Chem II Homework Page, Exam 4 Material

Homework Page Without Visible Answers

This page has all of the required homework for the material covered in the fourth exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12^th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Equilibrium (Ch. 15-17)

1. Use the equation ΔG° = -RTlnK to determine K for the following reaction at 25 °C. R is the gas constant: R = 8.31 J/mole-K.

   N₂(g) + 3H₂(g) ⇔ 2NH₃(g)

   Answer

   ΔG° = 2ΔG°_NH3(g) - ΔG°_N(g) - 3ΔG°_H2(g)

   ΔG° = 2(-16.66 kJ) - 0 - 3(0) = -33.32 kJ for two moles of NH₃

   ΔG° = -16.66 kJ/mole = -16,660 J/mole of NH₃

   lnK = -ΔG°/(RT) = (16,660 J/mole)/[(8.31 J/mole-K)(298 K)] = 6.73

   ⇒ K = e^6.73 = 835

2. The gaseous reaction

   2NO(g) + O₂(g) ⇔ 2NO₂(g)

   has K_c = 1.20 at 1000 K. At equilibrium it was found that there were 7.50 g NO and 26.67 g O₂ in a 250 mL container. What is the concentration of NO₂ at this equilibrium?

   Answer

   Find moles of each and then concentration of each:

   (7.50 g NO)(1 mole/30 g) = 0.25 mole NO; (0.25 mole NO)/(0.25 L) = 1.00 M NO

   (26.67 g O₂)(1 mole/32 g) = 0.83 mole; (0.83 mole O₂)/(0.25 L) = 3.33 M O₂

   For this reaction K_c = [NO₂]²/[NO]²[O₂]

   ⇒ 1.2 = [NO₂]²/(1²)(3.33) ⇒ [NO₂] = 2 M

3. A mixture of 0.03 moles of PCl₃ and 0.02 moles of Cl₂ react in a 100 mL container as follows:

   PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

   At equilibrium there are 0.19 moles of PCl₃ present in the container. (a) Calculate the equilibrium concentrations of PCl₃, Cl₂, and PCl₅. (b) Calculate K_c under these conditions.

   Answer

   PCl3(g) + Cl2(g) ⇔ PCl5(g)
   Init. Moles	0.03 mole	0.02 mole	0 mole
   Init. Conc.	0.3 M	0.2 M	0 M
   Equilib	0.3 - x	0.2 - x	x
   Equilib	0.19	?	?

   (a) Find x:   0.3 - x = 0.19;   x = 0.3 - 0.19 = 0.11

   ⇒ [PCl₅] = x = 0.11 M

   ⇒ [Cl₂] = 0.2 - x = 0.2 - 0.11 = 0.09 M

   [PCl₃] = 0.19 M (given in the problem)

   (b) K_c = [PCl₅]/{[PCl₃][Cl₂]} = (0.11 M)/{(0.19 M)(0.09 M)} = 6.43

4. Nitrogen gas, N₂(g), and water vapor react as follows:

   N₂(g) + 2H₂O(g) ⇔ 2NO(g) + 2H₂(g)

   When 0.2 moles of N₂ are combined with 0.3 moles of H₂O in a 1.0 L container and allowed to go to equilibrium, it is found that there are 0.169 moles of N₂ in the equilibrium mixture. What are the equilibrium concentrations of each of the materials and what is the value of the equilibrium constant, K_c?

   Answer

   N2(g) + H2O(g) ⇔ 2NO(g) + 2H2(g)
   Init. Moles	0.2 mole	0.3 mole	0 mole	0 mole
   Init. Conc.	0.2 M	0.3 M	0 M	0 M
   Equilib	0.2 - x	0.3 - 2x	2x	2x
   Equilib	0.169	?	?	?

   (a) Find x:   0.2 - x = 0.169;   x = 0.2 - 0.169 = 0.031

   ⇒ [H₂O] = 0.3 - 2x = 0.238 M

   ⇒ [H₂] = [NO] = 2x = (2)(0.031) = 0.062 M

   [N₂] = 0.169 M (given in the problem)

   (b) K_c = [NO]²[H₂]²/{[N₂][H₂O]²}

   K_c = (0.062 M)²(0.062 M)²/{(0.169 M)(0.238 M)²} = 1.54 x 10^-3

5. Consider the following reaction:

   2NO(g) + Cl₂(g) ⇔ 2NOCl(g)   K_c = 1.42 x 10⁷

   Calculate the concentration of NO at equilibrium when 3.0 moles of NO are reacted with 2.0 moles of Cl₂ in a 500 mL container.

