Ionic Equilibrium
9.1
SOLUTION
Section A
Level-I
1. The conjugate base of HCO3– is CO32–
2. (a)
Left
Right
+
Bronsted acid
NH4
H 3O +
Bronsted base
H2O
NH3
(b)
Bronsted acid HNO2
HCN
Bronsted base CN–
NO2–
3. [H+] = 0.001 M
\
[OH–] =
Kw
+
[H ]
4. [OH– ] = 7.5 ´ 10–3 M
[H+] =
= 1.0 ´ 10–11 M
1.0 ´ 10-14
= 1.33 ´ 10–12 M
7.5 ´ 10 -3
5. pH = 2.29 = – log [H+]
[H+] = 5.13 ´ 10–3M
6. p(OH) = 4.90
[OH–] = antilog (–4.90)
[OH–] = 1.26 ´ 10–5 M
7. [H+]A = 10–2 M
[H+]B = 10–4 M
[H+]A is 100 times greater than the [H+]B
8. pH = 3.0
[H+] = 10–3 M
1 L contains 10–3 mol
\ 0.1 L contains 10–4 mol
9. C = 0.02 M
a = 0.025
2
2
Ka = Ca 0.02 ´ (0.025)
1-a
0.975
= Ka = 1.28 ´ 10–5
Chapter-09 OLC.p65
1
12/7/07, 2:02 AM
9.2
Physical Chemistry for IIT-JEE
10. pH = 1.95
[H+] = antilog (–1.95)
[H+] = 0.0112 M =
Ka C
Ka =
(0.0112)2
C
Ka =
(0.0112)2
= 1.26 ´ 10–3
0.1
11. pH = 10.10
(a) [H+] = 7.94 ´ 10–11 M
and [OH–] =
(b) a =
a=
12.
1.0 ´ 10-14
= 1.26 ´ 10–4 M
7.94 ´ 10 -11
[OH - ]
C
1.26 ´ 10-4
0.012
a = 0.010
as base is just 1% ionized, it is a weak base
Ka. Kb = 1.0 ´ 10–14
Kb =
1.0 ´ 10-14
1.7 ´ 10-4
Kb = 5.88 ´ 10–11
13. (a) K2CO3 pH > 7 as it is a salt of strong base KOH and weak acid H2CO3.
(b) BaCl2, pH = 7, salt of strong base Ba[OH]2 and strong acid HCl
(c) Fe[NO3]3, pH < 7, salt of weak base Fe[OH]3 and strong acid HNO3.
XXV HCO3– + H3O+
14. H2CO3 + H2O WXX
[H3O+] = [HCO3–] = Ca =
=
Ka , C
4.2 ´ 10 -7 ´ 0.08
[H3O+] = [HCO3–] = 1.83 ´ 10–4 M
[H2CO3] = C(1 – a)
and
Chapter-09 OLC.p65
[HCO3- ]
C
1.83 ´ 10 -4
a=
=2.29 ´ 10–3
0.08
a=
2
12/7/07, 2:02 AM
9.3
Ionic Equilibrium
[H2CO3] = 0.0798 M
XXV CO 2– + H O+
HCO3– + H2O WXX
3
3
[CO32–] =
K a ´ [HCO3- ]
2
[CO32–] =
4.8 ´ 10 -11 ´ 1.83 ´ 10 -4
[CO32–] = 9.37 ´ 10–8 M
XXV In– + H+
15. HIn WXX
pH = pKa + log
[In]
[HIn]
5.6 = 6.795 + log
log
[In]
[HIn]
[HIn]
= 1.196
[In - ]
[HIn]
= 15. 70
[In - ]
as [HIn] > [In–], the colour of the indicator is due to its unionized form i.e. yellow.
16. (a) For CaF2
Ksp = 4x 3 where x is the molar solubility
1
(b)
æ 4.0 ´ 10-11 ö 3
–4
x= ç
÷ = 2.15 ´ 10 M
4
è
ø
[F –] = 2x
= 2 ´ 2.15 ´ 10–4
[F – ] = 4.30 ´ 10–4 M
17. (a) As Ksp (CuI) < Ksp (PbI2), CuI will precipitate first on adding NaI solution.
