Answers to Study Guide for CHEM 102 Test 4, Fall 2007

  1. Define the terms below:

    1. Bronsted-Lowry acid - a proton donor
    2. Bronsted-Lowry base - a proton acceptor
    3. Lewis acid - an electron pair acceptor
    4. Lewis base - an electron pair donor
    5. amphoteric - having both basic and acidic properties
    6. pH the -log[H+]
    7. pOH - the -log[OH-]

  2. What is the pH of an aqueous solution that is 0.00897 M in hydrogen ion (H+(aq)) concentration? What is the pOH?

    pH = -log [H+] = -log(0.00897) = -(-2.05) = 2.05
    pH + pOH = 14 so pOH = 14 -pH = 14 -2.05 =11.95

  3. What is the hydroxide ion concentration ( [OH-] ) of an aqueous solution that has a pH of 13.31? What is the hydrogen ion concentration ( [H+] )?

    pH = -log[H+] so [H+] = 10-pH = 10-13.31 = 4.90 x 10-14

    pOH = 14 - pH = 14 - 13.31 = 0.69
    pOH = -log[OH-] so [OH-] = 10-pOH = 10-0.69 = 0.204 M

  4. Put the acids in each part below in order from strongest to weakest.)

    1. HClO, HClO4, HClO2, HClO3

      HClO4> HClO3> HClO2> HClO

    2. H3SbO4, H3PO4, H3BiO4, H3AsO4

      H3PO4>H3AsO4>H3SbO4>H3BiO4

  5. Find the pOH and pH of the solutions below:

    1. a solution where [H+] = 3.86 x 10-16

      pH = 15.41, pOH = -1.41

    2. a solution where [OH-] = 6.22 x 10-6

      pOH = 5.21, pH = 8.79

    3. a solution where [OH-] = [H+]

      pH = 7, pOH = 7

    4. a solution where the hydroxide ion concentration is equal to double the hydrogen ion concentration

      The product of the concentrations must equal 1 x 10-14

      (2x)x = 1 x 10-14 = 2x2
      x2 = 5 x 10-15

      x = (5 x 10-15)1/2 = 7.07 x 10-8 M = [H+] pH = 7.15, pOH = 6.85

    5. a solution where [OH-] = 1.06 x 10-12

      pOH = 11.97, pH = 2.03

  6. What is the pH of a solution that is 4.850 M in NaNO2 and 0.750 M in HNO2? The KA for HNO2 is 4.5 x 10-4.
    HNO2(aq) º H+(aq) + NO2-(aq)

    Ka[HNO2]}/[NO2-] = {(4.5 x 10-4)(0.750)}/(4.850) = [H+]

    [H+] = 6.96 x 10-5

    pH = -log[H+] = - log(6.96 x 10-5) = 4.16

  7. What is the pH of a buffer solution that is 2.00 M in sodium benzoate (NaOOCC6H5) and 0.100 M in benzoic acid (HOOCC6H5)? The KA of benzoic acid is 6.3 x 10-5.

    HOOCC6H5(aq) <-----------> H+(aq) + -OOCC6H(aq)

    Ka = {[H+][-OOCC6H5]}/[HOOCC6H5]
    {Ka[HOOCC6H5]}/[-OOCC6H5] = [H+]
    {(6.3 x 10-5)(0.100)}/(2.00) = 3.15 x 10-6 = [H+]
    pH = -log[H+] = -log(3.15 x 10-6) = 5.50

  8. Using the buffer solution in the problem above, what is the pH if 15.00 mL of 2.50 M HCl(aq) is added to 75.00 mL of the buffer solution? What would be the pH if 25.00 mL of 1.45 M NaOH was added to 85.00 mL of the buffer solution in problem the problem above?

