The p H o f 0.10 M H 2 S ( a q ) has to be calculated.

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Answer 1

Question

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(4.79×10-6)=5.32

Therefore, pH of 1.1×10−3 M H2TeO4(aq) is 5.32.

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C 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00

Curved arrows were used to generate the significant resonance structure and labeled the most significant contribute. What are the errors in these resonance mechanisms. Draw out the correct resonance mechanisms with an brief explanation.

Answer 2

Question

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(4.79×10-6)=5.32

Therefore, pH of 1.1×10−3 M H2TeO4(aq) is 5.32.

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Answer 3

Question

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(4.79×10-6)=5.32

Therefore, pH of 1.1×10−3 M H2TeO4(aq) is 5.32.

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Answer 4

Question

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(4.79×10-6)=5.32

Therefore, pH of 1.1×10−3 M H2TeO4(aq) is 5.32.

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Answer 5

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(4.79×10-6)=5.32

Therefore, pH of 1.1×10−3 M H2TeO4(aq) is 5.32.

Answer 6

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

Answer 7

Chapter 6, Problem 6E.4E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2S(aq) has to be calculated.

Answer 8

(a)

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 concentration0.10−00Change 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)1.3×10−7=(x)(x)(0.10−x)(1.3×10−7)(0.10−x)=x2x2+1.3×10−7x−1.3×10−8=0x=(−1.3×10−7)±(1.3×10−7)2−(4)×(1)×(−1.3×10−8)2x=1.14×10−4[H3O+]=x=1.14×10-4

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(1.14×10−4)=3.94

Therefore, pH of 0.10 M H2S(aq) is 3.94.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of 0.15 M H2C4H4O6(aq) has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

H2C4H4O6 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2C4H4O6(aq)+H2O(l)⇌H3O+(aq)+HC4H4O6−(aq), Ka1=6.0×10−4HC4H4O6−(aq)+H2O(l)⇌H3O+(aq)+C4H4O62−(aq), Ka2=1.5×10−5

The second deprotonation can be ignored.

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

Ka1=[H3O+][HC4H4O6−][H2C4H4O6]

An equilibrium table can be set up as given below.

Composition(M)H2C4H4O6H2OH3O+HC4H4O6−Initial concentration0.15−00Change in concentration−x−+x+xEquilibrium concentration0.15−x−xx

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

Ka1=(x)(x)(0.15−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.15−x)6.0×10−4=(x)(x)(0.15−x)(6.0×10−4)(0.15−x)=x2x2+6.0×10−4x−9×10−5=0x=(−6.0×10−4)±(6.0×10−4)2−(4)×(1)×(−9×10−5)2x=9.2×10−3[H3O+]=x=9.2×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(9.2×10−3)=2.03

Therefore, pH of 0.15 M H2C4H4O6(aq) is 2.03.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(4.79×10-6)=5.32

Therefore, pH of 1.1×10−3 M H2TeO4(aq) is 5.32.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The pH of 1.1×10−3 M H2TeO4(aq) has to be calculated.

Answer 20

(c)

Expert Solution

Explanation of Solution

H2TeO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2TeO4(aq)+H2O(l)⇌H3O+(aq)+HTeO4−(aq), Ka1=2.1×10−8HTeO4−(aq)+H2O(l)⇌H3O+(aq)+TeO42−(aq), Ka2=6.5×10−12

The second deprotonation can be ignored.

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

Ka1=[H3O+][HTeO4−][H2TeO4]

An equilibrium table can be set up as given below.

Composition(M)H2TeO4H2OH3O+HTeO4−Initial concentration1.1×10−3−00Change in concentration−x−+x+xEquilibrium concentration1.1×10−3−x−xx

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

Ka1=(x)(x)(1.1×10−3−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(1.1×10−3−x)2.1×10−8=(x)(x)(1.1×10−3−x)(2.1×10−8)(1.1×10−3−x)=x2x2+2.1×10−8x−2.31×10−11=0x=(−2.1×10−8)±(2.1×10−8)2−(4)×(1)×(−2.31×10−11)2x=4.79×10−6[H3O+]=x=4.79×10-6