   Answer

   A large K_c means that the reaction will go to right in the complete reaction line below.

   2NO(g) + Cl2(g) ⇔ 2NOCl(g)
   Init. Moles	3.0 mole	2.0 mole	0 moles
   Init. Conc.	6 M	4 M	0 M
   Complete Rxn	0	1	6
   Equilib Conc	0 + 2x	1 + x	6 - 2x

   K_c = 1.42 x 10⁷ = [NOCl]²/{[NO]²[Cl₂]} = (6-2x)²/{(2x)²(1+x)}

   Assume x << 1

   ⇒ x = {36/[(4)(1.42 x 10⁷)]}^½ = 8.02 x 10⁻⁴

   ⇒ [NO] = 2x = 16 x 10⁻⁴ M

6. Consider the following reaction:

   2NOCl(g) ⇔ 2NO(g) + Cl₂(g)   K_c = 7.04 x 10^-8

   Calculate the concentration of NO at equilibrium when 3.0 moles of NOCl and 2.0 moles of Cl₂ are initially placed in a 500 mL container.

   Answer

   A small K_c means that the reaction will go to left in the complete reaction line below.

   2NOCl(g) ⇔ 2NO(g) + Cl2(g)
   Init. Moles	3.0 mole	0 mole	2.0 moles
   Init. Conc.	6 M	0 M	4 M
   Complete Rxn	6	0	4
   Equilib Conc	6 - 2x	0 + 2x	4 + x

   K_c = 7.04 x 10⁻⁸ = {[NO]²[Cl₂]}/[NOCl]² = {(2x)²(4+x)}/(6-2x)²

   Assume x << 4

   ⇒ x = {(36)(7.04 x 10⁻⁸}/[(4)(4)]}^½ = (1.584 x 10⁻⁷}^½ = 3.98 x 10⁻⁴

   ⇒ [NO] = 2x = 7.96 x 10⁻⁴ M

7. Consider the following reaction:

   2NOCl(g) ⇔ 2NO(g) + Cl₂(g)   K_c = 7.04 x 10^-8

   Calculate the concentration of NO at equilibrium when 3.0 moles of NOCl, 2.0 moles of Cl₂, and 2.0 moles of NO are initially placed in a 500 mL container.

   Answer

   A small K_c means that the reaction will go to left in the complete reaction line below.

   2NOCl(g) ⇔ 2NO(g) + Cl2(g)
   Init. Moles	3.0 mole	2.0 mole	2.0 moles
   Init. Conc.	6 M	4 M	4 M
   Complete Rxn	10	0	2
   Equilib Conc	10 - 2x	0 + 2x	2 + x

   K_c = 7.04 x 10⁻⁸ = {[NO]²[Cl₂]}/[NOCl]² = {(2x)²(2+x)}/(10-2x)²

   Assume x << 2   ⇒ 7.04 x 10⁻⁸ = (2x)²(2)/10 = (0.8)x²

   ⇒ x = {(7.04 x 10⁻⁸)/(0.8)}^½ = 2.97 x 10⁻⁴

   ⇒ [NO] = 2x = 5.9 x 10⁻⁴ M

8. Calculate the pH when 132 g of butanoic acid, HC₄H₇O₂, which has K_a = 1.5 x 10^-5, is placed in 300 mL of water.

   Answer

   K_a = 1.5 x 10^-5 means that the reaction will go toward the left in the complete reaction line below.

   HC4H7O2 ⇔ H+ + C4H7O2-
   Init. Moles	1.5 mole	0 mole	0 mole
   Init. Conc.	5 M	0 M	0 M
   Complete Rxn	5	0	0
   Equilib Conc	5 - x	x	x

   K_a = 1.5 x 10^-5 = [H⁺][C₄H₇O₂^-]/[HC₄H₇O₂] = x²/(5 - x)   Assume x << 5

   ⇒ x = [(5)(1.5 x 10^-5)]^½ = 0.00866 = [H⁺]

   ⇒ pH = -log(0.00866) = 2.06

9. Calculate the pH in a solution where 1.2 moles of formic acid, HCHO₂, and 2.4 moles of sodium formate, NaCHO₂, are mixed with water to make a total of 2.0 L of solution. For formic acid, K_a = 1.8 x 10^-4.

   Answer

   K_a = 1.8 x 10^-4 means that the reaction will go toward the left in the complete reaction line below. Notice that the Na⁺ ion is a spectator ion and doesn't show up in the equilibrium equation.

