ElectrochemistryHard
Question
Given the ionic equivalent conductivities for the following ions:
λoeq K+ = 73.5 cm2 ohm-1 eq-1
λoeq Al3+ = 149 cm2 ohm-1 eq-1
λoeq SO42 = 85.8 cm2 ohm-1 eq-1
The Λoeq for potash alum (K2SO4.Al2(SO4)3.24H2O) is:
λoeq K+ = 73.5 cm2 ohm-1 eq-1
λoeq Al3+ = 149 cm2 ohm-1 eq-1
λoeq SO42 = 85.8 cm2 ohm-1 eq-1
The Λoeq for potash alum (K2SO4.Al2(SO4)3.24H2O) is:
Options
A.- 245.92
B.348.3
C.368.2
D.108.52
Solution
(Eq = Charge on the ion/Total charge)
[A+] =
mol × 2 = 
mol = 
Eq
[Al3+] =
Eq
[SO42-] =
= 1Eq
Λoeq K2SO4.Al2(SO4)3. 24H2O
= λoeq(K+) + λoeq (Al+3) + λoeq(SO42-)
=
× 73.5 + 189 × 
+ 85.8 × 1
= 18.375 + 141.75 + 85.8
= 249.92
[A+] =
mol × 2 = mol = Eq[Al3+] =
Eq[SO42-] =
= 1EqΛoeq K2SO4.Al2(SO4)3. 24H2O
= λoeq(K+) + λoeq (Al+3) + λoeq(SO42-)
=
× 73.5 + 189 × + 85.8 × 1= 18.375 + 141.75 + 85.8
= 249.92
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