    Using the acid solution:

    Dilution problems:

    [HOOCC6H5] = {(0.100 M)(75.00 mL)}/(90.00 mL) = 0.0833 M
    [-OOCC6H5] = {(2.00 M)(75.00 mL)}/(90.00 mL) = 1.67 M

    Added acid:[H+] = {(2.50 M)(15.00 mL)}/(90.00 mL) = 0.417 M

    The added acid will decrease the concentration of the benzoate ion, -OOCC6H5,
    new [-OOCC6H5] = 1.67 - 0.417 = 1.253 M
    The added acid will increase the benzoic acid, HOOCC6H5,
    new [HOOCC6H5] = 0.0833 + 0.417 = 0.500 M
    Rearranging Ka = {[H+][-OOCC6H5]}/[HOOCC6H5] to solve for [H+] yields [H+] = {Ka[HOOCC6H5]}/[-OOCC6H5]
    [H+] = {(6.3 x 10-5)(0.500)}/(1.253) = 2.61 x 10-5
    pH = -log[H+] = -log(2.61 x 10-5) = 4.58

    For added base:
    Dilution problems:
    [-OOCC6H5] = {(2.00 M)(85.00 mL)}/(110.00 mL) = 1.545 M
    [HOOCC6H5] = {(0.100 M)(85.00 mL)}/(110.00 mL) = 0.0773 M
    added base:
    [OH-] = {(1.45 M)(25.00 mL)}/(110.00 mL) = 0.330 M
    The added base will decrease the acid concentration,
    0.0773 M - 0.330 M = -0.2527 M
    The negative number indicates that this buffer does not have enough capacity to handle this amount of base, leaving an excess of 0.2527 M in hydroxide. Therefore the pOH of the solution is:
    pOH = -log[OH-] = -log(0.2527) = 5.97
    pH = 14 - pOH = 14 - 5.97 = 8.03

  9. What is the percent ionization for the following solutions of acids?

    1. 2.50 M HC2H3O2, KA = 1.8 x 10-5

      HC2H3O2(aq) º H+(aq) + C2H3O2-(aq)
      KA = {[H+][C2H3O2-]}/[HC2H3O2] = x2/[HC2H3O2]
      KA[HC2H3O2] = x2 = (1.8 x 10-5)(2.50) = 4.5 x 10-5
      x = (4.5 x 10-5)1/2 = 6.71 x 10-3 = [H+] = [C2H3O2-]
      % ionization = {(6.71 x 10-3)/2.50} x 100 = 0.268%
      quadratic not needed

    2. 0.00500 M HCOOH, KA = 1.7 x 10-4

      18.4%, quadratic needed

    3. 0.1500 M C6H5OH, KA = 1.3 x 10-10

      0.00294%, quadratic not needed

    4. 0.000515 M HF(aq), KA = 7.1 x 10-4

      completely ionized!! might be small difference with quadratic

    5. 0.00180 M HC9H7O4, KA = 3.0 x 10-4

      40.8%, quadratic needed

  10. What is the sulfate concentration in a 14.00 M H2SO4(aq) solution? Ka1 = extremely large, ~100 % dissociation, KA2 = 1.3 x 10-2.

    H2SO4(aq) 6 H+(aq) + HSO4-(aq), [H+] = [HSO4-] = 14.00 M
    HSO4-(aq) º H+(aq) + SO42-(aq)

    KA2 = {[H+][SO42-]}/[HSO4-] but [H+] = [HSO4-], since second dissociation provides an insignificant amount of H+ compared to first dissociation thus KA2 = [SO42-] = 1.3 x 10-2.

  11. What is the arsenate concentration (AsO43-) in 3.00 M arsenic acid (H3AsO4)? KA1 = 6.0 x 10-3, KA2 = 1.05 x 10-7, KA3 = 3.0 x 10-12

    H3AsO4(aq) º H+(aq) + H2AsO4-(aq)
    % ionization calculation:
    KA1 = {[H+][H2SO4-]}/[H3AsO4] = x2/[H3AsO4]
    KA1[H3AsO4] = x2
    (6.0 x 10-3)(3.00) = 1.80 x 10-2 = x2
    (1.8 x 10-2)1/2 = x = 0.134
    % ionization = {0.134/3.00} x 100 = 4.47%, quadratic not needed, thus [H+] = [H2AsO4-] = 0.134 M

    H2AsO4-(aq)º H+(aq) + HAsO42-(aq)
    KA2 = {[H+][HAsO42-]}/[H2AsO4-] but [H+] = [H2AsO4-] so KA2 = [HAsO42-]

    HAsO42-(aq) º H+(aq) + AsO43-(aq)
    KA3 = {[H+][AsO43-]}/[HAsO42-] KA3[HAsO42-]/[H+] = [AsO43-] but [H+] = 0.134 and [HAsO42-] = KA2
    so KA3KA2/[H+] = [AsO43-]
    {(3.0 x 10-12)(1.05 x 10-7)}/(0.134) = 2.35 x 10-18 M = [AsO43-]

  12. What is the pH of a solution that is

    HYDROLYSIS!!