(b) [I –] required to precipitate CuI is
1.1 ´ 10-12
= 1.1 ´ 10–8 M
1.0 ´ 10-4
and [I–] required to precipitate PbI2 is
1
æ 7.1 ´ 10-9 ö 2
ç
÷
è 1.0 ´ 10-3 ø
[I –] = 2.66 ´ 10–3 M
Chapter-09 OLC.p65
3
12/7/07, 2:02 AM
9.4
Physical Chemistry for IIT-JEE
(c) Cu+ ion will precipitate first and it continuous to precipitate till [I –] reached 2.66 ´ 10–3
when Pb2+ also starts precipitating. Cu+ remaining in the solution is
[Cu+] =
[Cu+] =
K sp (CuI)
[I - ]
1.0 ´ 10-12
2.66 ´ 10-3
= 3.76 ´ 10–10
Percentage of Cu+ ion remaining in the solution is
=
3.76 ´ 10-10
´ 100
1.0 ´ 10-4
= 3.76 ´ 10–6 ´ 100
= 0.000376
XXV [Cu(NH ) ]2+
18. Cu2+(aq) + 4NH3(aq) WXX
3 4 (aq)
Since K f is very large (5.0 ´ 1013), the equilibrium lies to the right. We can assume that all Cu+ ion
converted into [Cu(NH3)4]2+ ion
The amount of NH3 consumed is 4 ´ 0.01 = 0.04 mol
The conc. of NH3 at equilibrium is 1.0 – 0.04 = 0.96 M and that of [Cu(NH3)4]2+ is 0.01 M (same as the
initial conc. of Cu2+).
Let the conc. of Cu2+ ion at equilibrium is x and write
Kf =
[Cu(NH3 )2+
4 ]
[C u 2+ ] [NH3 ]4
= 5.0 ´ 1013
0.01
= 5.0 ´ 1013
x (0.96) 4
Solving for x gives
=
x = 2.17 ´ 10–16 M = [Cu2+]
XXV Ag+(aq) + Br–
19. AgBr(s) WXX
(aq)
Ksp = [Ag+] [Br–] = 7.7 ´ 10–13
XXV Ag(S O ) 3–
Ag+(aq) + 2S2O32–(aq) WXX
2 3 2
Kf = 1.0 ´ 1013
overall reaction is
XXV Ag(S2O3)23– + Br–
AgBr(s) + 2S2O32–(aq) WXX
K = Ksp. Kf
= 7.7 ´ 10–13 ´ 1.0 ´ 1013
= 7.7
Let s be the solubility of AgBr in mol/L, then
Chapter-09 OLC.p65
4
12/7/07, 2:02 AM
9.5
Ionic Equilibrium
XXV Ag(S2O3)23– + Br–(aq)
AgBr(s) + 2S2O32–(aq) WXX
Initial conc. :
at equilibrium :
0.1
(0.1–2s)
K=
s2
(0.1 - 2 s) 2
0
s
0
s
s.s
= 7.7
(0.1 - 2 s) 2
= 7.7
s
= 2.77
(0.1 - 2 s)
solving for s gives
s = 0.042 M
at equilibrium, 0.042 mole of AgBr dissolves in 0.1 M Na2S2O3
Level-II
XXV C6H5O– + H3O+
1. C6H5OH(aq) + H2O(l) WXX
(aq)
(aq)
Ka = 1.0 ´ 10–10
and [C6H5O– ] = Ca =
and
a=
Ka ´ C =
Ka
=
C
1.0 ´ 10-10 ´ 0.05 = 2.24 ´ 10–6 M
1.0 ´ 10-10
= 4.47 ´ 10–5
0.05
In 0.01 M C6H5ONa solution
XXV C6H5O– + H3O+
C6H5OH(aq) + H2O(l) WXX
(aq)
(aq)
(0.05 – x)
C6H5O– Na–
0.01M
®
C6H5O–
0.01M
x
x
+
Na+
0.01M
[C6 H5 O - ] [H3 O+ ] ( x + 0.01) x
=
[C6 H5 OH]
(0.05 - x)
Solving for x, we obtain
Ka =
x = 5.0 ´ 10–10
5 ´ 10-10
x
=
= 1.0 ´ 10–8
0.05
C
where a¢ = degree of dissociation of phenol in 0.01 M C6H5O–Na+ solution.
and a ¢ =
Chapter-09 OLC.p65
5
12/7/07, 2:02 AM
9.6
Physical Chemistry for IIT-JEE
XXV HS– + H+
2. H2S WXX
Ka1 =
[HS- ] [H + ]
= Ca 2
[H 2S]
and [HS–] = Ca =
[HS–] =
K a1C
9.1 ´ 10-8 ´ 0.1 = 9.54 ´ 10–5 M
in 0.1 M HCl
XXV HS– + H+
H2S WXX
HCl
®
Cl–
(0.1 – x)
0.1 M
Ka1 =
H+
+
x
x
0.1 M
0.1 M
x(0.1 + x) 0.1 x
=x
=
(0.1 - x )
0.1
x = 9.1 ´ 10–8 M = [HS– ] in 0.1 M HCl
The concentration of H5 has decreased in 0.1 M HCl.
