The p H o f 0.010 M H 2 C O 3 ( a q ) has to be calculated.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 6, Problem 6E.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.61×10−4)=3.80

Therefore, pH of 0.20 M H2S(aq) is 3.80.

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

For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. ? NH2 MgBr Will the first product that forms in this reaction create a new CC bond? ○ Yes ○ No MgBr ? Will the first product that forms in this reaction create a new CC bond? O Yes O No Click and drag to start drawing a structure. :☐ G x c olo Ar HE

Predicting As the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C - C bond as its major product: H₂N O H 1. ? 2. H3O+ If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. 0 If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. فا Explanation Check Click and drag to start drawing a structure.

Highlight the chirality (or stereogenic) center(s) in the given compound. A compound may have one or more stereogenic centers. OH OH OH OH OH OH

Answer 2

Question

Chapter 6, Problem 6E.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.61×10−4)=3.80

Therefore, pH of 0.20 M H2S(aq) is 3.80.

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.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.61×10−4)=3.80

Therefore, pH of 0.20 M H2S(aq) is 3.80.

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.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.61×10−4)=3.80

Therefore, pH of 0.20 M H2S(aq) is 3.80.

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.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.61×10−4)=3.80

Therefore, pH of 0.20 M H2S(aq) is 3.80.

Answer 6

Chapter 6, Problem 6E.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

Answer 7

Chapter 6, Problem 6E.3E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.010 M H2CO3(aq) has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

H2CO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2CO3(aq)+H2O(l)⇌H3O+(aq)+HCO3−(aq), Ka1=4.3×10−7HCO3−(aq)+H2O(l)⇌H3O+(aq)+CO32−(aq), Ka2=5.6×10−11

The second deprotonation can be ignored.

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

Ka1=[H3O+][HCO3−][H2CO3]

An equilibrium table can be set up as given below.

Composition(M)H2CO3H2OH3O+HCO3−Initial concentration0.010−00Change in concentration−x−+x+xEquilibrium concentration0.010−x−xx

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

Ka1=(x)(x)(0.010−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.010−x)4.3×10−7=(x)(x)(0.010−x)(4.3×10−7)(0.010−x)=x2x2+4.3×10−7x−4.3×10−9=0x=(−4.3×10−7)±(4.3×10−7)2−(4)×(1)×(−4.3×10−9)2x=6.53×10−5[H3O+]=x=6.53×10-5

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(6.53×10−5)=4.18

Therefore, pH of 0.010 M H2CO3(aq) is 4.18.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M (COOH)2(aq) has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

(COOH)2 is a polyprotic acid. The first and second deprotonation reactions are given below.

CHEMICAL PRINCIPLES (LL) W/ACCESS, Chapter 6, Problem 6E.3E

The value of Ka1 is 5.9×10−2 and the value of Ka2 is 6.5×10−5.

The second deprotonation can be ignored.

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

Ka1=[H3O+][COOH−COO−][COOH−COOH]

An equilibrium table can be set up as given below.

Composition(M)(COOH)2H2OH3O+(COOH−COO−)Initial concentration0.10−00Change in concentration−x−+x+xEquilibrium concentration0.10−x−xx

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

Ka1=(x)(x)(0.10−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.10−x)5.9×10−2=(x)(x)(0.10−x)(5.9×10−2)(0.10−x)=x2x2+5.9×10−2x−5.9×10−3=0x=(−5.9×10−2)±(5.9×10−2)2−(4)×(1)×(−5.9×10−3)2x=0.052[H3O+]=x=0.052

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.052)=1.28

Therefore, pH of 0.10 M (COOH)2(aq) is 1.28.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.61×10−4)=3.80

Therefore, pH of 0.20 M H2S(aq) is 3.80.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The pH of 0.20 M H2S(aq) has to be calculated.

Answer 20

(c)

Expert Solution

Explanation of Solution

H2S is a polyprotic acid. The first and second deprotonation reactions are given below.

H2S(aq)+H2O(l)→H3O+(aq)+HS−(aq),Ka1=1.3×10−7HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq), Ka2=7.1×10−15

The second deprotonation can be ignored.

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

Ka1=[H3O+][HS−][H2S]

An equilibrium table can be set up as given below.

Composition(M)H2SH2OH3O+HS−Initial concentration0.20−00Change in concentration−x−+x+xEquilibrium concentration0.20−x−xx

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

Ka1=(x)(x)(0.20−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.20−x)1.3×10−7=(x)(x)(0.20−x)(1.3×10−7)(0.20−x)=x2x2+1.3×10−7x−2.6×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−2.6×10−8)2x=1.61×10−4[H3O+]=x=1.61×10-4