    1. 6.50 M in NaC2H3O2, HC2H3O2 KA = 1.8 x 10-5

      C2H3O2-(aq) + H2O º HC2H3O2(aq) + OH-(aq)
      KB = (1.0 x 10-14)/KA = (1.0 x 10-14)/(1.8 x 10-5)
      KB = 5.56 x 10-10
      KB = {[H2C2H3O2][OH-]}/[C2H3O2-] = x2/[C2H3O2-]
      KB[C2H3O2-] = x2 = (5.56 x 10-10)(6.50)
      x2 = 3.614 x 10-9<
      x = (3.614 x 10-9)1/2 = 6.01 x 10-5 = [OH-]
      pOH = -log[OH-] = -log(6.01 x 10-5) = 4.22
      pH = 14 -pOH = 14 -4.22 = 9.78

    2. 2.60 M in NaCN, HCN KA = 2.1 x 10-9

      same method as a, pH = 11.55

    3. 4.50 M in NH4Cl, NH3 KB = 1.8 x 10-5

      NH4+(aq) º H+(aq) + NH3(aq)
      KA = Kw/KB = (1.0 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10
      KA = {[H+][NH3]}/[NH4+] = x2/[NH4+]
      KA[NH4+] = x2 = (5.56 x 10-10)(4.50) = 2.502 x 10-9
      x = (2.502 x 10-9)1/2 = 5.00 x 10-5 = [H+]
      pH = -log[H+] = -log(5.00 x 10-5) = 4.30

    4. 2.34 M in NaIO3, HIO3 KA = 1.58 x 10-1

      same method as a, pH = 7.59

    5. 2.95 M in NaN3, HN3 KA = 1.9 x 10-5

      same method as a, pH = 9.60

  13. What is the pH of a buffer that is 3.93 M in NaCN and 0.951 M in HCN? If 10.00 mL of 0.1500 M HCl is added to 100.00 mL of this buffer what is the pH? If 15.00 mL of 0.6000 M NaOH is added to 95.00 mL of this buffer what is the pH?

    HCN(aq)º H+(aq) + CN-(aq) KA = 2.1 x 10-9

    KA = {[H+][CN-]}/[HCN] = 2.1 x 10-9
    KA[[HCN]/[CN-] = [H+] = {(2.1 x 10-9)(0.951)/(3.93)
    [H+] = 5.08 x 10-10
    pH = -log[H+] = -log(5.08 x 10-10) = 9.29

    Added acid problem:
    Dilutions:
    [CN-] = {(3.93 M)(100.00 mL)}/(110.00 mL) = 3.57 M
    [HCN] = {(0.951 M)(100.00 mL)}/(110.00 mL) = 0.865 M
    added acid:
    [H+] = {(0.1500 M)(10.00 mL)}/(110.00 mL) = 0.0136 M
    Added acid will decrease [CN-]
    new [CN-] = 3.57 - 0.0136 = 3.5564 M
    Added acid will increase [HCN]
    new [HCN] = 0.865 + 0.0136 = 0.8786 M
    KA[HCN]/[CN-] = [H+] = {(2.1 x 10-9)(0.8786)}/(3.5564)
    [H+] = 5.19 x 10-10, pH = 9.28

    Added base problem:
    Dilutions:
    [CN-] = {(3.93 M)(95.00 mL)}/(110.00 mL) = 3.39 M
    [HCN] = {(0.951 M)(95.00 mL)}/(110.00 mL) = 0.821 M
    added base:
    [OH-] = {(0.6000 M)(15.00 mL)}/(110.00 mL) = 0.0818 M
    Added base will decrease [HCN] and increase [CN-]
    new [CN-] = 3.39 + 0.0818 = 3.4718 M
    new [HCN] = 0.821 -0.0818 = 0.7392 M
    A[HCN]/[CN-] = {(2.1 x 10-9)(0.7392)}/(3.4718) = [H+]
    [H+] = 4.47 x 10-10, pH = 9.35

