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22/09/2021 2:05 pm
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The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol–1. What is the dissociation constant of acetic acid? Choose the correct option.
[∧°H+ = 350 S cm2mol–1
∧°HCH3COO- = 50 S cm2mol–1 ]
(1) 1.75 × 10–4 mol L–1
(2) 2.50 × 10–4 mol L–1
(3) 1.75 × 10–5 mol L–1
(4) 2.50 × 10–5 mol L–1
1 Answer
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22/09/2021 2:18 pm
Correct answer: (3) 1.75 × 10–5 mol L–1
Explanation:
∧°M(CH3COOH) = ∧°M(CH2COO-) + ∧°MH+
= 350 + 50 = 400 S cm2 mol–1
α = \(\frac{\wedge_M^C}{\wedge_M^0}\)
α = \(\frac{20}{400}\)
= 5 x 10–2
Ka(CH3COOH) = Cα2
= 0.007 x (5 x 10–2)2
= 1.75 x 10–5 mol L–1