Question

Exercise 18.102 Review Constants l Periodic Table Part A The solubility of AgCI(s) in water at 25 Cis 1.33 x 10 s mol/L and its AH of solution is 65.7 kJ/mol What is its solubility at 44.4°C? Express your answer using two significant figures x®X5b返返 6.52 10- Request Answer Prev ious Answers Submit Incorrect; Try Again; 14 attempts rema

Answer 1

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