General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 23, Problem 23.72P

(a)

Interpretation Introduction

Interpretation:

The solubility-product constant of AgCl(s) has to be calculated.

Concept Introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG.

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Equilibrium constant of a reaction can be determined from the value of ΔGrxno using the given formula,

  ΔGrxno=RTln K

Solubility product constant Ksp refers to the level in which the solute gets dissolved in a solution.  It explains the equilibrium of a solid with its corresponding ions in a solution.

It is the product of concentration of ions that are dissolved raised to power of their corresponding coefficient of stoichiometry.

  AcBdcA++dBKsp=[A+]c[B]d

(a)

Expert Solution
Check Mark

Answer to Problem 23.72P

Value of solubility-product constant is 2×1010.

Explanation of Solution

The given reaction is shown below,

  AgCl(s)Ag+(aq)+ Cl(aq)

  • Calculate the value of ΔGrxno:

Standard free energy of formation values is given below,

  ΔG°[AgCl(s)]=109.8 kJ/molΔG°[Ag+(aq)]=77.1 kJ/molΔG°[Cl(aq)]=131.2 kJ/mol

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Substitute the values as follows,

  ΔGrxno[(1)(77.1 kJ/mol)+(1)(131.2 kJ/mol)][(1)(109.8 kJ/mol)]=55.7 kJ/mol

Therefore, the value of ΔGrxno is 55.7 kJ/mol.

  • Calculate the value of the solubility-product constant:

For the above reaction, equilibrium constant can be written as below,

  K = [Ag+(aq)][Cl(aq)][AgCl(s)]

However, AgCl is a solid and therefore,

  K = [Ag+(aq)][Cl(aq)] = Ksp

Value of the solubility-product constant is determined as follows,

  T = 25°C(25+273) = 298 KΔGrxno=RTln K55.7 kJ/mol =(8.314×103 kJ.K1.mol1)(298 K)ln Kspln Ksp=55.7 kJ/mol(8.314×103 kJ.K1.mol1)(298 K) =22.482Ksp = 2×1010

Therefore, the value of the solubility-product constant is 2×1010.

(b)

Interpretation Introduction

Interpretation:

The solubility-product constant of AgBr(s) has to be calculated.

Concept Introduction:

Refer to (a).

(b)

Expert Solution
Check Mark

Answer to Problem 23.72P

Value of solubility-product constant is 5.4×1013.

Explanation of Solution

The given reaction is shown below,

  AgBr(s)Ag+(aq)+ Br(aq)

  • Calculate the value of ΔGrxno:

Standard free energy of formation values is given below,

  ΔG°[AgBr(s)]=96.9 kJ/molΔG°[Ag+(aq)]=77.1 kJ/molΔG°[Br(aq)]=104.0 kJ/mol

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Substitute the values as follows,

  ΔGrxno[(1)(77.1 kJ/mol)+(1)(104.0 kJ/mol)][(1)(96.9 kJ/mol)]=70 kJ/mol

Therefore, the value of ΔGrxno is 70 kJ/mol.

  • Calculate the value of the solubility-product constant:

For the above reaction, equilibrium constant can be written as below,

  K = [Ag+(aq)][Br(aq)][AgBr(s)]

However, AgCl is a solid and therefore,

  K = [Ag+(aq)][Br(aq)] = Ksp

Value of the solubility-product constant is determined as follows,

  T = 25°C(25+273) = 298 KΔGrxno=RTln K70 kJ/mol =(8.314×103 kJ.K1.mol1)(298 K)ln Kspln Ksp=70 kJ/mol(8.314×103 kJ.K1.mol1)(298 K) =28.25347Ksp = 5.4×1013

Therefore, the value of the solubility-product constant is 5.4×10-13.

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Chapter 23 Solutions

General Chemistry

Ch. 23 - Prob. 23.11PCh. 23 - Prob. 23.12PCh. 23 - Prob. 23.13PCh. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44PCh. 23 - Prob. 23.45PCh. 23 - Prob. 23.46PCh. 23 - Prob. 23.47PCh. 23 - Prob. 23.48PCh. 23 - Prob. 23.49PCh. 23 - Prob. 23.50PCh. 23 - Prob. 23.51PCh. 23 - Prob. 23.52PCh. 23 - Prob. 23.53PCh. 23 - Prob. 23.54PCh. 23 - Prob. 23.55PCh. 23 - Prob. 23.56PCh. 23 - Prob. 23.57PCh. 23 - Prob. 23.58PCh. 23 - Prob. 23.59PCh. 23 - Prob. 23.60PCh. 23 - Prob. 23.61PCh. 23 - Prob. 23.62PCh. 23 - Prob. 23.63PCh. 23 - Prob. 23.64PCh. 23 - Prob. 23.65PCh. 23 - Prob. 23.66PCh. 23 - Prob. 23.67PCh. 23 - Prob. 23.68PCh. 23 - Prob. 23.69PCh. 23 - Prob. 23.70PCh. 23 - Prob. 23.71PCh. 23 - Prob. 23.72PCh. 23 - Prob. 23.73PCh. 23 - Prob. 23.74PCh. 23 - Prob. 23.75PCh. 23 - Prob. 23.76PCh. 23 - Prob. 23.77PCh. 23 - Prob. 23.78PCh. 23 - Prob. 23.79PCh. 23 - Prob. 23.80PCh. 23 - Prob. 23.81PCh. 23 - Prob. 23.82PCh. 23 - Prob. 23.83PCh. 23 - Prob. 23.84PCh. 23 - Prob. 23.85PCh. 23 - Prob. 23.86PCh. 23 - Prob. 23.87PCh. 23 - Prob. 23.88P
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