   # The overall formation constant for HgI 4 2− is 1.0 × 10 30 . That is, 1.0 × 10 30 = [ H g I 4 2 − ] [ H g 2 + ] [ I − ] 4 What is the concentration of Hg 2+ in 500.0 mL of a solution that was originally 0.010 M Hg 2+ and 0.78 M I − ? The reaction is Hg 2+ ( a q ) + 4 I − ( a q ) ⇌ H g I 4 2 − ( a q ) ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 67E
Textbook Problem
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## The overall formation constant for HgI42− is 1.0 × 1030. That is, 1.0   ×   10 30   =   [ H g I 4 2 − ] [ H g 2 + ] [ I − ] 4 What is the concentration of Hg2+ in 500.0 mL of a solution that was originally 0.010 M Hg2+ and 0.78 M I−? The reaction is Hg 2+ ( a q )   +   4 I − ( a q )   ⇌   H g I 4 2 − ( a q )

Interpretation Introduction

Interpretation: The combination reaction of Hg2+ and I , the overall formation constant of HgI42 and concentration of Hg2+,I before the reaction is given. The concentration of Hg2+ in the given volume of a solution is to be calculated.

Concept introduction: At equilibrium, the equilibrium constant expression is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

### Explanation of Solution

To determine: The concentration of Hg2+ in 500mL of the given solution.

Given

Concentration of Hg2+ before reaction is 0.010M .

Concentration of I before reaction is 0.78M .

The overall formation constant for HgI42 is 1.0×1030 .

The reaction that takes place is,

Hg2+(aq)+4I(aq)HgI42(aq)

The reaction is assumed to be completed. The net reaction is,

Hg2++4IHgI42Beforereaction0.010M0.78M0Afterreaction00.78(4×0.010)=0

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