    HCN(aq)º H+(aq) + CN-(aq)
  14. Design buffers with the at pH = 5.00 with the following

    1. a system 1.00 M in HC2H3O2, KA = 1.8 x 10-5

      pH 5 means [H+] = 1.00 x 10-5
      HC2H3O2(aq) º H+(aq) + C2H3O2-(aq)
      KA = {[H+][C2H3O2-]}/[HC2H3O2]
      KA[HC2H3O2]/[H+] =
      {(1.8 x 10-5)(1.00)}/(1.00 x 10-5)=[C2H3O2-] = 1.80 M

    2. a system 0.750 M in NaCN, HCN KA = 2.1 x 10-9

      [HCN] = 1.575 x 10-4 M

    3. a system 1.00 M in NH3, KB = 1.8 x 10-5

      [NH4+] = 17985.6 M, a ridiculous answer, since the molarity of water in itself is the maximum at 55.55 M. You cannot make a buffer at this pH with these concentrations with this system .

    4. a system 1.00 M in NH4Cl, NH3 KB = 1.8 x 10-5

      [NH3] = 5.56 x 10-5 M. Although possible to make this buffer, it actually has almost no capacity.

    5. a system 1.00 M in NaC2H3O2, HC2H3O2 KA = 1.8 x 10-5

      [HC2H3O2] = 0.556 M

  15. What are the equilibrium concentrations of the reactants and products below if a solution was made that was initially 1.00 M in Cl3CCOOH(aq), 0.500 M in H+(aq) and 0.325 M in Cl3CCOO-(aq)? KA = 1.29 x 10-1

    Cl3CCOOH(aq) º H+(aq) + Cl3CCOO-(aq)

    KA = {[H+][Cl3CCOO-]}/[Cl3COOH]
    > KA = {[0.500 + x][0.325 + x]}/[1.00 - x]
    KA[1.00 - x] = [0.500 + x][0.325 + x]
    0.129 - 0.129x = x2 + 0.825x + 0.1625
    0 = x2 + 0.954x + 0.0335
    x = {-b + (b2 - 4ac)1/2}/2a
    x = {-0.954 + [(0.954)2 -4(1)(0.0335)]1/2}/2(1)
    x = {-0.954 + (0.910116 - 0.134)1/2}/2
    x = {-0.954 + (0.776116)1/2}/2
    x = {-0.954 + 0.881}/2
    x = -0.0365 or -0.9175
    The only root that makes sense is the first since the second gives negative concentrations for [H+] and [Cl3COO-]

    Thus the equilibrium concentrations are:

    [Cl3COOH] = 1.00 - (-0.0365) = 1.0365 = 1.04 M
    [Cl3COO-] = 0.325 + (-0.0365) = 0.2885 = 0.288 M
    [H+] = 0.500 + (-0.0365) = 0.4635 = 0.464 M

  16. What is the molar solubility of PbSO4 if the Ksp for PbSO4 is 1.3 x 10-8?

    PbSO4(s) º Pb2+(aq) + SO42-(aq), Ksp = [Pb2+][SO42-]

    Since one mole yields 1 mole of Pb2+ and 1 mole of SO42- Ksp = x2 therefore, x, the molar solubility is equal to
    (Ksp)1/2 = (1.3 x 10-8)1/2 = 1.14 x 10-4 moles/L

  17. What is Ksp of CaF2 if the molar solubility of CaF2 is 2.136 x 10-4 M?

    CaF2(s) º Ca2+(aq) + 2 F-(aq), Ksp = [Ca2+][F-]2

    therefore Ksp =(x)(2x)2 = 4x3 but x is equal to the molar solubility so Ksp = 4(2.136 x 10-4)3 = 3.898 x 10-11

  18. If the molar solubility of Fe(OH)2 is 5.85 x 10-6, what is Ksp for Fe(OH)2?

    Fe(OH)2(s) º Fe2+(aq) + 2 OH-(aq)

    Ksp = [Fe2+][OH-]2

    Using the same arguments as in the previsou problems, 5.85 x 10-6 moles of Fe2+ will be present, but twice as many moles (1.17 x 10-5) of OH-will be present.

