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.

Question

Want to see more full solutions like this?

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Use the literature Ka value of the acetic acid, and the data below to answer these questions. Note: You will not use the experimental titration graphs to answer the questions that follow. Group #1: Buffer pH = 4.35 Group #2: Buffer pH = 4.70 Group #3: Buffer pH = 5.00 Group #4: Buffer pH = 5.30 Use the Henderson-Hasselbalch equation, the buffer pH provided and the literature pKa value of acetic acid to perform the following: a) calculate the ratios of [acetate]/[acetic acid] for each of the 4 groups buffer solutions above. b) using the calculated ratios, which group solution will provide the best optimal buffer (Hint: what [acetate]/[acetic acid] ratio value is expected for an optimal buffer?) c) explain your choice

How would you prepare 1 liter of a 50 mM Phosphate buffer at pH 7.5 beginning with K3PO4 and 1 M HCl or 1 M NaOH? Please help and show calculations. Thank you

Draw the four most importantcontributing structures of the cation intermediate thatforms in the electrophilic chlorination of phenol,(C6H5OH) to form p-chlorophenol. Put a circle aroundthe best one. Can you please each step and also how you would approach a similar problem. Thank you!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Concept explainers

Videos

Interpretation:

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

Explanation of Solution

Want to see more full solutions like this?

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.

Concept explainers

Videos

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

Explanation of Solution

Want to see more full solutions like this?

Chapter 6 Solutions

Interpretation:

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