   # Calculate the solubility of AgCN( s ) ( K sp = 2.2 × 10 −12 ) in a solution containing 1.0 M H + . ( K a for HCN is 6.2 × 10 −10 .) ### 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 103CP
Textbook Problem
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## Calculate the solubility of AgCN(s) (Ksp = 2.2 × 10−12) in a solution containing 1.0 M H+. (Ka for HCN is 6.2 × 10−10.)

Interpretation Introduction

Interpretation: The solubility of AgCN in a solution containing 1MH+ is to be calculated.

Concept introduction: The solubility product is the mathematical product of a substance’s dissolved ion concentration raised to its power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water.

### Explanation of Solution

Explanation

To determine: The solubility of AgCN in 1MH+ .

The value of equilibrium constant K1 is 3.5×10-3_ .

The solubility equilibrium constant of AgCN is as follows:

AgCN(s)Ag+(aq)+CN(aq)Ksp=2.2×1012 (1)

The dissociation of HCN in its aqueous solution is as follows:

HCN(aq)H+(aq)+CN(aq)Ka=6.2×1010 (2)

The equation (2) is reversed as follows.

H+(aq)+CN(aq)HCN(aq)K=1Ka (3)

The equilibrium constant (K) for equation (3) is calculated as,

K=16.2×1010

Equation (1) and equation(3) are added to get the equation required to calculate the solubility of AgCN in 1MH+ . The resultant equation is,

AgCN(s)+H+(aq)Ag+(aq)+HCN(aq)K1=Ksp×K (4)

The equilibrium constant K1 is calculated as,

K1=Ksp×K

Substitute the values of K and Ksp in the above equation.

K1=2.2×1012×16.2×1010K1=3.5×103

The value of equilibrium constant K1 is 3.5×10-3_

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