    Ksp = (5.85 x 10-6)(1.17 x 10-5)2 = 8.01 x 10-16

  19. If the Ksp for Mn(OH)2 is 2.00 x 10-13, what is the molar solubility of Mn(OH)2?

    Mn(OH)2(s) º Mn2+(aq) + 2 OH-(aq)

    For every Mn(OH)2 that dissociates one Mn2+ is formed but two OH- s are formed thus the molar solubility may be calculated by using x for the Mn2+ concentration and 2x for the OH- concentration. The quantity "x" will be equal to the molar solubility.

    Ksp = x(2x)2 = 4x3
    Ksp/4 = x3
    (Ksp/4)1/3 = x
    (2.00 x 10-13/4)1/3 = x
    (5.00 x 10-14)1/3 = x = 3.68 x 10-5 mol/L

  20. What is the molar solubility of AgCl in 1.00 M in NaCN?

    KspAgCl = 1.6 x 10-10, KfAg(CN)2- = 1.0 x 1021

    (s) º Ag+(aq) + Cl-(aq)
    Ag+(aq) + 2 CN-(aq) º Ag(CN)2-(aq)
    Overall rxn: AgCl(s) + 2 CN-(aq)º Ag(CN)2-(aq) + Cl-(aq)

    K = KspKf = {[Ag(CN)2-][Cl-]}/[CN-]2
    K = (1.6 x 10-10)(1.0 x 1021) = 1.6 x 1011
    1.6 x 1011 =(s)(s)/(1.0-2s)2
    1.6 x 1011 = s2/(1.0-2s)2
    (1.6 x 1011)1/2 = 4.0 x 105 = s/(1.0 -2s)
    4.0 x 105 - (8.0 x 105)s = s
    4.0 x 105 = (8.0 x 105)s + s
    s = 0.500 moles/L for molar solubility

  21. What is the molar solubility of Zn(OH)2? What is its molar solubility in a solution buffered at pH 14? At pH 5? The solubility product for Zn(OH)2 is 2.1 x 10-14.

    Zn(OH)2(s) º Zn2+(aq) + 2 OH-(aq)

    Ksp = [Zn2+][OH-]2
    Ksp = x(2x)2 = 4x3
    Ksp/4 = 5.25 x 10-15 = x3
    x = (5.25 x 10-15)1/3 = 1.74 x 10-5 M = molar solubility

    In a pH 14 buffer pOH = 14-14 = 0 so [OH-] = 100 = 1.00
    so Ksp/[OH-]2 = [Zn2+] = (2.1 x 10-14)/(1.00)2 = 2.1 x 10-14 which is the molar solubility

    In pH 5 buffer, pOH = 14 -5 = 9, [OH-] = 1.00 x 10-9
    [Zn2+] = Ksp/[OH-]2 = (2.1 x 10-14)/(1.00 x 10-9)2
    [Zn2+] = 2.1 x 104 M (larger than maximum molarity, very soluble)

  22. Define:

    1. Ksp - the solubility product equal to the reaction quotient of the equilibrium expression of a salt of limited aqueous solubility

    2. Kf - formation constant for complex ions

    3. buffer - a solution that resist pH change

    4. buffer capacity - refers to how much added acid or base a buffer can absorb before coming non-functional

    5. Kd dissociation constant for a complex ion, the inverse of Kf, i.e. 1/Kf

    6. common ion effect - refers to the shift in equilibrium that occurs when an ion common to the equilibrium is added (i.e. the effect of LeChatelier's principle)

    7. pKa = -logKa, which during a pH titration of an acid is equal to the pH at the half-equivalence point

    8. pKb = -lofKb

    9. titration - a volumetric method for determining concentration of a material in solution

    10. molar soulubility - how many moles of a material can be dissolved in 1.00 L of water

    11. complex ion - an ion formed by the association of another species with an metal ion
      (eg. Ag+(aq) + 2 NH3(aq) <-------> Ag(NH3)2+(aq))

