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Chemistry: An Atoms First Approach

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

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 88AE
Textbook Problem
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Will a precipitate of Cd(OH)2 form if 1.0 mL of 1.0 M Cd(NO3)2 is added to 1.0 L of 5.0 M NH3?

Cd 2+ ( a q ) + 4 N H 3 ( a q ) C d ( N H 3 ) 4 2 + ( a q ) K = 1.0 × 10 7 Cd(OH) 2 ( s ) C d 2 + ( a q ) + 2 O H ( a q ) K sp = 5.9 × 10 15

Interpretation Introduction

Interpretation:

The reaction corresponding to the addition of 1.0mL 1.0M Cd(NO3)2 to 1.0L of 5.0M NH3. It is to be calculated if the formation of a precipitate of Cd(OH)2 takes place or not.

Concept introduction:

At equilibrium, the equilibrium constant expression is expressed by the formula,

  K=ConcentrationofproductsConcentrationofreactants

The number of moles is calculated by the formula,

  Numberofmoles=Volume(L)×Molarity

Explanation of Solution

To determine: If the formation of a precipitate of Cd(OH)2 takes place in the given reaction.

The number of moles of Cd(NO3)2 is 1×10-3mol_ and that of NH3 is 5.0mol_.

Given

The volume of Cd(NO3)2 is 1.0mL(0.001L).

The molarity of Cd(NO3)2 is 1.0M.

The volume of NH3 is 1.0L.

The molarity of NH3 is 5.0M.

The number of moles is calculated by the formula,

  Numberofmoles=Volume(L)×Molarity        (1)

Substitute the value of volume and molarity of Cd(NO3)2 in the equation (1).

  Numberofmoles=(0.001L)×(1.0M)=1×10-3mol_

Substitute the value of volume and molarity of NH3 in the equation (1).

  Numberofmoles=(1L)×(5.0M)=5.0mol_

The [Cd2+] is 1×10-3mol/L_ and that of NH3 is 5.0mol/L_.

The number of moles of Cd(NO3)2 is 1×103mol.

The number of moles of NH3 is 5.0mol.

The total volume of the solution =(0.001+1.0)L=1.001L

The concentration is calculated by the formula,

  Concentration=NumberofmolesVolumeofsolution(L)        (2)

Substitute the value, the number of moles [Cd2+] and volume of the solution in equation (2).

  Concentration=0.001mol1.001L=1×10-3mol/L_

Substitute the value, the number of moles NH3 and volume of the solution in equation (2).

  Concentration=5.0mol1.0L=5mol/L_

The [Cd2+] available in the solution is 1.6×10-13mol/L_

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Chapter 15 Solutions

Chemistry: An Atoms First Approach
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