What is the concentration of free Zn²⁺ if 0.020 M Zn²+ solution is mixed with an equal volume of 2.0 M NH₃? Kf for [Zn(NH₃)₄]²⁺ is 4.1 × 10⁸.

Question

Quizplus · Accepted Answer

First, we need to write the balanced equation for the reaction between ZnS and NHS:

ZnS + 2NHS → [Zn(NHS)2] + H2S

The equilibrium constant for this reaction can be written as:

Kf = [Zn(NHS)2][H2S] / [ZnS][NHS]^2

We are given that Kf = 4.1 × 10^-18 and [ZnS] = 0.020 M. We also know that the volume of the NHS solution added is equal to the volume of the ZnS solution, so the final concentration of NHS is 1.0 M.

Let x be the concentration of free ZnS in the solution after the reaction reaches equilibrium.

At equilibrium, the concentration of [Zn(NHS)2] and [H2S] are both equal to x, while the concentration of [NHS] is 1.0 - x.

Substituting these values into the equilibrium constant expression and solving for x, we get:

4.1 × 10^-18 = x^2 / (0.020 - x)^2

x = 2.88 × 10^-11 M

Therefore, the concentration of free ZnS in the solution is 2.88 × 10^-11 M. The closest choice is B, which is the correct answer.

What Is the Concentration of Free Zn2+ If 0

What Is the Concentration of Free Zn²⁺ If 0