The solubility of agcl (s) in water at 25 ∘c is 1.33*10-5mol/l and its δh∘ of solution is 65.7 kj/mol. what is the solubility at 47.7 c

According to this equation:

AgCl (s) ↔ Ag + (aq) + Cl - (aq)

so K1 = [Ag+][Cl-]

when [Ag+] = [Cl-] we can assume both = X

and when we have X the solubility = 1.33 x 10^-5 mol / L

by substitution:

∴ K1 = X^2

= (1.33 x 10^-5) ^2

= 1.77 x 10^-10

by using vant's Hoff equation:

ln (K2/K1) = (ΔH/R) * (1/T2-1/T1)

when ΔH = 65700 J / mol

R = 8.314

T1 = 25+273 = 298 K

T2 = 47.7 + 273 = 320.7

by substitution:

∴㏑ (K2/1.77 x 10^-10) = (65700/8.314) * (1/320.7 - 1 / 298)

by solving for K2

∴K2 = 2.7 x 10^-11

and when K2 = X^2

∴ the solubility X = √ (2.7 x 10^-11)

= 5.2 x 10^-6 mol/L