IONIC EQUILIBRIA

IONIC EQUILIBRIA
10 DISEMBER 2020


TUTORIAL






























































HAZMIRA SUZLIN BT AB HAMID
KOLEJ MATRIKULASI MELAKA

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

TUTORIAL 7


7.1: Acid & Base

















1.

a) Acid is a species that can donate proton to form conjugate base.
Base is a species that can accept proton to form conjugate acid.


-
b) i. CH3COO + HCN CH 3 COOH + CN
-
Conjugate
Base Acid Conjugate

Acid Base

ii. HCO 3 + HCO 3 - H 2 CO 3 + CO 3
2-
-
Base Acid Conjugate Conjugate

Acid Base

iii. HClO + CH 3 NH 2 CH 3 NH 3 + ClO
+
-
Acid Base Conjugate Conjugate

Acid Base

c) i. CO 3 ii.C 2 H 5 O iii. ClO 2
2-
-
-



















1

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA















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.
a) Ba(OH)2 is a strong base.

-
pOH = -log [OH ] = -log (0.0051 M) = 2.29
pH = 11.71


b) C5H5N is a weak base
-
+
C5H5N (aq) + H2O (l) C5H5NH (aq) + OH (aq)
− 3
[ ]i 1.23×10 M - 0 0
△[ ] -x M - +x +x

− 3
[ ]f 1.23×10 – x M - x x

[C H NH + ][OH − ]
Kb = 5 5
[C 5 H 5 N ]


x
( )( ) x
− 9
1.7×10 =
(1.23 10 − 3 − ) x
2

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

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

c) HI is a strong acid.
+
− 3
[H ] = [HI] = 5.04×10 M
− 3
pH = -log (5.04×10 M) = 2.30


d) HCN is a weak acid.

HCN(aq) + H2O (l) CN (aq) + H3O (aq)
-
+


[ ]i 0.10 M - 0 0
△[ ] -x M - +x +x


[ ]f 0.10 – x M - x x

  CN   H O 
−
+

 
Ka= 3
HCN 
( )( ) x
x
4.9×10 − 10 =
−
(0.10 x )
(assume, x <<0.10, thus 0.10–x = 0.10)
( )( ) x
x
4.9×10 − 10 =
(0.10 )
+
x = 7.0×10 M = [H3O ]
− 6
pH = -log(7.0 × 10 − 6 M )= 5.15










3

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA











NH3(aq) + H2O (l) NH4 (aq) + OH (aq)
+
-


[ ]i 0.01 M - 0 0
△[ ] -x M - +x +x


[ ]f 0.01 – x M - x x
4.
NH 
%α = 3 dissociate  100%
NH 
3 initial
x
4.2 =  100%
0.01 M
+
-
− 4
x = [NH4 ] = [OH ] = 4.2×10 M
− 3
[NH3] = 9.58×10 M

−
+
  NH   OH   ( 4.2 10 − 4 )( 4.2 10 − 4 )


4  
Kb = = = 1.84×10
− 5
NH 3  (9.58 10 − 3 )


− 4
pOH = -log (4.2×10 M) = 3.38
pH =10.62









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



4

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA









6.

Given, pH = 13.0
pOH = 1.0
-
-log [OH ] = 1.0
-
[OH ] = 0.1M

+
-
NaOH(aq) → Na (aq) + OH (aq)
[OH-] = [NaOH] = 0.1 M
nNaOH = 0.1 M × 0.250 L = 0.0250 mol


mass of NaOH = 0.0250 mol × 40 g/mol = 1.0 g
















7.
a) [OH ] =[NaOH]= 0.20 M
-
pOH = -log (0.20 M) = 0.70

pH = 13.3


b) nHCl = 0.5 M × 0.050 L = 0.0250 mol
0.025 mol
[HCl] = = 0.25 M
0.1 L

[H ] = [HCl] = 0.250 M
+
pH = -log (0.250) = 0.60




5

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

+
c) HF (aq) + H2O(l) F (aq) + H3O (aq)
-
[ ]I, M 0.60 0 0
[ ]c, M -x +x +x
[ ]e, M 0.60-x x x


