What is the concentration of Ag+ (aq) ion in 0.010 M AgNO3 that is also1.00 M NH3? Kf for Ag(NH3)2+ is 1.7 x 107.

To solve this problem, we need to use the concept of complex ion formation and equilibrium.

Step 1: Write the reaction equation The reaction of Ag+ with NH3 to form the complex ion Ag(NH3)2+ is as follows: Ag+ + 2NH3 ⇌ Ag(NH3)2+

Step 2: Write the expression for the formation constant (Kf) The formation constant (Kf) for this reaction is given by: Kf = [Ag(NH3)2+] / ([Ag+][NH3]^2)

Step 3: Set up the ICE table The ICE table for this reaction would look like this:

..........| Ag+ + 2NH3 | ⇌ | Ag(NH3)2+ Initial | 0.010 M | 1.00 M | 0 Change | -x | -2x | +x Equilibrium | 0.010-x | 1.00-2x | x

Step 4: Substitute the equilibrium concentrations into the Kf expression Substituting the equilibrium concentrations into the Kf expression gives: 1.7 x 10^7 = x / ((0.010-x)(1.00-2x)^2)

Step 5: Solve for x This is a quadratic equation in x, and solving it gives the equilibrium concentration of Ag+. However, because Kf is very large, we can make the assumption that x is very small compared to 0.010 and 1.00. This simplifies the equation to: 1.7 x 10^7 = x / ((0.010)(1.00)^2) Solving this equation gives x = 1.7 x 10^-9 M.

So, the concentration of Ag+ (aq) ion in 0.010 M AgNO3 that is also 1.00 M NH3 is approximately 1.7 x 10^-9 M.