5.0 mL of 12 M NH₃ is added to 500. mL of 0.050 M AgNO₃. What concentration of silver ion will exist after equilibrium is established [K_f for Ag(NH₃)₂⁺ is 1.5 * 10⁷.]

Question

Quizplus · Accepted Answer

First, we need to write the balanced equation for the reaction: 
AgNO3 + NHSO2C6H4CO2H -> Ag(NHSO2C6H4CO2H) + HNO3
Next, we set up an ICE table to determine the concentrations at equilibrium:

|         | AgNO3              | Ag(NHSO2C6H4CO2H) | 
|---------|--------------------|------------------|
| Initial | 0.050 M            | 0                  |
| Change  | -x                 | +x                 |
| Equil   | 0.050 M - x        | x                  |

Now we can set up the equilibrium expression and solve for x: 
Kf = [Ag(NHSO2C6H4CO2H)][HNO3] / [AgNO3] 
1.5 x 10^-17 = x^2 / (0.050 M - x) 
Since x is much smaller than 0.050 M, we can simplify the equation to: 
1.5 x 10^-17 = x^2 / 0.050 M 
x = 8.4 x 10^-7 M, which is the concentration of Ag+ ions at equilibrium. Therefore, the answer is B.

50 ML of 12 M NH3 Is Added to 500

50 ML of 12 M NH₃ Is Added to 500