+

 
 F  −   H O 
3
Ka =
 
HF
x 2
6.8×10 =
− 4
−
0.6 x
x = 0.0199 M = [H3O ]
+
pH = 1.69

+
d) CH3NH2(aq) + H2O(l) CH3NH3 (aq) + OH (aq)
-
[ ]I, M 0.5 - -
[ ]c, M -x +x +x
[ ]e, M 0.5-x x x


[CH NH + ][OH − ]
Kb = 3 3
[CH NH 2 ]
3
( )( ) x
x
4.4 × 10 =
− 4
−
(0.5 x )
x1 = 0.01461 M (choosen), x2 = -0.0151 M
-
x = 0.01461 M = [OH ]

pOH = 1.84; pH = 12.16














6

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA







8.

-
+
HF (aq) + H2O(l) F (aq) + H3O (aq)
[ ]I, M 0.49 0 0

[ ]c, M -x +x +x
[ ]f, M 0.49-x x x



pH = 1.88 = - log x

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


)
+
 F  −   H O   (0.0132 (0.0132)
 
3
− 4
Ka = = = 3.65×10
  (0.477 )
HF








9.

+
-
a) N2H4(aq) + H2O(l) N2H5 (aq) + OH (aq)
[ ]I, M 0.02 - -

[ ]c, M -x +x +x

[ ]f, M 0.02 – x x x

N H 
%α = 2 4 dissociate  100%

N H 4 initial
2
x
0.69% =  100%
0.02


x = [OH ] = 1.4×10 M
− 4
-
7

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

b)

−


+
  N H   OH  (1.4 10 − 4 )(1.4 10 − 4 )
5  

− 7
Kb = 2 = = 9.6×10
N H 4  0.01986
2

− 4
c) pOH = -log (1.4×10 M) = 3.85
pH = 10.15












10.
+
-
HBrO(aq) + H2O (l) BrO (aq + H3O (aq)
a)



[ ]i 0.0015 M - 0 0

△[ ] -x M - +x +x


[ ]f 0.0015 – x M - x x


HBrO 
%α = dissociate  100%
HBrO 
initial
x
0.137% =  100%
0.0015 M
− 6
x = 2.055×10 M


-
+
− 6
[BrO ] = [H3O ] = 2.055×10 M
[HBrO] = 1.498×10 M
− 3

−
+
  BrO   H O  (2.055 10 − 6 )(2.055 10 − 6 )

 
− 9
Ka = 3 = = 2.819×10
HBrO  (1.498 10 − 3 )

8

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

− 3
b) [HBrO] = 1.498×10 M
− 6
[BrO ] = 2.055×10 M
-

pH = -log (2.055×10 M) = 5.69
− 6
pOH = 8.31
-
− 9
[OH ] = 4.898×10 M


























11.

a) C6H5NH2< NH2OH < N2H4< NH3

b)

Compound Kb Conjugate Acid

−10
C6H5NH2 3.810 C6H5NH3 +
−6
N2H4 1.710 N2H5 +
−5
NH3 1.810 NH4 +

NH2OH 1.110 + NH3OH
−8



Ascending order of acidity:
+
+
+ +
NH4 < N2H5 < NH3OH < C6H5NH3




9

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA









12.

−
+
+
a) NaCN(aq) → Na (aq) CN (aq)
+
• Na (aq) comes from strong base and cannot react with water.
-
• CN (aq) comes from weak acid and can react with water to produce HCN and OH -
(aq).
−
−
( ) + ( ) ⇌ ( ) + ( )
2
-
• Since the hydrolysis producing OH ion, thus NaCN is basic salt.