To calculate the concentration of S2– ion
Initial conc.
at equilibrium
HS– ®
9.54 ´ 10–5
(9.54 ´ 10–5– x)
Ka2 =
\
1.2 ´ 10–13 =
H+
0
x
+
S2–
0
x
(H + ) (S2 - )
(HS– )
x2
x2
=
=x
9.54 ´ 10-5 - x
x
[S2–] = 1.2 ´ 10–13 M
3. Concentration of Sr (OH)2 =
19.23
= 0.158 mol/L
121.62
XXV Sr2+ + 2OH –
Sr(OH)2 (aq) WXX
(aq)
(aq)
[OH – ] = 2[Sr(OH)2] = 2 ´ 0.158 = 0.316 M
\
and
1.0 ´ 10-14
= 3.16 ´ 10–14 M
0.316
pH = – log 3.16 ´ 10–14
[H + ] =
pH = 13.50
Chapter-09 OLC.p65
6
12/7/07, 2:02 AM
9.7
Ionic Equilibrium
XXV CH CH COO– + H O+
4. CH3CH2COOH + H2O WXX
3
2
3
a=
Ka
1.32 ´ 10-5
=
= 0.016
C
0.05
[H3O+] = Ca = 0.05 ´ 0.016 = 8.0 ´ 10–4 M
and pH = – log(8.0 ´ 10–4) = 3.09
In 0.01 M HCl
Ka =
\
x (0.01 + x ) 0.01 x
= 1.32 ´ 10–5
=
(0.05 - x)
0.05
x = 6.6 ´ 10–5
-5
The degree of ionization = 6.6 ´ 10 = 1.32 ´ 10–3
0.05
–4
5. Ka = 4.5 ´ 10 and C = 0.04 M
XXV HNO2 + OH –
NO2– + H2O WXX
[OH –] = Ch
and
Kh =
Kw
Ch 2
=
= ch2
1- h
Ka
h = degree of hydrolysis
h=
\
Kw
=
Ka × C
1.0 ´ 10-14
4.5 ´ 10 -4 ´ 0.04
= 2.36 ´ 10–5
[OH–] = Ch = 0.04 ´ 2.36 ´ 10–5 = 9.43 ´ 10–7 M
p[OH] = – log[OH –] = 6.03
\
pH = 7.97
6. Pyridinium hydrochloride is a salt of weak base and strong acid.
Therefore,
pH = 7 –
\
3.44 = 7 –
1
[log C + pKb]
2
1
[log 0.02 + pKb]
2
1
(–1.70 + pKb)
2
pKb = 8.82
–3.56 =
taking antilog gives
Kb = 2.0 ´ 10–9
Chapter-09 OLC.p65
7
12/7/07, 2:02 AM
9.8
Physical Chemistry for IIT-JEE
7. (i) nca(OH)2 = 0.2 ´ 0.01 = 0.002
nOH– = 2 ´ nca(OH)2 = 0.004
and nH + = nHCl = 0.1 ´ 0.025 = 0.0025
H ion is completely consumed,
+
therefore nOH– left = 1.5 ´ 10–3
[OH –] = 0.0375 M
and
p[OH] = 1.43
and pH = 14 – 1.43 = 12.57
(ii) nH2SO4 = 0.01 ´ 0.01 = 0.0001
nH+ = 0.002
nCa(OH)2 = 0.01 ´ 0.01 = 0.0001
nOH– = 2 ´ 0.0001 = 0.0002
as nH+ = nOH –
\
pH = 7.0
(iii) nH2SO4 = 0.1 ´ 0.01 = 0.001
\
nH+ = 0.002
nOH – = 0.001
nH+ left = 0.001
\
and
[H + ] =
0.001
20 ´ 10 -3
= 0.05
pH = 1.30
8. C6H5COOH + NaOH ® C6H5COONa + H2O
nBenzoic acid = nNaOH = MNaOH VNaOH
= 0.2 ´ 50 ´ 10–3 = 0.01 mol
mass in grams
but n =
molar mass
Molar mass =
1.22
= 122 g/mp
0.01
9. Ksp(Ag2CrO4) > Ksp(AgBr)
