General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN: 9781305580343
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: Cengage Learning
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Chapter 17, Problem 17.122QP
Interpretation Introduction

Interpretation:

The quantity of water and ammonia required to dissolve silver chloride has to be determined.

Concept introduction:

  • The solubility product constant ( Ksp ) is defined as the equilibrium between compound and its ions in an aqueous solution.
  • Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
  • Ionic compound A3B Ksp= [A]3[B] .
  • Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

AB

Rate of forward reaction = Rate of reverse reactionkf[A]=kr[B]

On rearranging,

[A][B]=kfkr=Kc

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kc is the equilibrium constant.

Expert Solution & Answer
Check Mark

Answer to Problem 17.122QP

The quantity of water required to dissolve silver chloride is 5.2×103L

The quantity of 0.15M ammonia required to dissolve silver chloride is 9.4L

Explanation of Solution

To calculate: The quantity of water and ammonia required to dissolve silver chloride.

The number of moles of silver chloride is,

mol AgCl =10.0 g AgCl×1 mol AgCl143.3 g AgCl=0.06978 mol AgClmol AgCl = mol Ag+ (In water) = mol Ag(NH3)2+ (In NH3)The volume required to dissolve AgClis calculated using number of moles of silver chloride and concentration of silver ion.Ksp = 1.8 ×10-10=[Ag+][Cl-] = x2[Ag+]=x=1.8 ×10-10=1.34×10-5MVolume of water=0.06978 mol Ag+1.34×10-5mol Ag+/L=5.2×103L

Silver chloride reacts with ammonia to form silver-ammonia complex.  By combining the solubility equilibrium with complexation equilibrium to determine the total reaction.

When a chemical equation can be derived by taking sum of other equations, then the equilibrium constant of derived equation equals the product of the equilibrium constants of added equations.

AgCl(s)Ag+(aq)+Cl-(aq)Ksp = 1.8×10-10Ag+(aq) + 2NH3(aq)Ag(NH3)2+(aq)Kf = 1.7×107AgCl(s)+2NH3(aq)Ag(NH3)2+(aq)+Cl-(aq)Kc = Ksp×Kf=3.06×10-3

AgCl(s)+2NH3(aq)Ag(NH3)2+(aq)+Cl-(aq)Initial (M):                  0.15          00Change (M):             -2x                +x+xEquilibrium (M):        0.15 - 2x      xxKc =[Ag(NH3)2+][Cl-][NH3]2=(x)2(0.15 - 2x)2=3.06×10-3Takesquarerootonbothsidesweget,x(0.15 - 2x)=5.53×10-2x=(0.15 - 2x)×5.53×10-2=0.00829 - 0.1106xx=0.00746=[Ag(NH3)2+]The volume of ammonia required to dissolve AgClis calculated using number of moles of silver complex and concentration of silver complex.Volume of 0.15M ammonia=0.06978 mol Ag(NH3)2+0.00746 mol Ag(NH3)2+/L=9.4L

Due to complex formation and stability of complex Ag(NH3)2+ ion, the volume required to dissolve decreases from 5000 L to under 10 L

Conclusion

By using the solubility product expression, the volume of water and ammonia required to dissolve silver chloride were determined.

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Chapter 17 Solutions

General Chemistry - Standalone book (MindTap Course List)

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