Question

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?

Answer 1

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

Answer 2

The solubility of agcl(s) in water at 95°C is : 5.4 * 10⁻⁴  mol/L

Given data :

solubility of agcl(s) at 298 k ( 25°c ) = 1.33 * 10⁻⁵ mol/l

ΔH∘ of solution =  65.7 kJ/mol

Determine the solubility at 95°C

applying van't hoff equation

ln $(\frac{K_{2} }{k_{1} } ) =$  -  ΔH∘ / R * ( 1 / T₂ - 1 / T₁ )  --- ( 1 )

where : T₂ = 273 + 95 = 368 k , T₁ = 298 k

Substitute values into equation ( 1 )

ln ( Ksp / 1.7689 * 10⁻¹⁰ ) =  - 65.7 * 10³ / 8.314 * ( 1 / 368 - 1 / 298 )

therefore :

( Ksp / 1.7689 * 10⁻¹⁰ ) = -7.90 * 10³ * -0.00064

                                    = e^5.056

therefore:  Ksp = 1.7689 * 10⁻¹⁰ * 156.96

                         = 2.776 * 10⁻⁸

Given that:   Ksp = S² ( solubility )

solubility ( s ) at 95° C  = 5.4 * 10⁻⁴  mol/L

                                     = 5.4 * 10⁻⁴  mol/L

Hence we can conclude that The solubility of agcl(s) in water at 95°C is : 5.4 * 10⁻⁴  mol/L

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