   HCHO2 ⇔ H+ + CHO2-
   Init. Moles	1.2 mole	0 mole	2.4 mole
   Init. Conc.	0.6 M	0 M	1.2 M
   Complete Rxn	0.6	0	1.2
   Equilib Conc	0.6 - x	x	1.2 + x

   K_a = 1.8 x 10^-4 = [H⁺][CHO₂^-]/[HCHO₂] = (x)(1.2 + x)/(0.6 - x)

   Assume x << 0.6

   ⇒ x = [(0.6)(1.8 x 10^-4)]/1.2 = 9 x 10^-5 = [H⁺]

   ⇒ pH = -log(9 x 10^-5) = 4.05

10. Determine the concentration of all ions in a solution where solid calcium hydroxide, Ca(OH)₂, is in equilibrium with its ions. For calcium hydroxide K_sp = 6.5 x 10^-6.

    Answer

    K_sp = 6.5 x 10⁻⁶ means that there will not be very many ions in the solution. Because the solid (concentration set to one) is in equilibrium with the ions, we only need to consider what the condition is at equilibrium as shown in the table.

    Ca(OH)2(s) ⇔ Ca2+ + 2OH-
    Equilib Conc	-	x	2x

    K_sp = 6.5 x 10^-6 = [Ca²⁺][OH⁻]² = (x)(2x)² = 4x³

    ⇒ x = [(6.5 x 10^-6)/4]^⅓ = 0.0118

    ⇒ [Ca²⁺] = 0.0118 M and [OH^-] = 0.0236 M

11. An 8.5 mL sample of 1.0 M NaOH was needed to reach the equivalence point when titrating into 6.0 mL of acetic acid, HC₂H₃O₂. Acetic acid has K_a = 1.8 x 10⁻⁵. What is the pH of the original acetic acid solution?

    Answer

    (0.0085 L NaOH)(1.0 mole/L NaOH) = 0.0085 moles of OH⁻

    NaOH reacts with acetic acid in a one-to-one ratio. The NaOH reacts with both the dissociated and nondissociated hydrogen ions from the acetic acid solution, so the moles of NaOH used is the initial moles of HC₂H₃O₂ that would have been in the solution before any dissociation of the acid.

    K_a = 1.8 x 10^-5 means that the reaction will go toward the left in the complete reaction line below.

    HC2H3O2 ⇔ H+ + C2H3O2-
    Init. Moles	0.0085 mole	0 mole	0 mole
    Init. Conc.	1.417 M	0 M	0 M
    Complete Rxn	1.417	0	0
    Equilib Conc	1.417 - x	x	x

    K_a = 1.8 x 10^-5 = [H⁺][C₂H₃O₂^-]/[HC₂H₃O₂] = (x)(x)/(1.417 - x)

    Assume x << 1.417

    ⇒ x = [(1.417)(1.8 x 10^-5)]^½ = 0.005 = [H⁺]

    ⇒ pH = -log(0.005) = 2.3

12. Determine the pH after (a) 2 mL, (b) 3 mL, and (c) 4 mL of 0.5 M HCl has been titrated into 10.0 mL of 0.15 M NaOH.

    Answer (c)

    H⁺ + OH⁻ ⇔ H₂O   K_c = 1/K_w = 1 x 10¹⁴ = 1/([H⁺][OH⁻])

    After 4 mL HCl:
    (0.004 L HCl)(0.5 mole/L HCl) = 0.002 moles of H⁺
    ⇒ Initial [H⁺] = (0.002 moles H⁺)/(0.014 L soln) = 0.143 M HCl.

    (0.01 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0015 moles OH⁻)/(0.014 L soln) = 0.107 M OH⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    H+     +     OH−     ⇔ H2O
    Init. Moles	0.002 mole	0.0015 mole	-
    Init. Conc.	0.143 M	0.107 M	-
    Complete Rxn	0.036	0	-
    Equilib Conc	0.036 + x	x	-

    K_c = 1 x 10¹⁴ = 1/([H⁺][OH⁻]) = 1/{(0.036 + x)(x)}   Assume x << 0.036

    ⇒ [H⁺] = 0.036 M ⇒ pH = -log(0.036) = 1.44

    Answer (b)  |  

    H⁺ + OH⁻ ⇔ H₂O   K_c = 1/K_w = 1 x 10¹⁴ = 1/([H⁺][OH⁻])

    After 3 mL HCl:
    (0.003 L HCl)(0.5 mole/L HCl) = 0.0015 moles of H⁺
    ⇒ Initial [H⁺] = (0.0015 moles H⁺)/(0.013 L soln) = 0.115 M HCl.

