Homework Answers
In (s2/s1) = [ΔH/R] [(1/T1)] – (1/T2)] -- Eq(1)
Given that
Initial solubility, s1 = 1.33 x 10-5 mol/L
Final rsolubility , s2 = ?
Initial temperature T1 = 25oC = 25 + 273 K = 298 K
Final temperature T2 = 44.4 oC = 44.4 + 273 K = 317.4 K
ΔH = 65.7 kJ/mol = 65700 J/mol
substitute all these values in eq (1),
In (s2/s1) = [ΔH/R] [(1/T1)] – (1/T2)] -- Eq(1)
In (s2/1.33 x 10-5 mol/L)= [65700 J/mol/8.314 J/K/mol] [(1/298)] – (1/ 317.4)]
On simplification,
s2 = 6.72 x 10-5 mol/L
Therefore,
solubility of AgCl at 44.4oC = 6.72 x 10-5 mol/L
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