General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN: 9781305580343
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: Cengage Learning
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Textbook Question
Chapter 16, Problem 16.91QP

A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. Kb for CN is 2.0 × 10−5. Calculate the pH of the solution: a prior to the start of the titration; b after the addition of 15.0 mL of 0.200 M HCl; c at the equivalence point; d after the addition of 30.0 mL of 0.200 M HCl.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Answer to Problem 16.91QP

The pH prior to the start of the titration is 11.15

Explanation of Solution

To Calculate: The pH prior to the start of the titration

Given data:

The volume of NaCN = 50.0 mL

The concentration of NaCN = 0.100 M

The concentration of hydrochloric acid = 0.200 M

The Kb value for CN is 2.0×105

pH prior to the start of the titration

Construct an equilibrium table for the hydrolysis of cyanide ion from NaCN as follows,

     CN-  +   H2O        HCN    +   OH
Initial (M)

0.100

x

0.100-x

0.00 0.00
Change (M) +x +x
Equilibrium (M) x x

The Kb value for CN is 2.0×105

Now substitute equilibrium concentrations into the equilibrium-constant expression.

      Kb =(x)2(0.100x)Assume 0.100 is very small than x and neglect it in the denominator   2.0×105 (x)2(0.100)      x =1.414×103 M

Here, x gives the concentration of hydroxide ion 1.414×103 M

Finally calculate pOH and then the pH as follows,

pOH =-log[OH-] =-log(1.414×103) =2.849

The pH is calculated as follows,

pH + pOH  = 14 pH =14 - pOH =14 - 2.849 =11.15

Conclusion

The pH prior to the start of the titration is 11.15

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Answer to Problem 16.91QP

The pH after the addition of 15 mL of 0.200 M HCl is 9.12

Explanation of Solution

To Calculate: The pH after the addition of 15 mL of 0.200 M HCl

After the addition of 15.0 mL of 0.200 M HCl , the reaction will be as follows

CN + H3O+ HCN + H2O

At this point, the total volume is, 50.0 mL + 15.0 mL = 65.0 mL

Now, the moles of CN- present at the initial and the moles of HCl added are:

mol CN-  = M×V =0.100 M×50×103 L =5.00×103 molmol HCl  = M×V =0.200 M×15×103 L =3.00×103 mol

After the reaction, the moles of HCN will be equal to the moles of HCl added.

The moles of CN- present are:

mol CN-  = 5.000×103 mol - 3.000×103 =2.000×103 mol

The concentrations are:

[CN-] =2.000×103 mol CN-65.0×103 L =0.03076 M[HCN] =3.000×103 mol HCN65.0×103 L =0.04615 M

Solve for [OH-] by Substituting the concentrations into the Kb expression

     Kb =[HCN][OH-][CN-]  2.0×105 =(0.04615+x)(x)(0.03076x) (0.04615)(x)(0.03076) =(0.04615)(x)(0.03076) x =(0.03076)(2.0×105)(0.04615) =1.33×105 M

The pH is calculated from pOH as follows,

pOH =-log(1.33×10-5) =4.875pH =14 - pOH =14-4.875 =9.12

Conclusion

The pH after the addition of 15 mL of 0.200 M HCl is 9.12

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Answer to Problem 16.91QP

The pH at the equivalence point is 5.24

Explanation of Solution

To Calculate: The pH at the equivalence point

Calculate the volume of HCl added to reach the equivalence point

The volume of HCl added is calculated as follows,

Volume of HCl Mbase VbaseMacid = (0.100 M)(50.0 mL)0.200 M = 25.0 mL

Hence, the total volume is as follows,

Total volume = 50.0 mL + 25.0 mL =75.0 mL = 0.075 L

The equilibrium reaction is,

HCN + H2 H3O+ + CN-

[HCN] =5.000×103 mol HCN75.0×103 L =0.0666 M

The value of Ka is calculated from Kb as follows,

Ka =KwKb =1.00×10-142.0×10-5 =5.00×1010

Ka =[CN-][H3O+][HCN]5.00×1010 =x20.0666x x =5.773×106 M

Here, x gives the hydronium ion concentration.

pH =-log[H+] =-log(5.773×10-6) =5.24

Conclusion

The pH at the equivalence point is 5.24

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Answer to Problem 16.91QP

The pH after the addition of 30.0 mL of 0.200 M HCl is 1.903

Explanation of Solution

To Calculate: The pH after the addition of 30 mL of 0.200 M HCl

Calculate the moles of acid added.