+
−
+
b) N H Cl(aq) → 2 5 N H (aq) Cl (aq)
2
5
+
• N2H5 (aq) comes from weak base and can react with water.
+
+
( ) + ( ) ⇌ ( ) + ( )
2 4
2
3
2 5
-
• Cl (aq) comes from strong acid and cannot react with water.
+
• Since the hydrolysis producing H3O ion, thus N2H5Cl is acidic salt.

+
+
c) K SO (aq) → 2 4 2K (aq) SO 4 2− (aq)
+
• 2K (aq) comes from strong base and cannot react with water.
2-
• SO4 (aq) comes from strong acid and cannot react with water.
• Thus K2SO4 is neutral salt.

















10

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA








13.

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



b) ( ) + ⇌ − ( ) + + ( )
3
3
3
2 ( )
CH COONa (aq ) → CH COO − (aq ) + Na + (aq )
3
3
-
The presence of large amount of CH3COO ions from CH3COONa suppresses the
dissociation of CH3COOH. Hence, in the mixture contains large amount of undissociated
-
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.
)
CH COO − (aq ) + H O + (aq → CH COOH (aq ) + H O ( ) l
3
3
3
2
-
When small amount of base is added, the OH will be neutralized by CH3COOH (base) to
-
form CH3COO ions.
)
CH COOH (aq ) + OH − (aq → CH COO − (aq ) + H O ( ) l
3
3
2
+
-
In both cases, the added H3O and OH are effectively removed. Hence, the pH of the
solution is not much affected.
























11

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA
















14.

a) n NH3 = (0.40 M) (0.3 L) = 0.12 mol

0.12 mol
new [NH3] = = 0.24 M
0.5 L



n NH4Cl = (0.45 M) (0.2 L) = 0.09 mol

0.09 mol
new [NH4Cl] = = 0.18 M
0.5 L



  NH  4  + (0.18 M )
− 5
pOH = - log Kb + log = - log (1.8×10 ) + log = 4.6
NH 3  (0.24 M )

pH = 9.4



+
-
b) i. NH4 (aq) + OH (aq) → NH3(aq) + H2O(l)
− 4
ni, mol 0.09 1×10 0.12
− 4
− 4
nc, mol -1×10 -1×10 − 4 +1×10
nf, mol 0.0899 0 0.1201

[ ], M 0.179 0 0.2397


  NH  4  + (0.179 M )
− 5
pOH = - log Kb + log = - log (1.8×10 ) + log = 4.6
NH 3  (0.2397 M )

pH = 9.4



12

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

+
+
ii. NH3(aq) + H3O (aq) → NH4 (aq) + H2O(l)
− 4
ni, mol 0.12 1×10 0.09 -
− 4

− 4
− 4
nc, mol -1×10 -1×10 1×10 -
nf, mol 0.1199 0 0.091
[ ], M 0.239 0 0.182



  NH  4  + (0.182 M )
− 5
pOH = - log Kb + log = - log (1.8×10 ) + log = 4.6
NH 3  (0.239 M )

pH = 9.4














15.

-
-
CH3COOH (aq) + OH (aq) → CH3COO (aq) + H2O(l)
− 4
ni, mol 0.10 1×10 0.15
− 4
− 4
nc, mol -1×10 -1×10 − 4 +1×10
nf, mol 0.0999 0 0.1501

[ ], M 0.0998 0 0.1499



  CH COO  −  (0.1499 M )
− 5
pH = - log Ka + log 3 = - log (1.8×10 ) + log = 4.93
CH COOH  (0.0998 M )
3










13

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA
















16.
(b), (c), and (d)

















17.

a) n HNO3 = (0.20 M) (0.15 L) = 0.03 mol

n NaOH = (0.20 M) (0.075 L) = 0.015 mol


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

ninitial 0.030 mol 0.015 mol - -

△ n -0.015 mol -0.015 mol +0.015 mol

nfinal 0.015 mol 0 0.015 mol


Since NaNO3 does not undergoes hydrolysis, pH depends on dissociation of strong acid
(HNO3)

