CHEMISTRY TEXTBOOK

Problem 5.2 : Calculate the molar

conductivity of AgI at zero concentration

if the molar conductivities of NaI, AgNO 3

and NaNO 3 at zero concentration are

respectively, 126.9, 133.4 and 121.5 Ω-1

cm^2 mol-1.

Solution :

According to Kohrausch law,

i. ∧ 0 (NaI) = λ^0 Na⊕ + λ^0 I

= 126.9 Ω-1 cm^2 mol-1

ii. ∧ 0 (AgNO 3 ) = λ^0 Ag⊕ + λ^0 NO 3

= 133.4 Ω-1 cm^2 mol-1

iii. ∧ 0 (NaNO 3 ) = λ^0 Na⊕ + λ^0 NO 3

= 121.5 Ω-1 cm^2 mol-1

Eq. (i) + eq. (ii) - eq. (iii) gives

∧ 0 (NaI) + ∧ 0 (AgNO 3 ) - ∧ 0 (NaNO 3 )

= λ^0 Na⊕ + λ^0 I + λ^0 Ag⊕ + λ^0 NO 3 - λ^0 Na⊕ - λ^0 NO 3

= λ^0 Ag⊕ + λ^0 I

= ∧ 0 (AgI)

= 126.9 Ω-1 cm^2 mol-1 + 133.4 Ω-1 cm^2 mol-1

(^) - 121.5 Ω-1 cm (^2) mol-1
= 138.8 Ω-1 cm^2 mol-1
Problem 5.4 : The molar conductivity of
0.01M acetic acid at 25^0 C is 16.5 Ω-1 cm^2
mol-1. Calculate its degree of dissociation in
0.01 M solution and dissociation constant
if molar conductivity of acetic acid at zero
concentration is 390.7 Ω-1 cm^2 mol-1.
Solution :
The degree of dissociation,
∝ =
∧c
∧ 0
= 16.6 Ω
-1 cm (^2) mol-1
390.7 Ω-1 cm^2 mol-1
= 0.0422
Ka =
∝^2 c
1- ∝ =
(0.0422)^2 × 0.01
(1 - 0.0422)
= 1.85 × 10 -5
Try this...
Calculate ∧ 0 (CH 2 ClOOH)
if ∧ 0 values for HCl, KCl and
CH 2 ClCOOK are repectively, 4.261,
1.499 and 1.132 Ω-1 cm^2 mol-1.
Problem 5.3 : Calculate molar conductivities
at zero concentration for CaCl 2 and Na 2 SO 4.
Given : molar ionic conductivitis of Ca^2 ⊕,
Cl, Na⊕ and SO 42 ions are respectively,
104, 76.4, 50.1 and 159.6 Ω-1 cm^2 mol-1.
Solution :
According to Kohrausch law,
i. ∧ 0 (CaCl 2 ) = λ^0 Ca 2 ⊕ + 2λ^0 Cl
= 104 Ω-1 cm^2 mol-1 + 2 × 76.4 Ω-1 cm^2 mol-1
= 256.8 Ω-1 cm^2 mol-1
ii. ∧ 0 (Na 2 SO 4 ) = 2λ^0 Na⊕ + λ^0 SO 42
= 2 × 50.1 Ω-1 cm^2 mol-1

• 159.6 Ω-1 cm^2 mol-1
  = 259.8 Ω-1 cm^2 mol-1
  5.3.8 Molar conductivity and degree of
  dissociation of weak electrolytes : The
  degree of dissociation (∝) of weak electrolyte
  is related to its molar conductivity at a given
  concentration c by the equation,
  ∝ =
  ∧c
  ∧ 0 (5.12)
  where ∧c is the molar conductivity of
  weak electrolyte at concentration c ; ∧ 0 is
  molar conductivity at zero concentration.
  = λ^0 H⊕ + λ^0 CH 3 COO = ∧ 0 (CH 3 COOH)
  Thus,
  ∧ 0 (CH 3 COOH) = ∧ 0 (HCl) + ∧ 0 (CH 3 COONa)

• ∧ 0 (NaCl)

Because ∧ 0 values of strong electrolytes,
HCl, CH 3 COONa and NaCl, can be determined
by extrapolation method, the ∧ 0 of acetic acid
can be obtained.