The molar concentrations of H 2 S O 3 , H S O 3 − , S O 3 2 − , H 3 O + , a n d O H − in 0.170 M N a 2 S O 3 ( a q ) have to be calculated.

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Answer 1

Question

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

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Answer 2

Question

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

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Answer 3

Question

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

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Answer 4

Question

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

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Answer 5

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

Answer 6

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

Answer 7

Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Answer 8

Expert Solution & Answer

Explanation of Solution

The first protonation equilibrium is given below,

SO32−(aq)+H2O(l)⇌OH−(aq)+HSO3−(aq)

The expression for equilibrium constant for the first protonation reaction can be written as shown below,

Kb1=[OH−][HSO3−][SO32−]=KwKa2=1.0×10−141.2×10−7=8.3×10-8

An equilibrium table can be set up as given below.

Composition(M)SO32−H2OOH−HSO3−Initial concentration0.170−00Change in concentration−x−+x+xEquilibrium concentration0.170−x−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb1=(x)(x)(0.170−x)

Now, the above expression can be solved for x.

Kb1=(x)(x)(0.170−x)8.3×10−8=(x)(x)(0.170−x)(8.3×10−8)(0.170)=x2[Assuming x≪0.170]x=1.2×10-4[OH−]=x=1.2×10-4 M[HSO3−]=x=1.2×10-4 M[SO32−]=0.170−x≈0.170 M

The second protonation reaction is given below,

HSO3−(aq)+H2O(l)⇌OH−(aq)+H2SO3(aq)

The expression for equilibrium constant for the second protonation reaction can be written as shown below,

Kb2=[OH−][HSO3][HSO3−]=KwKa1=1.0×10−141.5×10−2=6.7×10-13

An equilibrium table can be set up as given below.

Composition(M)HSO3−H2OOH−H2SO3Initial concentration1.2×10−4−1.2×10−40Change in concentration−x−+x+xEquilibrium concentration1.2×10−4−x−1.2×10−4+xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)

Now, the above expression can be solved for x.

Kb2=(x)(1.2×10−4+x)(1.2×10−4−x)6.7×10−13=(1.2×10−4)(x)(1.2×10−4)[Assuming x≪1.2×10−4]x=6.7×10-13[OH−]=1.2×10−4+x≈1.2×10-4M[H2SO3]=x=6.7×10-13 M

The concentration of H3O+ can be calculated as given below.

[H3O+]=Kw[OH−]=1.0×10−141.2×10−4=8.3×10-11M

Therefore, the molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) can be summarized as follows, in order of decreasing concentration:

Species:SO32−OH−HSO3−H3O+H2SO3Concentration(M)0.1701.2×10−41.2×10−48.3×10−116.7×10−13

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

The molar concentrations of H 2 S O 3 , H S O 3 − , S O 3 2 − , H 3 O + , a n d O H − in 0.170 M N a 2 S O 3 ( a q ) have to be calculated.

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Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Explanation of Solution

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Chapter 6 Solutions

The molar concentrations of H 2 S O 3 , H S O 3 − , S O 3 2 − , H 3 O + , a n d O H − in 0.170 M N a 2 S O 3 ( a q ) have to be calculated.

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Videos

Interpretation: The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.

Explanation of Solution

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Chapter 6 Solutions

Interpretation:

The molar concentrations of H2SO3, HSO3−, SO32−, H3O+, and OH− in 0.170M Na2SO3(aq) have to be calculated.