+
[H ] = [HNO3]
0.015 mol
+
[H ] = = 0.067 M
0.225 L
+
pH = -log [H ] = -log 0.067 M = 1.2


14

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

b) n CH3COOH = (0.90 M) (0.025 L) = 0.022 mol

n NaOH = (0.45 M) (0.025 L) = 0.011 mol



CH3COOH(aq) + NaOH(aq) → H2O(l) + CH3COONa(aq)
ninitial 0.022 mol 0.011 mol - 0

△ 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). Thus, the pH is
determine by the pH of buffer.
 0.011 mol 
  CH COO  −    0.05 L  
− 5
pH = - log Ka + log 3 = - log (1.8×10 ) + log = 4.7
CH COOH   0.011 mol 
3
 
 0.05 L 
− 3
c) n HCl = (0.30 M) (0.025 L) = 7.5×10 mol
− 3
n NH3 = (0.10 M) (0.050 L) = 5.0×10 mol

HCl(aq) + NH3(aq) → NH4Cl(aq)

− 3
− 3
ninitial 7.5×10 mol 5.0×10 mol 0
− 3
− 3
− 3
△ n -5.0×10 mol -5.0×10 mol +5.0×10 mol
− 3
− 3
nfinal 2.5×10 mol 0 5.0×10 mol

The solution contain strong acid and salt. The pH is only determine by HCl because the
+
hydrolysis of NH4Cl produce less amount of H ion.

-3
2.5 10 mol
[H+] = [HCl] = = 0.0333 M
0.075 L
pH = 1.47







15

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA












18.

CH3COO (aq) + H (aq) → CH3COOH(aq)
+
−


ni 0.05 mol x mol 0
△n -x mol -x mol +x mol


nf 0.05 – x mol 0 x mol

  CH COO  − 
pH = - log Ka + log 3
CH COOH 
3

 0.05 x  −
 
− 5
4.74 = - log (1.75×10 ) + log  v 
 x 
 
 v 
x = n H = n HCl = 0.0256 mol
+
mass HCl = 0.93 g




















16

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA














19.
a) n NaOH = (0.100 M) (0.01 L) = 1×10 mol
− 3
− 3
n CH3COOH = (0.100 M) (0.05 L) = 5×10 mol


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

ni 5×10 mol 1×10 mol 0 -
− 3
− 3

− 3
− 3
− 3
△n -1×10 mol -1×10 mol +1×10 mol -

− 3
− 3
nf 4×10 mol 0 1×10 mol -

The solution contains acidic buffer solution. The pH is depends on pH of buffer solution.

−
4 10 mol
3
[CH3COOH] = = 0.067 M
0.06 L
3

−
1 10 mol
[CH3COO ] = = 0.017 M
−
0.06 L
  CH COO  −  (0.017 M )
pH = - log Ka + log 3 = - log (1.75×10 ) + log = 4.14
− 5
CH COOH  (0.067 M )
3

+
b) n H = n HCl = (0.100 M) (0.001 L) = 1×10 mol
− 4

+
CH3COO (aq) + H (aq) → CH3COOH(aq)
−

− 4
− 3
− 3
ni 1×10 mol 1×10 mol 4×10 mol
− 4
− 4
− 4
△n -1×10 mol -1×10 mol +1×10 mol
− 3
− 4
nf 9×10 mol 0 4.1×10 mol

−
3

4.1 10 mol
[CH3COOH] = = 0.067 M
0.061 L
17

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

−

4
9 10 mol
[CH3COONa] = = 0.015 M
0.061 L
 CH COO  − (0.015 M )
− 5
pH = - log Ka + log  3  = - log (1.75×10 ) + log = 4.09
CH COOH  (0.067 M )
3

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.







































18

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

7.2 Acid –base Titration



































1.
a) Titrant: HCl, Analyte: Ammonia


b) Possible indicator: methyl orange/methyl red/ chlorophenol blue/ bromophenol blue


