What is the concentration of Ag+(aq) ion in 1.00L of solution that is 0.010 M Ag

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Ag(NH3)2+ (aq) ? Ag+ (aq) + 2 NH3 (aq) 1/Kf = 5.9x10^-8 1/Kf = [Ag(NH3)2+] / [Ag+] [NH3]² Since we are assuming the reverse reaction, we assume [Ag(NH3)2+] = 0.01M initially, and it loses x in going to equilibrium. [Ag+] goes from 0 to x, and [NH3] gains x from 0.99 M. Substituting: 5.9x10^-8 = (0.01-x) / (x) (0.99+x)² We can see x will be very small, so we can ignore it in the numerator and the squared parentheses in the denominator. The equation becomes: 5.9x10^-8 = (0.01) / (x) (0.99)² Solving, x = [Ag+] = 5.8x10^-6 M, or pAg+ = 5.3. and AgCl--->Ag+ Cl- Ksp= 1.6x10^-10 Ag + 2NH3 ---> Ag(NH3)2 Kf= 1.7x10^7 net reaction: AgCl + 2NH3 ----> Cl- + Ag(NH3)2 so Ksp * Kf = .00272 = net equilbrium constant so, for the formation of the complex, you would use the net equilibrium constant for the values of your ICE chart of the formation of the Ag(NH3)2 and solve for x Kf = [Ag(NH3)2] / [Ag][NH3]^2 so Knet = [Ag(NH3)2+][Cl-] / [NH3]^2 Knet = 0.00272 = x^2 / 1 - 2x x^2 + 0.00544x - 0.00272 = 0 x = [AgNH3)2+]] = molar solubility of Ag(NH3)2+ = 0.0495M = 5x10^-2M.....A

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