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Determine the solubility (in g/L) of AgCl in a 0.25M solution ity in g/L) of AgCl in a 0.25M solution of NaCl. (Ksp (AgCl) = 1.7 x 10-10, MM = 143.4g/mol) A. 1.9 x 10-4 B. 3.5 x 10-9 C. 6.8 x 10 -10 D. 9.7 x 10-8

Respuesta :

Answer:

D) S AgCl = 9.7 E-8 g/L

Explanation:

AgCl ↔ Ag+  +  Cl-

 S           S        S + 0.25

NaCl ↔     Na+     +   Cl-

0.25M     0.25M      0.25M

⇒ Ksp = 1.7 E-10 = [ Ag+ ] * [ Cl- ]  = S * ( S + 0.25 )

⇒ 1.7 E-10 = S² + 0.25 S

⇒ 0 = - 1.7 E-10 + 0.25 S + S²

⇒ S = 6.8 E-10 mol/L * ( 143.4 g/mol ) = 9.751 E-8 g/L ≅ 9.7 E-8 g/L