The solubility of agcl (s) in water at 25°c is 1.33 x 10-5 mol/l and its δh° of solution is 65.7 kj/mol. what is its solubility at 95°c?

When AgCl (s) ↔ Ag + (aq) + Cl - (aq)

∴ K1 = [Ag+][Cl-]

when the solubility of AgCl = 1.33 x 10^-5

∴ K1 = X1^2 = (1.33 x 10^-5) ^2 = 1.77 x10^-10

by using vant's Hoff equation:

㏑ (K2/K1) = - (ΔH/R) * (1/T2 - 1 / T1)

when T1 = 25 + 273 = 298 K

T2 = 95 + 273 = 368 K

ΔH = 65.7Kkj/mol

R = 8.3145

by substitution:

㏑ (K2 / 1.77 x 10 ^-10) = (65.7 / 8.3145) * (1/368 - 1 / 298)

∴K2 = 1.76 x 10^-10