The p H o f 1.0 × 10 − 4 M H 3 B O 3 ( a q ) has to be calculated.

Question

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Answer 1

Question

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.032)=1.49

Therefore, pH of 0.10 M H2SO3(aq) is 1.49.

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Answer 2

Question

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.032)=1.49

Therefore, pH of 0.10 M H2SO3(aq) is 1.49.

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Answer 3

Question

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.032)=1.49

Therefore, pH of 0.10 M H2SO3(aq) is 1.49.

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Answer 4

Question

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.032)=1.49

Therefore, pH of 0.10 M H2SO3(aq) is 1.49.

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Answer 5

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.032)=1.49

Therefore, pH of 0.10 M H2SO3(aq) is 1.49.

Answer 6

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

Answer 7

Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.0×10−4 M H3BO3(aq) has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

H3BO3 is a monoprotic acid. The deprotonation reaction is given below.

H3BO3(aq)+H2O(l)⇌H3O+(aq)+H2BO3−(aq), Ka=7.3×10−10

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

Ka=[H3O+][H2BO3−][H3BO3]

An equilibrium table can be set up as given below.

Composition(M)H3BO3H2OH3O+H2BO3−Initial concentration1.0×10−4−00Change in concentration−x−+x+xEquilibrium concentration1.0×10−4−x−xx

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

Ka=(x)(x)(1.0×10−4−x)

Now, the above expression can be solved for x.

Ka=(x)(x)(1.0×10−4−x)7.3×10−10=(x)(x)(1.0×10−4−x)(7.3×10−10)(1.0×10−4−x)=x2x2+7.3×10−10x−7.3×10−14=0x=(−7.3×10−10)±(7.3×10−10)2−(4)×(1)×(−7.3×10−14)2x=2.7×10−7[H3O+]=x=2.7×10-7

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(2.7×10−7)=6.56

Therefore, pH of 1.0×10−4 M H3BO3(aq) is 6.56.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of 0.015 M H3PO4(aq) has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

H3PO4 is a polyprotic acid. The first and second deprotonation reactions are given below.

H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq), Ka1=7.6×10−3H2PO4−(aq)+H2O(l)⇌H3O+(aq)+HPO42−(aq), Ka2=6.2×10−8

The second deprotonation can be ignored.

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

Ka1=[H3O+][H2PO4−][H3PO4]

An equilibrium table can be set up as given below.

Composition(M)H3PO4H2OH3O+H2PO4−Initial concentration0.015−00Change in concentration−x−+x+xEquilibrium concentration0.015−x−xx

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

Ka1=(x)(x)(0.015−x)

Now, the above expression can be solved for x.

Ka1=(x)(x)(0.015−x)7.6×10−3=(x)(x)(0.015−x)(7.6×10−3)(0.015−x)=x2x2+7.6×10−3x−1.14×10−4=0x=(−7.6×10−3)±(7.6×10−3)2−(4)×(1)×(−1.14×10−4)2x=7.53×10−3[H3O+]=x=7.53×10-3

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(7.53×10-3)=2.12

Therefore, pH of 0.015 M H3PO4(aq) is 2.12.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032

The pH of the solution can now be calculated.

pH=−log[H3O+]=−log(0.032)=1.49

Therefore, pH of 0.10 M H2SO3(aq) is 1.49.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M H2SO3(aq) has to be calculated.

Answer 20

(c)

Expert Solution

Explanation of Solution

H2SO3 is a polyprotic acid. The first and second deprotonation reactions are given below.

H2SO3(aq)+H2O(l)⇌H3O+(aq)+HSO3−(aq), Ka1=1.5×10−2HSO3−(aq)+H2O(l)⇌H3O+(aq)+SO32−(aq), Ka2=1.2×10−7

The second deprotonation can be ignored.

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

Ka1=[H3O+][HSO3−][H2SO3]

An equilibrium table can be set up as given below.

Composition(M)H2SO3H2OH3O+HSO3−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)1.5×10−2=(x)(x)(0.10−x)(1.5×10−2)(0.10−x)=x2x2+1.5×10−2x−1.5×10−3=0x=(−1.5×10−2)±(1.5×10−2)2−(4)×(1)×(−1.5×10−3)2x=0.032[H3O+]=x=0.032