaching 11.12

11

pH H7
H

3
1
H

25
Volume of NH3 added (ml)

b) At equivalence point: 25 mL

All NH3 has been neutralized by HCl. The solution only contain NH4Cl.
NH4+ will hydrolysed water to form H3O+.

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)

∴The salt form is acidic salt. pH< 7.

After addition 35 mL NH3

NH3(aq) + HCl(aq) → NH4Cl(aq)

ni 3.5 x 10-3 2.5 x 10-3 -

nc -2.5 x 10-3 -2.5 x 10-3 +2.5 x 10-3

nf 1 x 10-3 0 2.5 x 10-3

[ ] 0.0167 0.0417

Buffer solution is form.

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SESSION 2020/2021 TOPIC 7: IONIC EQUILIBRA
CHEMISTRY SK015

( )pOH = − log 1.810−5 + log 0.0417 = 5.14
0.0167

pH = 8.86

5. a) Solubility:
The maximum amount of solute that can be dissolved in a given quantity of solvent to
form saturated solution of a given temperature.
Solubility Product:
The product of the concentration of ions each raised to the power of its stoichiometric
coefficient in the equilibrium equation.

b) Solubility = 0.506 = 0.0162 M
0.1 311.9

Ag2SO4(s) 2Ag+(aq) + SO42-(aq)
2(0.0162) 0.0162

Ksp = (0.0324)2(0.0162) = 1.71 x 10-5

c) Ag2SO4(s) 2Ag+(aq) + SO42-(aq)

(0.01) x

1.71 x 10-5 = (0.01)2 x

x = 0.17 M

141