The p H o f 8.50 × 10 − 5 M a n d 7.37 × 10 − 6 M H C N ( a q ) has to be calculated.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 6, Problem 6F.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−1.36X−0.0049=0X=1.166, so x=1.166×10−7[H3O+]=1.166×10−7M

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

pH=−log[H3O+]=−log(1.166×10−7)=6.93

Therefore, the pH of 7.37×10−6MHCN(aq) is 6.93.

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

1. Which one(s) of these can be oxidized with CrO3 ? (could be more than one) a) triphenylmethanol b) 2-pentanol c) Ethyl alcohol d) CH3 2. Write in all the product(s) of this reaction. Label them as "major" or "minor". 2-methyl-2-hexanol H2SO4, heat

3) Determine if the pairs are constitutional isomers, enantiomers, diastereomers, or mesocompounds. (4 points)

In the decomposition reaction in solution B → C, only species C absorbs UV radiation, but neither B nor the solvent absorbs. If we call At the absorbance measured at any time, A0 the absorbance at the beginning of the reaction, and A∞ the absorbance at the end of the reaction, which of the expressions is valid? We assume that Beer's law is fulfilled.

Answer 2

Question

Chapter 6, Problem 6F.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−1.36X−0.0049=0X=1.166, so x=1.166×10−7[H3O+]=1.166×10−7M

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

pH=−log[H3O+]=−log(1.166×10−7)=6.93

Therefore, the pH of 7.37×10−6MHCN(aq) is 6.93.

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.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−1.36X−0.0049=0X=1.166, so x=1.166×10−7[H3O+]=1.166×10−7M

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

pH=−log[H3O+]=−log(1.166×10−7)=6.93

Therefore, the pH of 7.37×10−6MHCN(aq) is 6.93.

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.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−1.36X−0.0049=0X=1.166, so x=1.166×10−7[H3O+]=1.166×10−7M

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

pH=−log[H3O+]=−log(1.166×10−7)=6.93

Therefore, the pH of 7.37×10−6MHCN(aq) is 6.93.

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.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−1.36X−0.0049=0X=1.166, so x=1.166×10−7[H3O+]=1.166×10−7M

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

pH=−log[H3O+]=−log(1.166×10−7)=6.93

Therefore, the pH of 7.37×10−6MHCN(aq) is 6.93.

Answer 6

Chapter 6, Problem 6F.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

Answer 7

Chapter 6, Problem 6F.9E

(a)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(aq) has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

Calculation of pH of 8.50×10−5M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration8.50×10−500Change in concentration−x+x+xEquilibrium concentration8.50×10−5−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(8.5×10−5−x)x2=(4.9×10−10)(8.5×10−5) [Assuming x≪8.5×10−5]x=(4.9×10−10)(8.5×10−5)=2.04×10−7[H3O+]=2.04×10-7M

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

pH=−log[H3O+]=−log(2.04×10−7)=6.69

Therefore, the pH of 8.50×10−5M HCN(aq) is 6.69.

Calculation of pH of 7.37×10−6M HCN(aq):

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

HCN(aq)+H2O(l)⇌H3O+(aq)+CN−(aq)Ka=[H3O+][CN−][HCN]=4.9×10−10

An equilibrium table can be set up as given below.

Composition(M)HCNH3O+CN−Initial concentration7.37×10−600Change in concentration−x+x+xEquilibrium concentration7.37×10−6−xxx

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

Ka=[H3O+][CN−][HCN]=4.9×10−104.9×10−10=(x)(x)(7.37×10−6−x)x2=(4.9×10−10)(7.37×10−6) [Assuming x≪7.37×10−6]x=(4.9×10−10)(7.37×10−6)=2.04×10−7[H3O+]=6.00×10-8M

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

pH=−log[H3O+]=−log(6.00×10−8)=7.22

Therefore, the pH of 7.37×10−6M HCN(aq) is 7.22.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−1.36X−0.0049=0X=1.166, so x=1.166×10−7[H3O+]=1.166×10−7M

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

pH=−log[H3O+]=−log(1.166×10−7)=6.93

Therefore, the pH of 7.37×10−6MHCN(aq) is 6.93.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of 8.50×10−5M and 7.37×10−6MHCN(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 8.50×10−5M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(8.50×10−5)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(5.165×10−14)x−(4.9×10−24)=0

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

X3+0.0049X2−5.165X−0.0049=0X=2.272, so x=2.272×10−7[H3O+]=2.272×10−7M

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

pH=−log[H3O+]=−log(2.272×10−7)=6.64

Therefore, the pH of 8.50×10−5MHCN(aq) is 6.64.

Calculation of pH of7.37×10−6M HCN(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+(4.9×10−10)x2−[(1.0×10−14)+(4.9×10−10)×(7.37×10−6)]x−(1.0×10−14)×(4.9×10−10)=0x3+(4.9×10−10)x2−(1.36×10−14)x−(4.9×10−24)=0