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

Question

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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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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