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 12th 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. The gaseous reaction

    2NO(g) + O2(g) 2NO2

    has Kc = 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 Kc = [NO2]²/[NO]²[O2]

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

  2. 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 0.019 moles of PCl3 present in the container. (a) Calculate the equilibrium concentrations of PCl3, Cl2, and PCl5. (b) Calculate Kc under these conditions.
    Answer
    PCl3(g) + Cl2(g) PCl5(g)
    Init. Moles0.03 mole0.02 mole0 mole
    Init. Conc.0.3 M0.2 M0 M
    Equilib0.3 - x0.2 - xx
    Equilib0.19??

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

  3. 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 H2 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, Kc?
    Answer
    N2(g) + H2O(g) 2NO(g) + 2H2(g)
    Init. Moles0.2 mole0.3 mole0 mole0 mole
    Init. Conc.0.2 M0.3 M0 M0 M
    Equilib0.2 - x0.3 - 2x2x2x
    Equilib0.169???

    (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) Kc = [NO]²[H2]²/{[N2][H2O]²}

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

  4. Consider the following reaction:

    2NO(g) + Cl2(g) ⇔ 2NOCl(g)   Kc = 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 Kc means that the reaction will go to right in the complete reaction line below.

    2NO(g) + Cl2(g) ⇔ 2NOCl(g)
    Init. Moles3.0 mole2.0 mole0 moles
    Init. Conc.6 M4 M0 M
    Complete Rxn016
    Equilib Conc0 + 2x1 + x6 - 2x

    Kc = 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

  5. Consider the following reaction:

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

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

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

    2NOCl(g) ⇔ 2NO(g) + Cl2(g)
    Init. Moles3.0 mole0 mole2.0 moles
    Init. Conc.6 M0 M4 M
    Complete Rxn604
    Equilib Conc6 - 2x0 + 2x4 + x

    Kc = 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

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

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

    HC4H7O2 H+ + C4H7O2-
    Init. Moles1.5 mole0 mole0 mole
    Init. Conc.5 M0 M0 M
    Complete Rxn500
    Equilib Conc5 - xxx

    Ka = 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

  7. 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, Ka = 1.8 x 10-4.
    Answer

    Ka = 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. Moles1.2 mole0 mole2.4 mole
    Init. Conc.0.6 M0 M1.2 M
    Complete Rxn0.601.2
    Equilib Conc0.6 - xx1.2 + x

    Ka = 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

  8. 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 Ksp = 6.5 x 10-6.
    Answer

    Ksp = 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-x2x

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

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

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

  9. 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 Ka = 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.

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

    HC2H3O2 H+ + C2H3O2-
    Init. Moles0.0085 mole0 mole0 mole
    Init. Conc.1.417 M0 M0 M
    Complete Rxn1.41700
    Equilib Conc1.417 - xxx

    Ka = 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

  10. Determine the pH after (a) 2 mL, (b) 3 mL, and (c) 4 mL of 0.5 M HCl has been titrated into 1.0 mL of 1.5 M NaOH.
    Answer (c)

    H+ + OH H2O   Kc = 1/Kw = 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)(1.5 mole/L NaOH) = 0.0015 moles of OH
    Initial [OH] = (0.0015 moles OH)/(0.014 L soln) = 0.107 M OH.

    A large Kc 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. Moles0.002 mole0.0015 mole-
    Init. Conc.0.143 M0.107 M-
    Complete Rxn0.0360-
    Equilib Conc0.036 + xx-

    Kc = 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   Kc = 1/Kw = 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)(1.5 mole/L NaOH) = 0.0015 moles of OH
    Initial [OH] = (0.0015 moles OH)/(0.013 L soln) = 0.115 M OH.

    A large Kc 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. Moles0.0015 mole0.0015 mole-
    Init. Conc.0.115 M0.115 M-
    Complete Rxn000-
    Equilib Concxx-

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

    pH = -log(10−7) = 7

    Answer (a)  |  

    H+ + OH H2O   Kc = 1/Kw = 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)(1.5 mole/L NaOH) = 0.0015 moles of OH
    Initial [OH] = (0.0015 moles OH)/(0.012 L soln) = 0.125 M OH.

    A large Kc 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. Moles0.001 mole0.0015 mole-
    Init. Conc.0.0833 M0.125 M-
    Complete Rxn00.0417-
    Equilib Concx0.0417 + x-

    Kc = 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- 1.38 = 12.62

  11. 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 Ka = 1.3 x 10−10
    Answer (c)

    HC6H5O + OH H2O + C6H5O   Kc = Ka/Kw = 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 Kc 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. Moles0.0084 mole.0096 mole-0 moles
    Init. Conc.0.227 M0.259 M-0
    Complete Rxn00.03246-0.227
    Equilib Concx0.03246 + x-0.227 - x

    Kc = 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   Kc = Ka/Kw = 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 Kc 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. Moles0.0084 mole.0084 mole-0 moles
    Init. Conc.0.24 M0.24 M-0
    Complete Rxn00-0.24
    Equilib Concxx-0.24 - x

    Kc = 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   Kc = Ka/Kw = 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 Kc 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. Moles0.0084 mole.0072 mole-0 moles
    Init. Conc.0.255 M0.218 M-0
    Complete Rxn0.0370-0.255
    Equilib Conc0.037 + xx-0.218 - x

    Kc = 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

  12. Calculate the pH for addition of 0.15 M NaOH to a solution iitially containing 15 mL of 0.1 M HCN (Ka=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   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.003 mole-0.002 mole
    Init. Conc.0.0333 M0.0667 M-0.0444 M
    Complete Rxn00.0334-0.0333
    Equilib Concx0.0334 + x-0.0333 - x

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

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

    pH = 14 - 1.47 = 12.52

    Answer (f)  |  

    HCN + OH H2O + CN   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.00225 mole-0.002 mole
    Init. Conc.0.0375 M0.0563 M-0.05 M
    Complete Rxn00.0188-0.0375
    Equilib Concx0.0188 + x-0.05 - x

    Kc = 4.9 x 104 = (0.5 - x)/{(x)(0.0188 + x)}   Assume x << 0.0188

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

    pH = 14 - 1.73 = 12.27

    Answer (e)  |  

    HCN + OH H2O + CN   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.0018 mole-0.002 mole
    Init. Conc.0.0405 M0.0563 M-0.0541 M
    Complete Rxn00.0158-0.0405
    Equilib Concx0.0158 + x-0.0405 - x

    Kc = 4.9 x 104 = (0.0405 - 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   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.0015 mole-0.002 mole
    Init. Conc.0.0429 M0.0429 M-0.0571 M
    Complete Rxn00-0.1
    Equilib Concxx-0.1 - x

    Kc = 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   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.0012 mole-0.002 mole
    Init. Conc.0.0455 M0.0364 M-0.0606 M
    Complete Rxn0.00910-0.097
    Equilib Conc0.0091 + xx-0.097 - x

    Kc = 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   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.00075 mole-0.002 mole
    Init. Conc.0.05 M0.025 M-0.0667 M
    Complete Rxn0.0250-0.0917
    Equilib Conc0.025 + xx-0.0917 - x

    Kc = 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   Kc = Ka/Kw = 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 Kc 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. Moles0.0015 mole0.000 mole-0.002 mole
    Init. Conc.0.06 M0 M-0.08 M
    Complete Rxn0.060-0.08
    Equilib Conc0.06 + xx-0.08 - x

    Kc = 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