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

+
NH (aq) HCl(aq) → NH Cl(aq)
4
3
1 mol of NH3 ≡ 1 mol HCl
(0.025 L )(0.5 M )
[HCl] = = 0.5 M
0.025 L






19

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA













2.
a) pH of 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.

n HNO3 = (0.1 M) (0.1 L) = 0.01 mol

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

− 3
ninitial 0.01 mol 9.5×10 mol 0 -
− 3
− 3
− 3
△ n -9.5×10 mol -9.5×10 mol +9.5×10 mol -
− 4
− 3
nfinal 5.0×10 mol 0 9.5×10 mol -

Solution contain excess HNO3 and neutral salt NaNO3. pH is determined by HNO3.


-4
5.0 10 mol
+
− 3
[HNO3] = [H ] = = 2.56×10 M
0.195 L
− 3
pH = -log (2.56×10 M) = 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 cannot react with water.
• Therefore the salt solution is neutral.
pH = 7






20

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

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

n NaOH = (0.1 M) (0.105 L) = 0.0105 mol

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


ninitial 0.01 mol 0.0105mol 0 -



△ n -0.01mol -0.01mol +0.01mol -
− 4

nfinal 0 5.0×10 mol 0.01mol -

Solution consist of NaOH and NaNO3. The pH is determine by NaOH only because the
-
hydrolysis of NaNO3 producing less amount of OH ion.


-4
5.0 10 mol
− 3
-
[NaOH] = [OH ] = = 2.44×10 M
0.205 L

− 3
pOH = - log (2.44×10 M) =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.
• pH jump
pH 3 – 7
slightly before and after equivalence volume

• at the equivalence point,
All HBr has been neutralized by NH3. The solution only contains NH4Br.
+
+
NH4 comes from weak base. Thus can react with water to produce H3O ion.
+
+
+
+
NH (aq) H O(l) NH (aq) H O (aq)
4
3
3
2
-
Br comes from strong acid and cannot react with water.
21

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

Therefore the salt solution is acidic.
pH < 7 at 25C


(0.1 M )(0.025 L )
Volume equivalence = = 12.50 mL
0.2 M


• Final pH
-
+
NH3(aq) + H2O(aq) ⇌ NH4 (aq) + OH (aq)
[ ]initial 0.02 M - 0 0

△[ ] -x - +x +x

[ ]final 0.02 M - x - x X

  NH   OH 
+
−
4  

Ka =
NH 3 
x
( )( ) x
− 5
1.8×10 =
−
(0.02 M x )
Since Ka is too small, therefore, assume that 0.02 M – x ≈ 0.02 M
x
( )( ) x
1.8=
(0.02 M )
− 4
-
x = [OH ] = 6×10 M

− 4
pOH = -log (6×10 M) = 3.2
pH = 10.8


After the equivalence point the pH gradually increases and plateaus off at pH ̴ 10.8
upon continued addition of 0.20 M NH3.


















22

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

pH

14
13
12
11
10
9
8
7
6
5 pH at equivalence point < 7
4
3
2
1
0 V NH , mL
12.5 3



b)
• before the addition of 0.20 M NaOH, the solution only contains HF.

-
+
HF(aq) + H2O(l) ⇌ F (aq) + H3O (aq)
[ ]I, M 0.20 - 0.00 0.00
[ ], M -x - +x +x

[ ]eq, M 0.20-x - x X


+
 F  −   H O  
 
3
Ka =
 
HF
x
( )( ) x
− 4
6.8×10 =
−
(0.02 M x )
Since Ka is too small, therefore, assume that 0.02 M – x ≈ 0.02 M
( )( ) x
x
− 4
6.8×10 =
(0.02 M )
+
[H3O ]= x = 0.0117 M

pH = 1.93


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

23

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA


• At equivalence point,

All HF has been neutralized by NaOH. The solution only contains NaF.
+
Na comes from strong base and cannot react with water.
-
-
F comes from weak acid and can react with water producing OH ion.
−
−
+
+
F (aq) H O(l) HF(aq) OH (aq)
2
Therefore the salt solution is basic.
pH > 7 at 25C

