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The solubility of AgCl(s) in water at 25 degrees Celsius is 1.33 x 10^-5 mol/L and its delta H of solution is 65.7 kJ/mol. What is its solubility at 45 degrees Celsius?
Express your answer using two significant figures.
S_{\rm AgCl} =____.
\rm M
Find Molarity of AgCl(s) at 45 degrees Celsius?

Respuesta :

Answer:

S AgCl(45°C) = 3.09 E-5 mol/L

Explanation:

  • AgCl(s) ↔ Ag + + Cl-

          S             S          S

∴ S = 1.33 E-5 mol/L

⇒Ksp1 (25°C) = S² = 1.8 E-10

Vant Hoff:

  • Ln K1/K2 = - ΔH/R ( 1/T1 - 1/T2)

∴ T1 = 25°C = 298.15K

∴T2 = 45°C = 318.15 K

∴ R = 8.314 E-3 KJ/mol.K

⇒ K1/K2 = e ( -ΔH/R (1/T1 - 1/T2))

⇒ K1/K2 = e ( - (65.7/8.314E-3) ( 1/298.15 - 1/318.15))

⇒ K1/K2 = e ( - 1.666 )

⇒ K1/K2 = 0.1889

⇒ K2 = K1/0.1889 = 1.8 E-10 / 0.1889 = 9.525 E-10 (45°C)

⇒ Ksp(45°C) = S² = 9.525 E-10

⇒ S = √9.525 E-10 = 3.0863 E-5 mol/L

Molarity (M):

M ≡ mol/L

M AgCl = S = 3.09 E-5 M

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