  23. What is the solubility of AgCl in 5.00 M NH3?
    AgCl(s)
    K - s

    <--------->
    Ag+(aq)
    s

    +
    Cl-(aq)
    s

    Ksp = 1.8 x 10-10
    Ag+(aq)
    s
    +
    2 NH3(aq)
    5.00 - 2s
    <-------->
    		 Ag(NH3)2+(aq)
    s
    Kf = 1.7 x 107


    KC = KspKf = (1.8 x 10-10)(1.7 x 107) = 3.06 x 10-3

    KC = [Ag+][Cl-][Ag(NH3)2+]/[Ag+][NH3]2

    KC = [Cl-][Ag(NH3)2+]/[NH3]2

    KC = s2/(5.00 - 2s)2

    (KC)1/2 = (s2/(5.00 - 2s)2)1/2 = s/(5.00 - 2s) = (3.06 x 10-3)1/2 = 5.53 x 10-2

    s/(5.00 - 2s) = 5.53 x 10-2

    s = (5.00 -2s)(5.53 x 10-2

    s = 0.2765 - 0.1106s

    s + 0.1106 s = 0.2765

    1.1106s = 0.2765

    s = 0.2765/1.1106 = 0.249 M

  24. Explain how the pKA of an acid may be determined with a pH titration

    In general a weak acid may be reperesented by the equation:

    HA(aq) <--------> H+(aq) + A-(aq)


    thus KA = [H+][A-]/[HA]

    But at the half-equivalence point, half of the the HA has been consumed, and the concentrations of [HA] and [A-] will be equal

    thus at the half-equivalence point KA = [H+]

    so at the half equivlance point -logKA = -log[H+]

    and pKA = pH

  25. Are the solutions below acidic or basic?

    1. 3.00 M Na+F- - contains the conjugate base of the weak acid HF, solution will be basic

    2. 3.00 M Na+C2H3O2- - contains the conjugate base of the weak acid HC2H3O2, solution will be basic

    3. 3.00 M NH4+Cl- - contains the conjugate acid of the weak base NH3, solution will be acidic

    4. 3.00 M CH3CH2NH3+Cl- - contains the conjugate acid of the weak base CH3CH2NH2, solution will be acidic.

  26. What is the molar solubility of AgCl in 1.00 M in NaCN?

    KspAgCl = 1.6 x 10-10, KfAg(CN)2- = 1.0 x 1021

    (s) º Ag+(aq) + Cl-(aq)
    Ag+(aq) + 2 CN-(aq) º Ag(CN)2-(aq)
    Overall rxn: AgCl(s) + 2 CN-(aq)º Ag(CN)2-(aq) + Cl-(aq)

    K = KspKf = {[Ag(CN)2-][Cl-]}/[CN-]2
    K = (1.6 x 10-10)(1.0 x 1021) = 1.6 x 1011
    1.6 x 1011 =(s)(s)/(1.0-2s)2
    1.6 x 1011 = s2/(1.0-2s)2
    (1.6 x 1011)1/2 = 4.0 x 105 = s/(1.0 -2s)
    4.0 x 105 - (8.0 x 105)s = s
    4.0 x 105 = (8.0 x 105)s + s
    s = 0.500 moles/L for molar solubility

  27. What is the molar solubility of Zn(OH)2? What is its molar solubility in a solution buffered at pH 14? At pH 5? The solubility product for Zn(OH)2 is 2.1 x 10-14.

    Zn(OH)2(s) º Zn2+(aq) + 2 OH-(aq)

    Ksp = [Zn2+][OH-]2
    Ksp = x(2x)2 = 4x3
    Ksp/4 = 5.25 x 10-15 = x3
    x = (5.25 x 10-15)1/3 = 1.74 x 10-5 M = molar solubility

    In a pH 14 buffer pOH = 14-14 = 0 so [OH-] = 100 = 1.00
    so Ksp/[OH-]2 = [Zn2+] = (2.1 x 10-14)/(1.00)2 = 2.1 x 10-14 which is the molar solubility

    In pH 5 buffer, pOH = 14 -5 = 9, [OH-] = 1.00 x 10-9
    [Zn2+] = Ksp/[OH-]2 = (2.1 x 10-14)/(1.00 x 10-9)2
    [Zn2+] = 2.1 x 104 M (larger than maximum molarity, very soluble)