The p H o f 1.89 × 10 − 5 M a n d 9.64 × 10 − 7 M H C l O ( a q ) has to be calculated.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 6, Problem 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−3.892X−0.03=0X=1.96, so x=1.96×10−7[H3O+]=1.96×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.96×10−7)=6.70

Therefore, the pH of 9.64×10−7M HClO(aq) is 6.70.

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

None

Dr. Mendel asked his BIOL 260 class what their height was and what their parent's heights were. He plotted that data in the graph below to determine if height was a heritable trait. A. Is height a heritable trait? If yes, what is the heritability value? (2 pts) B. If the phenotypic variation is 30, what is the variation due to additive alleles? (2 pts) Offspring Height (Inches) 75 67.5 60 52.5 y = 0.9264x + 4.8519 55 60 65 MidParent Height (Inches) 70 75 12pt v V Paragraph B IUA > AT2 v V

Experiment: Each team will be provided with 5g of a mixture of acetanilide and salicylic acid. You will divide it into three 1.5 g portions in separate 125 mL Erlenmeyer flasks savıng some for melting point analysis. Dissolve the mixture in each flask in ~60mL of DI water by heating to boiling on a hotplate. Take the flasks off the hotplate once you have a clear solution and let them stand on the bench top for 5 mins and then allow them to cool as described below. Sample A-Let the first sample cool slowly to room temperature by letting it stand on your lab bench, with occasional stirring to promote crystallization. Sample B-Cool the second sample 1n a tap-water bath to 10-15 °C Sample C-Cool the third sample in an ice-bath to 0-2 °C Results: weight after recrystalization and melting point temp. A=0.624g,102-115° B=0.765g, 80-105° C=1.135g, 77-108 What is the percent yield of A,B, and C.

Answer 2

Question

Chapter 6, Problem 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−3.892X−0.03=0X=1.96, so x=1.96×10−7[H3O+]=1.96×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.96×10−7)=6.70

Therefore, the pH of 9.64×10−7M HClO(aq) is 6.70.

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 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−3.892X−0.03=0X=1.96, so x=1.96×10−7[H3O+]=1.96×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.96×10−7)=6.70

Therefore, the pH of 9.64×10−7M HClO(aq) is 6.70.

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 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−3.892X−0.03=0X=1.96, so x=1.96×10−7[H3O+]=1.96×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.96×10−7)=6.70

Therefore, the pH of 9.64×10−7M HClO(aq) is 6.70.

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 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−3.892X−0.03=0X=1.96, so x=1.96×10−7[H3O+]=1.96×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.96×10−7)=6.70

Therefore, the pH of 9.64×10−7M HClO(aq) is 6.70.

Answer 6

Chapter 6, Problem 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

Answer 7

Chapter 6, Problem 6F.10E

(a)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration1.89×10−500Change in concentration−x+x+xEquilibrium concentration1.89×10−5−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(1.89×10−5−x)x2=(3.0×10−8)(1.89×10−5) [Assuming x≪1.89×10−5]x=(3.0×10−8)(1.89×10−5)=7.53×10−7[H3O+]=7.53×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.53×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

HClO is a weak acid. Its pH can be calculated as shown below.

HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)Ka=[H3O+][ClO−][HClO]=3.0×10−8

An equilibrium table can be set up as given below.

Composition(M)HClOH3O+ClO−Initial concentration9.64×10−700Change in concentration−x+x+xEquilibrium concentration9.64×10−7−xxx

Now, the above equilibrium expression can be solved for x.

Ka=[H3O+][ClO−][HClO]=3.0×10−83.0×10−8=(x)(x)(9.64×10−7−x)x2=(3.0×10−8)(9.64×10−7) [Assuming x≪9.64×10−7]x=(3.0×10−8)(9.64×10−7)=1.7×10−7[H3O+]=1.7×10-7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.7×10−7)=6.77

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.77.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−3.892X−0.03=0X=1.96, so x=1.96×10−7[H3O+]=1.96×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(1.96×10−7)=6.70

Therefore, the pH of 9.64×10−7M HClO(aq) is 6.70.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of 1.89×10−5M and 9.64×10−7MHClO(aq) has to be calculated by taking into account the autoprotolysis of water.

Concept Introduction:

The contribution of autoprotolysis of water to the calculation of pH of very dilute solutions of weak acid must be taken into consideration if the hydronium ion concentration is less than 10−6M.

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

Answer 14

(b)

Expert Solution

Explanation of Solution

Calculation of pH of 1.89×10−5M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(1.89×10−5)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(5.77×10−13)x−(3.0×10−22)=0

Write x=X×10−7 and divide the resulting equation by 10−21.

X3+0.03X2−57.7X−0.03=0X=7.6, so x=7.6×10−7[H3O+]=7.6×10−7M

Now, the pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(7.6×10−7)=6.12

Therefore, the pH of 1.89×10−5M HClO(aq) is 6.12.

Calculation of pH of 9.64×10−7M HClO(aq):

For very dilute solutions of weak acid, the given below equation should be used.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0

By substituting x for [H3O+] in the above equation, the equation can be solved for x.

[H3O+]3+Ka[H3O+]2−(KW+Ka[HA]Initial)[H3O+]−KaKW=0x3+(3.0×10−8)x2−[(1.0×10−14)+(3.0×10−8)×(9.64×10−7)]x−(1.0×10−14)×(3.0×10−8)=0x3+(3.0×10−8)x2−(3.892×10−14)x−(3.0×10−22)=0 Write x=X×10−7 and divide the resulting equation by 10−21.