    (0.01 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0015 moles OH⁻)/(0.013 L soln) = .115 M OH⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    H+     +     OH−     ⇔ H2O
    Init. Moles	0.0015 mole	0.0015 mole	-
    Init. Conc.	0.115 M	0.115 M	-
    Complete Rxn	0	0	-
    Equilib Conc	x	x	-

    K_c = 1 x 10¹⁴ = [H⁺][OH⁻] = 1/x² ⇒ x = 1 x 10⁻⁷ = [H⁺]

    ⇒ pH = -log(10⁻⁷) = 7

    Answer (a)  |  

    H⁺ + OH⁻ ⇔ H₂O   K_c = 1/K_w = 1 x 10¹⁴ = 1/([H⁺][OH⁻])

    After 2 mL HCl:
    (0.002 L HCl)(0.5 mole/L HCl) = 0.001 moles of H⁺
    ⇒ Initial [H⁺] = (0.001 moles H⁺)/(0.012 L soln) = 0.0833 M HCl.

    (0.01 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0015 moles OH⁻)/(0.012 L soln) = 0.125 M OH⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    H+     +     OH−     ⇔ H2O
    Init. Moles	0.001 mole	0.0015 mole	-
    Init. Conc.	0.0833 M	0.125 M	-
    Complete Rxn	0	0.0417	-
    Equilib Conc	x	0.0417 + x	-

    K_c = 1 x 10¹⁴ = 1/([H⁺][OH⁻]) = 1/{(x)(0.0417 + x)}   Assume x << 0.0417

    ⇒ [OH⁻] = 0.0417 M ⇒ pOH = -log(0.0417) = 1.38 ⇒ pH = 14- 0.067 = 12.62

13. Calculate the pH after (a) 12 mL, (b) 14 mL, and (c) 16 mL of 0.6 M NaOH has been titrated into 21 mL of 0.4 M phenol, HC₆H₅O. For phenol K_a = 1.3 x 10⁻¹⁰

    Answer (c)

    HC₆H₅O + OH⁻ ⇔ H₂O + C₆H₅O⁻   K_c = K_a/K_w = 1.3 x 10⁴ = [C₆H₅O⁻]/([HC₆H₅O][OH⁻])

    After 16 mL NaOH:
    (0.016 L NaOH)(0.6 mole/L NaOH) = 0.0096 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0096 moles OH⁻)/(0.037 L soln) = 0.259 M OH⁻.

    (0.021 L HC₆H₅O)(0.4 mole/L NaOH) = 0.0084 moles of HC₆H₅O
    ⇒ Initial [HC₆H₅O] = (0.0084 moles HC₆H₅O)/(0.037 L soln) = 0.227 M OH⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HC6H5O   +   OH−   ⇔   H2O + C6H5O−
    Init. Moles	0.0084 mole	.0096 mole	-	0 moles
    Init. Conc.	0.227 M	0.259 M	-	0
    Complete Rxn	0	0.03246	-	0.227
    Equilib Conc	x	0.03246 + x	-	0.227 - x

    K_c = 1.3 x 10⁴ = (0.227 - x)/{(x)(0.03246 + x)}   Assume x << 0.03246

    ⇒ [OH⁻] = 0.03246 M ⇒ pOH = -log(0.03246) = 1.49

    ⇒ pH = 14 - 1.49 = 12.5

    Answer (b)  |  

    HC₆H₅O + OH⁻ ⇔ H₂O + C₆H₅O⁻   K_c = K_a/K_w = 1.3 x 10⁴ = [C₆H₅O⁻]/([HC₆H₅O][OH⁻])

    After 14 mL NaOH:
    (0.014 L NaOH)(0.6 mole/L NaOH) = 0.0084 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0084 moles OH⁻)/(0.035 L soln) = 0.24 M OH⁻.

