J.R. S. answered • 03/03/21
Ph.D. University Professor with 10+ years Tutoring Experience
AgCl(s) ==> Ag+(aq) + Cl-(aq) ... Ksp = 1.80x10-10
Ag+(aq) + 2NH3(aq) ==> Ag(NH3)2+(aq) ... Kf = 1.7x107
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AgCl(s) + 2NH3(aq) ==> Ag(NH3)2+ + Cl- ... Keq = Ksp x Kf = (1.80x10-10)(1.7x107) = 3.1x10-3
Keq = [Ag(NH3)2+] [Cl-] / [NH3]2
3.1x10-3 = (x)(x) / (0.560 -2x)2 .... I have omitted x from denominator to avoid using quadratic
x2 = (3.1x10-3)(0.560)2 = 1.74x10-3
x = 4.2x10-2 M (NOTE: this is greater than 5% of 0.560 M so we shouldn't have omitted it in the step above. I'll leave it to you to go back and repeat the calculations using the quadratic.
J.R. S.
05/07/22
Mackenna Q.
When you set up the equation: 3.1x10-3 = (x)(x) / (0.560 - x)2 Shouldn't it be 3.1x10-3 = (x)(x) / (0.560 - 2x)2?05/07/22