(0.2 M )(0.025 L )
Volume equivalence = = 25 mL
0.2 M
• after the equivalence point
pOH = - log (0.10 M) = 1
pH = 13

the pH gradually increases and plateaus off at pH ̴13 upon continued addition of 0.10
M NaOH.

pH
14
13
12
11
10
9 pH at equivalence point > 7
8
7
6
5
4
3
2
1
0 V NaOH, mL
25












24

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA




















4.

a) NaOH(aq) + HNO3(aq) → NaNO3(aq) + 2H2O(l)


b) Mole of NaOH = 0.125 mol
0.125 mol
[NaOH] = = 0.625 M
0.2 L


c) Mole of NaOH neutralized = (0.625 M) (0.020 L) = 0.0125 mol


d) From equation,
1 mol NaOH Ξ 1mol HNO3

n HNO3 = 0.0125 mol
0.125 mol
[HNO3] = = 0.500 M
0.25 L


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.










25

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA

7.3: Solubility Equilibria













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

b)

+
2-
Ag2SO4(s) ⇌ 2Ag (aq) + SO4 (aq)
-x M +2x M +x M


−2
x =1.510 M
3
− 5
Ksp = [Ag ] [SO4 ] = (2x) (x) = 4x = 4(1.510 M) = 1.4×10 M 3
2-
2
+ 2
3
−2
c)
2+
2-
CaSO4(s) ⇌ Ca (aq) + SO4 (aq)
-x M +x M +x M

2-
Ksp = [Ca ] [SO4 ]
2+
2.410 = (x)(x)
−4
− 2
Molar solubility = x = 1.5×10 M




















26

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA












2.
a)

-
2+
Ca(IO3)2(s) ⇌ Ca (aq) + 2IO3 (aq)
-x M +x M +2x M

Ksp = [Ca ] [IO3 ]
- 2
2+
2
− 7
6.3×10 = (x)(2x)
3
= 4x
− 3
Solubilty = x = 5.4×10 M

b)
-
2+
Ca(NO3)2(aq) → Ca (aq) + 2NO3 (aq)
-0.10 M +0.10 M +0.20 M


-
2+
Ca(IO3)2(s) ⇌ Ca (aq) + 2IO3 (aq)
-x M 0.10 + x M +2x M

- 2
2+
Ksp = [Ca ] [IO3 ]
− 7
2
6.3×10 = (x + 0.1) (2x)
assume x<< 0.1, x+ 0.1 = 0.1
2
= 0.1(4x )
Solubility = x = 1.3×10 M
− 3

c) Solubility of Ca(IO3)2 in pure water > in 0.10 M Ca(NO3)2.
• In Ca(NO3)2 solution, Ca (common ion) is present.
2+
• The equilibrium position shifts backward.
• Solubility of Ca(IO3)2 decrease.
2+
• The present of common ion, Ca reduces the solubility of Ca(IO3)2.











27

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA












3.
a) i.
-
2+
PbF2(s) ⇌ Pb (aq) + 2F (aq)
-s M +s M +2s M


- 2
2+
Ks p= [Pb ] [2F ]
3.6×10 = (s)(2s)
2
− 8
3
= 4s
− 3
s= 2.1×10 M
o
− 3
The solubility of lead (II) fluoride, PbF2 at 25 C in pure water is 2.1×10 M.

ii.

NaF(aq) → Na (aq) + F (aq)
-
+
-0.10 M +0.10 M +0.10 M


2+
-
PbF2(s) ⇌ Pb (aq) + 2F (aq)
-s M +s M 0.10+2s M

- 2
2+
Ksp= [Pb ] [F ]
− 8
3.6×10 = (s)(2s+0.10) 2
assume s<< 0.1, s+ 0.1 = 0.1
− 6
s= 3.6×10 M
The solubility of lead (II) fluoride, PbF2 at 25 C in NaF solution is 3.6×10 M.
o
− 6

o
b) The solubility of lead (II) fluoride, PbF2 at 25 C in NaF solution has decreased when
compared to the solubility of Lead (II) fluoride, PbF2 at 25 C in pure water due to
o
common ion effect.