\ Ag2CrO4 is more soluble
XXV 2Ag+ + CrO 2–
Ag2CrO4(s) WXX
4 (aq)
(aq)
Ksp = 4s3 and
1
æ 1.1 ´ 10-12 ö 3
–5
\ s= ç
÷ = 6.50 ´ 10 M
4
è
ø
XXV Ag+(aq) + Br–(aq)
AgBr(s) WXX
Chapter-09 OLC.p65
8
12/7/07, 2:02 AM
9.9
Ionic Equilibrium
Ksp = (s¢ )2
1
S¢ = (50 ´ 10–13) 2 = 7.07 ´ 10–7 M
The ratio of their molarities
S 6.50 ´ 10 -5
=
= 91.92
S ¢ 7.07 ´ 10-7
10. Suppose the volume of each solution a 1L, then
=
nNaIO3 = 0.002 ´ 1 = 0.002 mol
nIO–3 – = 0.002 mol and
nCu2+ = nCu(ClO3)2 = 0.002
Total volume after the mixing of two solution = 2L
\
[IO3–] = 0.001 M and
[Cu2+] = 0.001 M
Then Q = [IO–3 ]2 [Cu2+ ] = (0.001)2 ´ 0.001
Q = 1.0 ´ 10–9
As Q < Ksp, the reaction will proceed in the forward direction. No precipitate of copper iodate will form
11. Suppose S is the molar solubility of silver benzoate in water then
XXV C6H5COO– + Ag+
C6H5COOAg(s) WXX
(aq)
(aq)
Ksp = S2
and S =
2.5 ´ 10-13 = 5.0 ´ 10–7 M
If the solubility of salt of weak acid of ionization constant Ka is S¢, then Ksp and Ka, S¢ are related to
each other as
æ
ö
Ka
Ksp = S¢ 2 ç
+ ÷
è K a + [H ] ø
and
1
ì
ü2
ï
ï
K sp
ï
ï
S¢ = í
ý
öï
ka
ïæ
ï çè k +[H + ] ÷ø ï
î a
þ
Given pH = 3.19 and [H+] = 6.46 ´ 10–4 M
1
\
Chapter-09 OLC.p65
9
ì
ü2
ï
ï
2.5 ´ 10 -13
ï
ï
S= í
ý
5
ùï
6.46 ´ 10
ïé
ï êê 6.46 ´ 10-5 + 6.46 ´ 10-4 úú ï
ûþ
îë
12/7/07, 2:02 AM
9.10
Physical Chemistry for IIT-JEE
S¢ = 1.658 ´ 10–6 M
\ The ratio
1.658 ´ 10-6
S¢
=
= 3.32
5.0 ´ 10-7
S
Silver benzoate is 3.32 times more soluble in buffer of pH 3.19 than in water.
12. Let the concentration of FeSO4
Solution =
n Fe 2+
V
n
and the concentration of Na2S =
s2
–
V
After mixing
[Fe2+] = [S2–] =
n Fe 2 +
2V
=
nS22V
XXV Fe2+ + S2–
WXX
(aq)
(aq)
FeS(s)
Ksp = [Fe2+ ] [S2–] = [Fe2+ ]2
æn + ö
Ksp = ç Fe 2 ÷
è 2V ø
n Fe 2+
2V
2
= 6.3 ´ 10–18
= 2.5099 ´ 10–9 M
[Fe2+]ini =
n Fe 2+
= [S2– ] ini
2V
= 2 ´ 2.5099 ´ 10–9 = 5.02 ´ 10– 9 M
XXV Ca2+ + SO42–(aq)
13. CaSO4 (s) WXX
(aq)
Ksp = S2
and
S=
nCaSO4 =
9.1 ´ 10 -6 = 3.017 ´ 10–3 M
mass
1
=
mol
molar mass 136
but nCaSO4 = MCaSO4 VCaSO4
VCaSO4 =
1
136 ´ 3.017 ´ 10 -3
VCaSO4 = 2.44 L
Chapter-09 OLC.p65
10
12/7/07, 2:02 AM