Q No. 70.d
1. What is the pH of
a solution containing 0.5 M propionic acid and 0.5 M sodium propionate? The Ka of propionic acid is 1.34 x
10-5.
Ka of propionic acid = 1.34 x 10–5
∴ pKa = – log Ka = – log (1.34 x 10–5)
= 5 – log 1.34
= 5 – 0.1271
pKa = 4.8729
pH = pKa
+ log [salt] / [acid] ... Henderson - Hasselbalch equation
= 4.8729 + log 0.5 / 0.5
pH = 4.8729
Or
The
dissociation equilibrium of propionic acid will be
C2H5COOH
⇌
C2H5COO– + H+
Ka = [C2H5COO–][ H+ ] / [C2H5COOH
]
= 0.5 x [ H+] / 0.5
Ka = [H+]
∴ pH = – log [H+] = –
log Ka
= – log (1.34 x 10–5)
= 5 – log 1.34
= 5 – 0.1271
∴ pH =
4.8729
2. Find the pH of a
buffer solution containing 0.30 mole per litre of CH3COONa and 0.15 mole per litre of CH3COOH. Ka for acetic
acid is 1.8 x 10-5.
Ka = 1.8 x 10–5
pKa = – log Ka
pKa = – log (1.8 x 10–5)
= 5 – log 1.8
= 5 – log 1.8
= 5- 0.2553
pKa = 4.7447
pH = pKa
+ log[salt] / [acid]
... Henderson - Hasselbalch equation
= 4.7447 + log 0.30 / 0.15
pH
= 4.7447 + log 2
=
4.7447 + 0.3010
pH = 5.0457
pH = 5.0457
3. Calculate the pH of the buffer solution containing 0.04 M NH4Cl and 0.02 M NH4OH. For NH4OH Kb is 1.8 x 10-5.
Kb = 1.8 x 10–5
pKb = – log Kb
pKb = – log (1.8 x 10–5)
= 5 – log 1.8
= 5- 0.2553
pKa = 4.7447
pOH = pKb
+ log[salt] / [base]
... Henderson - Hasselbalch equation
= 4.7447 + log 0.04 / 0.02
pOH
= 4.7447 + log2
= 4.7447 + 0.3010 = 5.0457
pH + pOH = 14.00
pH = 14.00 – pOH
pH = 14.00 – 5.0457
pH = 8.9543
4. Equivalent conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 M acetic acid the equivalent conductance is 5.2 mho.cm2.gm.equiv–1. Calculate degree of dissociation, H+ ion concentration and dissociation constant of the acid.
4. Equivalent conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 M acetic acid the equivalent conductance is 5.2 mho.cm2.gm.equiv–1. Calculate degree of dissociation, H+ ion concentration and dissociation constant of the acid.
Equivalent
conductivity of acetic acid at infinite dilution, λ∞ = 390.7
Equivalent
conductance of acetic acid, λc = 5.2
mho.cm2.gm.equiv–1
Concentration
of acetic acid, C = 0.1 M
Degree of
dissociation, a = λc / λ∞
= 5.2
/ 390.7
=
0.01333
= 1.33 x 10-2 = 1.33%
CH3COOH
⇌
H+ + CH3COO–
C
(1 – a) Ca Ca
∴
H+ ion concentration,
[H+] = Ca
= 0.1 x 0.0133
= 0.00133
= 1.33 x 10-2
M
Dissociation
constant of acetic acid, K = a2C / 1 - a
=
(0.0133)2 x 0.1 / (1
- 0.0133)
= 0.000017689 / 0.9867
= 1.79 x 10-5 M
5. Calculate / Find
the pH of a buffer solution containing 0.2 / 0.20 mole per litre CH3COONa and 0.15 mole per litre CH3COOH. Ka for acetic acid is 1.8x10-5.
Ka = 1.8 x 10–5
pKa = – log Ka
pKa = – log (1.8 x 10–5)
= 5 – log 1.8
= 5- 0.2553
pKa = 4.7447
pH = pKa
+ log [salt] / [acid] ... Henderson - Hasselbalch equation
= 4.7447 + log 0.20 / 0.15
pH
= 4.7447 + log 4 / 3
pH
= 4.7447 + log 4 - log 3
pH
= 4.7447 + 0.6021 – 0.4771
pH = 4.8697
6. An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being deposited?
6. An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being deposited?
Wt. of Copper / Wt. of
Iodine = Eqvt. wt. Copper (31.7) / Eqvt. wt. of Iodine
(127)
1.25 / x = 31.7 / 127
Wt. of Iodine, x
= 1.25 x 127 / 31.7
Hence, Wt. of Iodine, x = 5.0 g
Also,
Wt. of
Copper / Wt. of Silver = Eqvt. wt. of Copper (31.7) / Eqvt. wt. of Silver
(108)
1.25 / y = 31.7 / 108
Wt. of
Silver, y = 108 x 1.25 / 317
Wt. of Silver, y = 4.26 g
Or
31.7
g of Copper (1g eqvt) is liberated by = 96,495 coulomb
1.25
g of Copper is liberated by = 96,495 x 1.25 / 31.7 coulomb
Quantity
of electricity = 3805 coulomb
96,495
coulomb deposits 127 g of Iodine
∴ 3805 coulomb deposits = 127 x 3805 /
96,495
= 5.0 g of Iodine
96,495
coulomb deposits 108 g of Silver
∴ 3805 coulomb deposits = 108 x 3805 /
96,495
= 4.26 g of Silver
7. The
equivalent conductance of HCl, CH3COONa and NaCl at
infinite dilution are 426.16, 91.0 and 126.45 ohm-l cm2 (gram equivalent)-1 respectively. Calculate the equivalent
conductance / λ∞ of acetic acid.