    (0.021 L HC₆H₅O)(0.4 mole/L NaOH) = 0.0084 moles of HC₆H₅O
    ⇒ Initial [HC₆H₅O] = (0.0084 moles HC₆H₅O)/(0.035 L soln) = 0.24 M OH⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HC6H5O   +   OH−   ⇔   H2O + C6H5O−
    Init. Moles	0.0084 mole	.0084 mole	-	0 moles
    Init. Conc.	0.24 M	0.24 M	-	0
    Complete Rxn	0	0	-	0.24
    Equilib Conc	x	x	-	0.24 - x

    K_c = 1.3 x 10⁴ = (0.24 - x)/{(x)(x)}   Assume x << 0.24

    ⇒ x = [(0.24)/(1.3 x 10⁴)]^½ = 0.0043 M = [OH⁻] ⇒ pOH = -log(0.0043) = 2.37

    ⇒ pH = 14 - 2.37 = 11.63

    Answer (a)  |  

    HC₆H₅O + OH⁻ ⇔ H₂O + C₆H₅O⁻   K_c = K_a/K_w = 1.3 x 10⁴ = [C₆H₅O⁻]/([HC₆H₅O][OH⁻])

    After 12 mL NaOH:
    (0.012 L NaOH)(0.6 mole/L NaOH) = 0.0072 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0072 moles OH⁻)/(0.033 L soln) = 0.218 M OH⁻.

    (0.021 L HC₆H₅O)(0.4 mole/L NaOH) = 0.0084 moles of HC₆H₅O
    ⇒ Initial [HC₆H₅O] = (0.0084 moles HC₆H₅O)/(0.033 L soln) = 0.255 M OH⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HC6H5O   +   OH−   ⇔   H2O + C6H5O−
    Init. Moles	0.0084 mole	.0072 mole	-	0 moles
    Init. Conc.	0.255 M	0.218 M	-	0
    Complete Rxn	0.037	0	-	0.255
    Equilib Conc	0.037 + x	x	-	0.218 - x

    K_c = 1.3 x 10⁴ = (0.218 - x)/{(x)(0.037 + x)}   Assume x << 0.0367

    x = (0.218)/{(0.037)(1.3 x 10⁴)} = 0.000453 = [OH⁻]

    ⇒ pOH = -log(0.000453) = 3.34   ⇒ pH = 14 - 3.34 = 10.66

14. Calculate the pH for addition of 0.15 M NaOH to a solution iitially containing 15 mL of 0.1 M HCN (K_a=4.9x10⁻¹⁰) and 10 mL of 0.2 M KCN. Do the calculation for (a) 0 mL, (b) 5 mL, (c) 8 mL, (d) 10 mL, (e) 12 mL, (f) 15 mL, and (g) 20 mL of added NaOH.

    Answer (g)

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 20 mL NaOH:

    (0.02 L NaOH)(0.15 mole/L NaOH) = 0.003 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.003 moles OH⁻)/(0.045 L soln) = 0.0667 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.045 L soln) = 0.0333 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.045 L soln) = 0.0444 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN   +   OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.003 mole	-	0.002 mole
    Init. Conc.	0.0333 M	0.0667 M	-	0.0444 M
    Complete Rxn	0	0.0334	-	0.0777
    Equilib Conc	x	0.0334 + x	-	0.0777 - x

    K_c = 4.9 x 10⁴ = (0.0777 - x)/{(x)(0.0334 + x)}   Assume x << 0.0333

    ⇒ [OH⁻] = 0.00334 M ⇒ pOH = -log(0.0334) = 1.47

    ⇒ pH = 14 - 1.47 = 12.5

    Answer (f)  |  

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 15 mL NaOH:

    (0.015 L NaOH)(0.15 mole/L NaOH) = 0.00225 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.00225 moles OH⁻)/(0.04 L soln) = 0.0563 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.04 L soln) = 0.0375 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.04 L soln) = 0.05 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN     +     OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.00225 mole	-	0.002 mole
    Init. Conc.	0.0375 M	0.0563 M	-	0.05 M
    Complete Rxn	0	0.0188	-	0.0875
    Equilib Conc	x	0.0188 + x	-	0.0875 - x

    K_c = 4.9 x 10⁴ = (0.0875 - x)/{(x)(0.0188 + x)}   Assume x << 0.0188

    ⇒ [OH⁻] = 0.0188 M ⇒ pOH = -log(0.0188) = 1.73

    ⇒ pH = 14 - 1.73 = 12.3

    Answer (e)  |  

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 12 mL NaOH:

    (0.012 L NaOH)(0.15 mole/L NaOH) = 0.0018 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0018 moles OH⁻)/(0.037 L soln) = 0.0487 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.037 L soln) = 0.0405 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.037 L soln) = 0.0541 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN     +     OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.0018 mole	-	0.002 mole
    Init. Conc.	0.0405 M	0.0563 M	-	0.0541 M
    Complete Rxn	0	0.0158	-	0.0946
    Equilib Conc	x	0.0158 + x	-	0.0946 - x