28

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA












4.
a) i.

2-
+
Ag2CrO4(s) ⇌ 2Ag (aq) + CrO4 (aq)
2x x

+ 2
2-
Ksp = [Ag ] [CrO4 ]
2
9.0×10 − 12 = (2x) (x)
x = 1.3×10 M (solubility in pure water)
− 4

ii.

2-
Ag2CrO4(s) ⇌ 2Ag (aq) + CrO4 (aq)
+
2x x +0.005 M


2-
+ 2
Ksp = [Ag ] [CrO4 ]
9.0×10 − 12 = (2x) (x+0.005 M)
2
− 5
x = 2.1×10 M (solubility in 0.005M K2CrO4)

b) • Solubility of Ag2CrO4 in pure water > in 0.005 M K2CrO4.

2-
• In K2CrO4 solution, [CrO4 ] is increased.
• The equilibrium position will shift backward.

• Solubility of Ag2CrO4 will decrease.
• The addition of common ion will reduce the solubility of Ag2CrO4.












29

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA









5.

n BaCl2 = (0.004 M) (0.2 L) = 8×10 mol
− 4
−

4
(8 10 mol )
New [BaCl2] = = 1×10 M
− 4
(0.8 L )

− 3
n K2SO4 = (0.008 M)(0.6 L) = 4.8×10 mol

3
−
(4.8 10 mol )
New [K2SO4] = = 6×10 M
− 3
(0.8 L )

BaCl2(aq) → Ba (aq) + 2Cl (aq)
2+
-

-1×10 M +1×10 M +2×10 M
− 4
− 4
− 4


2-
+
K2SO4(aq) → 2K (aq) + SO4 (aq)
− 3
− 3
-6×10 M +0.012 M +6×10 M


2-
2+
BaSO4 (s) ⇌ Ba (aq) + SO4 (aq)

− 4
− 3
1×10 M 6×10 M

− 6
2-
2+
− 3
− 4
Q = [Ba ] [SO4 ] = (1×10 M) ( 6×10 M) = 6×10 M 2

Q > K
• Solution is supersaturated.
• Precipitate will form.







30

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA














6.
-
+
a) C6H5COO , H3O
1.0 g
b) n C6H5COOH dissolved = = 8.20×10 mol
− 3
(122 g/mol )


-3
8.20 10 mol
[C6H5COOH] = = 0.0164 M
0.5 L
+
-
C6H5COOH (aq) + H2O (l) ⇌ C6H5COO (aq) + H3O (aq)
ninitial 0.0164 M - 0 0

△ n -x M - +x +x

nfinal 0.0164 – x M - x X


−
+
 C H COO   H O 


 
Ka = 6 5 3
C H COOH 
6
5
( )( ) x x 2
x
− 5
6.5×10 = =
0.0164 0.0164
− 3
x = [H3O ] = 1.03×10 M
+
pH = 3.0

C H COOH  1.03 10 − 3

c) %α = 6 5 dissociate  100% =  100 = 3.14%
C H COOH  initial 0.0164
5
6









31

CHEMISTRY SK015
Suggested Answer TOPIC 7: IONIC EQUILIBRA














7.
(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.
0.506
(b) Solubility = = 0.0162 M

0.1 311.9
2-
+
Ag2SO4(s) 2Ag (aq) + SO4 (aq)
2(0.0162) 0.0162
2
-5
Ksp = (0.0324) (0.0162) = 1.71 x 10
+
(c) Ag2SO4(s) 2Ag (aq) + SO4 (aq)
2-
(0.01) x
2
-5
1.71 x 10 = (0.01) x
x = 0.17 M = concentration os sodium sulphate needed.





























32