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.

Question

Want to see more full solutions like this?

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.

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

13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! Google

Print Last Name, First Name Initial Statifically more chances to abstract one of these 6H 11. (10pts total) Consider the radical chlorination of 1,3-diethylcyclohexane depicted below. 4 4th total • 6H total 래 • 4H total 21 total ZH 2H Statistical H < 3° C-H weakest - product abstraction here bund leads to thermo favored a) (6pts) How many unique mono-chlorinated products can be formed and what are the structures for the thermodynamically and statistically favored products? Product 6 Number of Unique Mono-Chlorinated Products Thermodynamically Favored Product Statistically Favored Product b) (4pts) Draw the arrow pushing mechanism for the FIRST propagation step (p-1) for the formation of the thermodynamically favored product. Only draw the p-1 step. You do not need to include lone pairs of electrons. No enthalpy calculation necessary H H-Cl Waterfox

10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation → depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O II HA H CH3O-H H ①

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.

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 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.

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 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.

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 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.

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

Want to see more full solutions like this?

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

Want to see more full solutions like this?

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.