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

Question

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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.

EBK CHEMICAL PRINCIPLES, 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.

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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.

EBK CHEMICAL PRINCIPLES, 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.

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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.

EBK CHEMICAL PRINCIPLES, 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.

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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.

EBK CHEMICAL PRINCIPLES, 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.

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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.

EBK CHEMICAL PRINCIPLES, 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.

EBK CHEMICAL PRINCIPLES, 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.

EBK CHEMICAL PRINCIPLES, 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