    K_c = 4.9 x 10⁴ = (0.0946 - x)/{(x)(0.0158 + x)}   Assume x << 0.0158

    ⇒ [OH⁻] = 0.0158 M ⇒ pOH = -log(0.0158) = 1.8

    ⇒ pH = 14 - 1.73 = 12.2

    Answer (d)  |  

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 10 mL NaOH:

    (0.010 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.0015 moles OH⁻)/(0.035 L soln) = 0.0429 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.035 L soln) = 0.0429 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.035 L soln) = 0.0571 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN     +     OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.0015 mole	-	0.002 mole
    Init. Conc.	0.0429 M	0.0429 M	-	0.0571 M
    Complete Rxn	0	0	-	0.1
    Equilib Conc	x	x	-	0.1 - x

    K_c = 4.9 x 10⁴ = (0.1 - x)/{(x)(x)}   Assume x << 0.1

    ⇒ [OH⁻] = x = (0.1/49000)^½ = 0.00143 M ⇒ pOH = -log(0.00143) = 2.85

    ⇒ pH = 14 - 2.85 = 11.2

    Answer (c)  |  

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 8 mL NaOH:

    (0.008 L NaOH)(0.15 mole/L NaOH) = 0.00012 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.00012 moles OH⁻)/(0.033 L soln) = 0.0364 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.033 L soln) = 0.0455 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.033 L soln) = 0.0606 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN     +     OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.0012 mole	-	0.002 mole
    Init. Conc.	0.0455 M	0.0364 M	-	0.0606 M
    Complete Rxn	0.0091	0	-	0.097
    Equilib Conc	0.0091 + x	x	-	0.097 - x

    K_c = 4.9 x 10⁴ = (0.097 - x)/{(0.0091 + x)(x)}   Assume x << 0.0091

    ⇒ [OH⁻] = x = 0.097/{(0.0091)(49000)} = .00022 M; pOH = 3.66

    ⇒ pH = 14 - 3.66 = 10.34

    Answer (b)  |  

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 5 mL NaOH:

    (0.005 L NaOH)(0.15 mole/L NaOH) = 0.00075 moles of OH⁻
    ⇒ Initial [OH⁻] = (0.00075 moles OH⁻)/(0.030 L soln) = 0.025 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.030 L soln) = 0.05 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.030 L soln) = 0.0667 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN     +     OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.00075 mole	-	0.002 mole
    Init. Conc.	0.05 M	0.025 M	-	0.0667 M
    Complete Rxn	0.025	0	-	0.0917
    Equilib Conc	0.025 + x	x	-	0.0917 - x

    K_c = 4.9 x 10⁴ = (0.0917 - x)/{(0.025 + x)(x)}   Assume x << 0.025

    ⇒ [OH⁻] = x = 0.0917/{(0.025)(49000)} = 0.00007 M; pOH = -log[OH⁻] = 4.13

    ⇒ pH = 14 - 4.13 = 9.87

    Answer (a)  |  

    HCN + OH⁻ ⇔ H₂O + CN⁻   K_c = K_a/K_w = 4.9 x 10⁴ = [CN⁻]/([HCN][OH⁻])

    After 0 mL NaOH:

    (0 L NaOH)(0.15 mole/L NaOH) = 0 moles of OH⁻
    ⇒ Initial [OH⁻] = (0 moles OH⁻)/(0.025 L soln) = 0 M OH⁻.

    (0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
    ⇒ Initial [HCN] = (0.0015 moles HCN)/(0.025 L soln) = 0.06 M HCN.

    (0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN⁻
    ⇒ Initial [CN⁻] = (0.002 moles CN⁻)/(0.025 L soln) = 0.08 M CN⁻.

    A large K_c means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.

    HCN     +     OH−   ⇔   H2O   +   CN−
    Init. Moles	0.0015 mole	0.000 mole	-	0.002 mole
    Init. Conc.	0.06 M	0 M	-	0.08 M
    Complete Rxn	0.06	0	-	0.08
    Equilib Conc	0.06 + x	x	-	0.08 - x

    K_c = 4.9 x 10⁴ = (0.08 - x)/{(0.06 + x)(x)}   Assume x << 0.06

    ⇒ [OH⁻] = x = 0.08/{(0.06)(49000)} = 0.00003 M; pOH = -log[OH⁻] = 4.57

    ⇒ pH = 14 - 4.57 = 9.43