Question

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

Answer 1

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