Question

Calculate the equilibrium concentration of Zn²+ in a solution initially with 0.150 M Zn²+ and 2.50 M CN⁻.

a) 0.10 M
b) 0.12 M
c) 0.15 M
d) 0.20 M

Answer 1

The equilibrium concentration of Zn²+ in the solution is b) 0.12 M

Step-by-step explanation

To calculate this, we use the formation constant (Kf) of the complex ion formed between Zn²+ and CN⁻. The balanced chemical equation is Zn²+ + 4CN⁻ ⇌ Zn(CN)₄²⁻.

The formation constant expression is Kf = [Zn(CN)₄²⁻] / [Zn²+][CN⁻]⁴. At equilibrium, assuming complete dissociation, the change in concentration of Zn²+ is x, and for CN⁻, it's 4x. Substituting these values into the expression, we get Kf = x / (0.150 - x)(2.50 - 4x)⁴.

Given Kf = 1.6 × 10²⁰, solving for x gives approximately 0.12 M as the equilibrium concentration of Zn²+, corresponding to option (b). This signifies a significant portion of Zn²+ ions forming the Zn(CN)₄²⁻ complex at equilibrium. OPTION B