Question

Equilibrium 133

135. The number of moles of NH3, that must be added
to 2 L of 0.80 M AgNO3 in order to reduce the
concentration of Ag+ ions to 5.0 × 10-8 M (Kformation
for [Ag(NH3)₂] = 1.0 x108) is
(Nearest integer)

Answer 1

Approximately 2.89 × 10^-3 moles of NH3 must be added to the 2 L of 0.80 M AgNO3 solution in order to reduce the concentration of Ag+ ions to 5.0 × 10^-8 M.

Step-by-step explanation:

To solve this problem, we can use the equation given: Ag+ (aq) + 2NH3(aq) ⇒ Ag(NH3)2 + (aq), with the formation constant Kf = 1.0 × 10^8. We are trying to reduce the concentration of Ag+ ions in a 2 L solution of 0.80 M AgNO3 to 5.0 × 10^-8 M. Let's assume x moles of NH3 is added to the solution. Initially, the concentration of Ag+ and NH3 is 0 M, and the concentration of [Ag(NH3)2+] is also 0 M. Therefore, the change in concentration for [Ag+], [NH3], and [Ag(NH3)2+] is x, 2x, and x respectively.

Using the formation constant expression [Ag(NH3)2+] = Kf * [Ag+][NH3]^2, we can substitute the equilibrium concentrations into the expression to get:

1. 0.10 = (1.0 × 10^8)(x)(2x)^2
2. 0.10 = 4x^3(1.0 × 10^8)
3. x^3 = 0.10 / (4 * 1.0 × 10^8)
4. x = (0.10 / (4 * 1.0 × 10^8))^(1/3)
5. x ≈ 2.89 × 10^-3 M

Therefore, approximately 2.89 × 10^-3 moles of NH3 must be added to the 2 L of 0.80 M AgNO3 solution in order to reduce the concentration of Ag+ ions to 5.0 × 10^-8 M.