The p H o f 0.00500 M N H 4 [ B ( O H ) 4 ] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

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Answer 1

Question

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

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Answer 2

Question

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

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Answer 3

Question

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

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Answer 4

Question

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

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Answer 5

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

Answer 6

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

Answer 7

Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Answer 8

Expert Solution & Answer

Answer to Problem 121QRT

The pH of 0.00500 M NH4[B(OH)4] (a) by using the simplifying assumption regarding x is 10.46, (b) by solving a quadratic equation is 10.45 and (c) by using the method of successive approximations is 10.45.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The solution contains NH4+ and [B(OH)4]− ions. Ka for NH4+ is 5.6×10−10 and Kb for [B(OH)4]− is 1.7×10−5. As Kb>Ka, the solution is basic.

The equation for the equilibrium can be written as given below.

B(OH)4−(aq)+H2O(l)⇌B(OH)3H2O(aq)+OH−(aq)

The basic ionization constant for the above equation can be written as given below.

Kb=[B(OH)3(H2O)][OH−][B(OH)3−]

A table can be set up as given below.

B(OH)4−(aq)B(OH)3H2O(aq)OH−(aq)Initial conc.(M)0.0050001.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.00500−xxx

The concentration of OH− comes from its auto ionization that is 1.0×10−7. It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(x)(0.00500−x)=1.7×10−5

(a) Assuming x is very small, it can be written as 0.00500−x≅0.00500.

Then,

x2=(1.7×10−5)(0.00500)x=2.9×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.9×10−4)=3.54.pH=14.00−pOH=14.00−3.54=10.46.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.46.

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

x2=(1.7×10−5)(0.00500−x)x2+(1.7×10−5)x−8.5×10−8=0x=2.8×10−4=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

(c) using the method of successive approximations, it can be written as given below.

0.00500−x=0.00500−2.9×10−4M=0.00471 M.

Now,

x2=(1.7×10−5)(0.00471 )x=2.8×10−4M=[OH−]

The pH of the solution can be calculated as follows,

pOH=−log[OH−]=−log(2.8×10−4)=3.55.pH=14.00−pOH=14.00−3.55=10.45.

Therefore, the pH of 0.00500 M NH4[B(OH)4] is 10.45.

Answer 12

Ch. 14.1 - Identify each molecule or ion as a Brnsted-Lowry...Ch. 14.1 - Using Le Chatelier’s Principle Use Le Chatelier’s...Ch. 14.1 - Prob. 14.3E Ch. 14.1 - Complete the table. Ch. 14.1 - Prob. 14.4E Ch. 14.1 - Prob. 14.5E Ch. 14.2 - Prob. 14.6CE Ch. 14.2 - Prob. 14.7E Ch. 14.3 - Prob. 14.2PSP Ch. 14.3 - Prob. 14.3PSP

The p H o f 0.00500 M N H 4 [ B ( O H ) 4 ] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

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Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Answer to Problem 121QRT

Explanation of Solution

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Chapter 14 Solutions

The p H o f 0.00500 M N H 4 [ B ( O H ) 4 ] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Concept explainers

Videos

Interpretation: The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Answer to Problem 121QRT

Explanation of Solution

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Chapter 14 Solutions

Interpretation:

The pH of 0.00500 M NH4[B(OH)4] has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.