Consider a solution made by mixing 500.0 mL of 4.0 M NH 3 and 500.0 mL of 0.40 M AgNO 3 .Ag + reacts with NH 3 to form AgNH 3 + and Ag(NH 3 ) 2 + : Ag + + NH 3 AgNH 3 + K 1 = 2.1 $\times$ 10 3 AgNH 3 + + NH 3 Ag(NH 3 ) 2 + K 2 = 8.2 $\times$10 3 -The concentration of Ag(NH₃)₂⁺ at equilibrium is: A)2.0 M B)0.40 M C)0.20 M D)1.0 $\times$ 10^-³ M E)none of these

Question

Quizplus · Accepted Answer

This is a simple equilibrium problem. The first step in solving an equilibrium problem is to write the balanced chemical equation: Ag⁺ + NH₃ ⇌ AgNH₃⁺ AgNH₃⁺ + NH₃ ⇌ Ag(NH₃)₂⁺ From the equation, we can see that AgNH₃⁺ is produced from Ag⁺ and NH₃ at a rate proportional to the concentrations of Ag⁺ and NH₃: rate1 = k1[Ag⁺][NH₃] Similarly, Ag(NH₃)₂⁺ is produced from AgNH₃⁺ and NH₃ at a rate proportional to the concentrations of AgNH₃⁺ and NH₃: rate2 = k2[AgNH₃⁺][NH₃] The concentrations of Ag⁺ and AgNH₃⁺ are given, so we can calculate their initial concentrations: [Ag⁺] = 0.40 M [AgNH₃⁺] = 0 M (since it is being produced) The initial concentration of NH₃ is also given, so we can calculate its initial concentration: [NH₃] = 4.0 M At equilibrium, the rates of the forward and reverse reactions are equal: rate1 = rate2 k1[Ag⁺][NH₃] = k2[AgNH₃⁺][NH₃] We are given the values of k1 and k2. Plugging in the values and solving for [Ag(NH₃)₂⁺] gives: [Ag(NH₃)₂⁺] = k1[Ag⁺][NH₃]/k2 [Ag(NH₃)₂⁺] = (2.1 × 10-10) (0.40 M) (4.0 M) / (8.2 × 10-10) [Ag(NH₃)₂⁺] ≈ 0.20 M Therefore, the concentration of Ag(NH₃)₂⁺ at equilibrium is 0.20 M, so the answer is C.

Consider a Solution Made by Mixing 500 ×\times× 103 AgNH3+ + NH3 Ag(NH3)2+ K2 =

Consider a Solution Made by Mixing 500 $\times$ 10³
AgNH₃⁺ + NH₃
Ag(NH₃)₂⁺
K₂ =