(0.200 M)×0.030 L = 6.000×10-3 mol

Moles of acid remaining  = 6.000×10-3 mol - 5.000×10-3 mol = 1.000×10-3 mol

The total volume after the addition of 30.0 mL of HCl is: 50.0 mL + 30.0 mL = 80.0 mL

The hydronium ion concentration and the pH are:

[H3O+] =1.000×10-3 mol80.0×10-3 L =0.01250 M

The pH is calculated as follows,

pH =log[H+] =log(0.01250) =1.903

Conclusion

The pH after the addition of 30.0 mL of 0.200 M HCl is 1.903

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Chapter 16 Solutions

General Chemistry - Standalone book (MindTap Course List)

Ch. 16.4 - Benzoic acid, HC7H5O2, and its salts are used as...Ch. 16.4 - Which of the following aqueous solutions has the...Ch. 16.5 - The chemical equation for the hydrolysis of...Ch. 16.5 - What is the concentration of formate ion, CHO2, in...Ch. 16.5 - One liter of solution was prepared by dissolving...Ch. 16.6 - What is the pH of a buffer prepared by adding 30.0...Ch. 16.6 - Suppose you add 50.0 mL of 0.10 M sodium hydroxide...Ch. 16.6 - Prob. 16.5CCCh. 16.6 - The beaker on the left below represents a buffer...Ch. 16.7 - What is the pH of a solution in which 15 mL of...Ch. 16.7 - What is the pH at the equivalence point when 25 mL...Ch. 16.7 - Prob. 16.16ECh. 16 - Write an equation for the ionization of hydrogen...Ch. 16 - Prob. 16.2QPCh. 16 - Briefly describe two methods for determining Ka...Ch. 16 - Describe how the degree of ionization of a weak...Ch. 16 - Prob. 16.5QPCh. 16 - Phosphorous acid, H2PHO3, is a diprotic acid....Ch. 16 - Prob. 16.7QPCh. 16 - Write the equation for the ionization of aniline,...Ch. 16 - Which of the following is the strongest base: NH3,...Ch. 16 - Do you expect a solution of anilinium chloride...Ch. 16 - Prob. 16.11QPCh. 16 - The pH of 0.10 M CH3NH2 (methylamine) is 11.8....Ch. 16 - Define the term buffer. Give an example.Ch. 16 - What is meant by the capacity of a buffer?...Ch. 16 - Prob. 16.15QPCh. 16 - If the pH is 8.0 at the equivalence point for the...Ch. 16 - Which of the following salts would produce the...Ch. 16 - If you mix 0.10 mol of NH3 and 0.10 mol of HCl in...Ch. 16 - Hydrogen sulfide, H2S, is a very weak diprotic...Ch. 16 - If 20.0 mL of a 0.10 M NaOH solution is added to a...Ch. 16 - Aqueous Solutions of Acids, Bases, and Salts a For...Ch. 16 - The pH of Mixtures of Acid, Base, and Salt...Ch. 16 - Which of the following beakers best represents a...Ch. 16 - You have 0.10-mol samples of three acids...Ch. 16 - Prob. 16.25QPCh. 16 - You have the following solutions, all of the same...Ch. 16 - Prob. 16.27QPCh. 16 - A chemist prepares dilute solutions of equal molar...Ch. 16 - Prob. 16.29QPCh. 16 - Prob. 16.30QPCh. 16 - You are given the following acidbase titration...Ch. 16 - The three flasks shown below depict the titration...Ch. 16 - Write chemical equations for the acid ionizations...Ch. 16 - Write chemical equations for the acid ionizations...Ch. 16 - Acrylic acid, whose formula is HC3H3O2 or...Ch. 16 - Heavy metal azides, which are salts of hydrazoic...Ch. 16 - Boric acid, B(OH)3, is used as a mild antiseptic....Ch. 16 - Formic acid, HCHO2, is used to make methyl formate...Ch. 16 - C6H4NH2COOH, para-aminobenzoic acid (PABA), is...Ch. 16 - Barbituric acid. 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