λ∞
CH3COOH = λ∞ CH3COONa
+ λ∞ HCl – λ∞ NaCl
λ∞
CH3COOH = 91.0 + 426.16 – 126.45 = 517.16 - 126.45
λ∞ CH3COOH
= 390.71
8. 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrochemical equivalent of copper?
8. 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrochemical equivalent of copper?
Current, I = 0.2
ampere
Time, t = 50 minutes = 50 ´ 60
seconds
Quantity
of electricity used, Q = I x t = 0.2 x 50 x 60 = 600 coulombs
Amount
of copper deposited by 600 coulombs, m = 0.1978 g
Amount
of copper deposited by 1 coulomb, Z = 0.1978 / 600g = 0.0003296 g
Electrochemical
equivalent of copper, Z = m / I . t
=0.0003296
gC–1
= 3.296 x 10–4 gC–1
= 3.296 x 10–4 gC–1
= 3.206 x 10–7 kgC–1
Or
Electrochemical
equivalent of copper, Z = m / I . t
(or) Z = m / Q
= 0.1978 g /
0.2 amp x 50 x 60 sec
= 0.1978 g / 600 C [ ∵ amp . sec = C
=0.0003296
= 3.296 x 10–4 gC–1
= 3.296 x 10–7 kgC–1
9. Calculate the pH of 0.1
M acetic acid / CH3COOH solution. Dissociation constant of acetic
acid is 1.8 x 10-5 M.
For
weak acids, [H+]
= √Ka x C
= √1.8 x 10-5 x 0.1 = √1.8 x 10-6
= 1.34 x
10–3 M
∴ pH = – log [H+]
= – log (1.34 x 10–3)
= 3 – log1.34
= 3 – 0.1271
∴ pH =
2.8729
10. 0.04 N solution of a weak acid has a specific conductance 4 x10–4 mho.cm–1. The degree of dissociation of acid at this dilution is 0.05. Calculate the equivalent conductance of weak acid at infinite solution.
10. 0.04 N solution of a weak acid has a specific conductance 4 x10–4 mho.cm–1. The degree of dissociation of acid at this dilution is 0.05. Calculate the equivalent conductance of weak acid at infinite solution.
Specific
conductance, κ = 4 x 10-4 mho.cm-1
Equivalent conductance
of weak acid, λc = κ .1000
/ C
= 4 x 10-4 x 1000 /
0.04
λc = 10 mho.cm2.eq-1
Degree
of dissociation, α = 0.05
α = λc / λ∞
α = λc / λ∞
Equivalent conductance
of weak acid at infinite solution, λ∞ = λc / α
λ∞ = 10 / 0.05
∴
λ∞ = 200 mho.cm2.gm.equiv.-1
11.Ionic conductances at infinite dilution of Al3+ and SO42– are 189 ohm–1cm2gm.equiv-1 and 160 ohm-1cm2gm.equiv.-1. Calculate equivalent and molar conductance of the electrolyte at infinite dilution.
11.Ionic conductances at infinite dilution of Al3+ and SO42– are 189 ohm–1cm2gm.equiv-1 and 160 ohm-1cm2gm.equiv.-1. Calculate equivalent and molar conductance of the electrolyte at infinite dilution.
Equivalent conductance
of the electrolyte at infinite dilution,
λ∞ = 1/n+. λA+ + 1/m- . λB–
Where,
λ∞+ and λ∞–
are the cationic and anionic equivalent conductances at infinite dilution
n+ and m– correspond the valency of cations and anions
The
electrolyte is Al2(SO4)3
λ∞
Al2(SO4)3 = 1/3 λ∞ Al3+ + 1/2 λ∞ SO42–
λ∞
Al2(SO4)3 = 189 / 3 + 160 / 2
= 63 + 80
= 63 + 80
= 143 mho cm2 gm.equiv–1
Molar conductance of
the electrolyte at infinite dilution,
μ∞ = γ+
μ∞+ + γ- μ∞-
Where,
μ∞- and μ∞+are
the ionic conductances at infinite dilution
γ+ = number of cations and γ- = number of anions
μ∞
Al2(SO4)3 = 2 μ∞+ + 3 μ∞-
= (2 x 189) + (3 x 160)
= 378 + 180
= 858 mho cm2 mol–1
= (2 x 189) + (3 x 160)
= 378 + 180
= 858 mho cm2 mol–1
12. What current strength in amperes will be required to liberate 10 g
of iodine from potassium iodide solution in one hour?
127
g of iodine (1g eqvt) is liberated by = 96,495 coulomb
10 g of iodine is liberated by =
(96,495 x 10 / 127) coulomb
Let
the current strength be = I
Time
(in seconds) = 1 x 60 x 60 = 3600 seconds
The
quantity of electricity used, Q = I x Time (in seconds)
Curernt strength, I = Q / t
= 96,495 coulomb
x 10 / 127 x 3600 seconds
= 2.11 coulomb . second–1
or =
2.11ampere [ ∵ coulomb . second–1 = ampere
