Calculate the solubility at 25 degrees Celsius in grams per liter of AgI in a so

Question

Calculate the solubility at 25 degrees Celsius in grams per liter of AgI in a solution that is 0.15 M in potassium iodide and 0.25M in ammonia. State any equilibrium constants you use.

Explanation / Answer

Ag+ + 2NH3 = [Ag(NH3)2]+ (Keq) AgI = [Ag+] + [I-] (Ksp) adding AgI + 2NH3 ----> I- + Ag(NH3)2 (K) .25 - 2x .15+x x K = Ksp * Keq = 1.5X10-16*1.7x10^7=2.55X10^-8 Keq = [Ag(NH3)2] / [Ag][NH3]^2 so K = [Ag(NH3)2+][I-] / [NH3]^2 K = x(.15+x) / (.25 - 2x)^2 .15+x~.15 .25-2x~.25 x=(2.55*.25*.25)/.15X10^-8 =1.0625X10^-8 moles per litre solubility in gms per litre = 1.0625X10^-8Xmol wt of AgI =2.49X10^-6 gm/litre

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