Calculate the molar concentration of uncomplexed Zn^2+ in a solution that contains .20 mole of Zn(NH3)4^2+ per liter.....?

2016-07-20 4:28 pm
And .0116 M NH3 at equilibrium? The overall Kf for Zn(NH3)4^2+ is 3.8 X 10^9.

How do I do this problem?


A) 2.9 X 10^-3
B) 8.8 X 10^-3
C) 6.7 X 10^-4
D) 2.0 X 10^-13

回答 (1)

2016-07-20 6:38 pm
Zn²⁺(aq) + 4NH₃(aq) ⇌ Zn(NH₃)₄²⁺(aq) .... Kf = 3.8 × 10⁹

Kf = 3.8 × 10⁹
[Zn(NH₃)₄²⁺] / {[Zn²⁺] [NH₃]⁴} = 3.8 × 10⁹
(0.20) / {[Zn²⁺] (0.0116)⁴} = 3.8 × 10⁹
[Zn²⁺] = (0.20) / {0.0116⁴ × (3.8 × 10⁹)} M = 2.9 × 10⁻³ M

...... The answer is: A) 2.9 × 10⁻³


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