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: The Quest for Insight, 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

Calculate the pH, pOH, and fraction of solute protonated or deprotonated in the following aqueous solutions: (a) 0.120 M CH;CH(OH)COOH(aq) (lactic acid), (b) 1.4× 10“M CH;CH(OH) COOH(aq), (c) 0.15 M NH,CI(aq), (d) 0.15 m NaCH;CO,(aq), and (e) 0.112 m (CH;),N(aq) (trimethylamine).

Use the values for the successive acidity constants for oxalic acid in the given table to calculate the molar concentrations of (COOH)2 , HOOCCO2-, (CO2)22-, H3O+ , and OH- in 0. 15 M (COOH)2(aq).

6D.17 Calculate the pH of (a) 0.63 M NaCH3 CO₂ (aq); (b) 0.65 M KCN(aq).

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: The Quest for Insight, 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: The Quest for Insight, 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: The Quest for Insight, 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: The Quest for Insight, 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: The Quest for Insight, 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: The Quest for Insight, 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