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  • 7/25/2019 Unit 8 Homework Answers[1]

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    AP Chemistry

    Unit 8- Homework Problems

    Kaand Kb

    Acid & Base Theory1. What is the definition of an Arrhenius acid? H+releaser2. What is the definition of an Arrhenius base?OH-releaser

    3. What is the definition of a Bronsted acid? H+donator4. What is the definition of a Bronsted base? H+acceptor

    5. What is the conjugate base of:a. HNO2 NO2

    -1b. H2SO3 HSO3

    -1

    c. NH4+1 NH3

    d. HCN CN-1

    6. What is the conjugate acid of:a. Cl

    -1 HCl

    b. HPO4-2

    H2PO4-1

    c. OH-1 H2Od. N2H4 N2H5

    +1

    7. In each equation below, identify each species as the acid, base, conjugate acid or conjugate base .a. HNO3 + PO4

    -3 NO3

    -1 + HPO4

    -2

    Acid Base CB CAb. N3

    -1 + H2SO4 HSO4-1 + HN3

    Base Acid CB CA

    c. NH3 + H2PO3-1 NH4

    +1 + HPO3-2

    Base Acid CA CB

    d. H2C2O4 + OH-1

    H2O + HC2O4-1

    Acid Base CA CB8. For each species below, describe whether a solution of it would be acidic, basic, or neutral.

    a. HCl Acid h. CaCl2 Neutralb. KOH Base i. Na2CO3 Base

    c. NaCl Neutral j. LiBr Neutrald. NaC2H3O2 Base k. KF Basee. NH4Cl Acid l. NaNO2 Base

    f. Na3PO4 Base m. K2SO3 Baseg. KNO3 Neutral n. KI Neutral

    9. What is an amphiprotic substance? List 4 of them.A substance that can act as either a Bronsted acid or Bronsted base

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    Auto-Ionization of water, pH, and strong acids1. What is the equation for the auto-ionization of water? 2 H2O (l) H3O

    +1 (aq) + OH-1(aq)

    2. What is the equation for Kwand what is its value? Kw= [H3O+][OH-1] = 1x10-14

    3. Because the pH scale is a logarithmic scale, every change in 1 is really a change in how much?104. Using the pH square and the equations learned in class, fill in the blank spots in the chart below:

    pH [H3O

    +1

    ] [OH

    -1

    ] pOH3.5 3.16x10-4 3.16x10-11 10.5

    8.04 9.2x10-9

    1.10x10-6

    5.96

    3.67 2.14x10-4 4.7x10-11 10.33

    5.8 1.58x10-6 6.31x10-9 8.2

    5. According to the leveling effect:

    a. What is the strongest acid species in water? H3O+1

    b. What is the strongest base species in water? OH-16. What is the % dissociation of strong acids and bases in water? 100%

    7. List 6 strong acids. HCl, HBr, HI, HNO3, HClO4, H2SO48. List 6 strong bases. LiOH, NaOH, KOH, RbOH, CsOH, Ba(OH)2, Sr(OH)2

    Weak acids & bases, Ka, Kb, and pH1. How are weak acids and bases different from strong ones? Weak acids dissociate

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    6. For a 0.055 M solution of HClO (Ka= 2.9x10-8) calculate:

    a. pHHA (aq) + H2O (l) H3O

    +1(aq) + A-1(aq)

    I 0.055 0 0C -x +x +xE 0.055 x x

    2.9x10-8

    = x2

    /0.055 x = 3.99x10-5

    pH = -log (3.99x10-5

    ) = 4.40b. % dissociation

    3.99x10-5x 100% = 0.0725%0.055

    7. For a 0.0034 M solution of C5H5N (Kb= 1.7x10-9) calculate:

    a. pHB (aq) + H2O (l) OH

    -1(aq) + HB+1(aq)

    I 0.0034 0 0C -x +x +x

    E 0.0034 x x1.7x10

    -9= x

    2/0.0034 x = 2.40x10

    -6 pOH = -log (2.40x10

    -6) = 5.62 pH = 8.38

    b. % dissociation2.40x10-6x 100% = 0.071%

    0.0034

    8. For a 0.15 M solution of H2C6H6O6(Ka1= 8x10-5Ka2= 1.6 x10

    -12) calculate:

    a. pHH2A (aq) + H2O (l) H3O

    +1(aq) + HA-1(aq)

    I 0.15 0 0C -x +x +xE 0.15 x x

    8x10-5= x2/0.15 x = 3.46x10-3 pH = -log (3.46x10-3) = 2.46b. Equilibrium concentrations of:

    i) HC6H6O6-1

    x = 3.46x10

    -3

    ii) C6H6O6-2 Ka2= 1.6x10

    -12

    9. For a 0.025 M solution of diprotic base, B (Kb1= 4x10-6

    , Kb2= 8x10-10

    ) calculate:a. pHB (aq) + H2O (l) OH

    -1(aq) + HB+1(aq)

    I 0.025 0 0C -x +x +x

    E 0.025 x x4x10-6= x2/0.025 x = 3.16x10-4 pOH = -log (3.16x10-4) = 3.5 pH = 10.5

    b. Equilibrium concentrations of:i) HB+ x = 3.16x10-4

    ii) H2B

    +2

    Kb2= 8x10

    -10

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    Acid-Base InteractionsStrong-Strong

    1. What is the pH when 50 mL of 0.1 M HCl is mixed with 50 mL of 0.1 M NaOH?Acid = (50/100)(0.1 M) = 0.05 M Base = (50/100)(0.1 M) = 0.05 M

    Equal amounts of strong acid/base so should be neutral

    H3O+1(aq) + OH-1(l) 2 H2O

    I 0.05 0.05C -0.05 -0.05

    E 0 0pH = 7

    2. What is the pH when 250 mL of 0.2 M HNO3is mixed with 500 mL of 0.1 M KOH?Acid = (250/750)(0.2 M) = 0.0667 M Base = (500/750)(0.1 M) = 0.0667 M

    Equal amounts of strong acid/base so should be neutral

    H3O+1(aq) + OH-1(l) 2 H

    2O

    I 0.0667 0.0667C -0.0667 -0.0667

    E 0 0pH = 7

    3. What is the pH when 100 mL of 0.001 M HBr is mixed with 50 mL of 0.001 M NaOH?

    Acid = (100/150)(0.001 M) = 6.67x10-4M Base = (50/150)(0.001 M) = 3.33x10-4M

    More acid than base so should be acidic

    H3O+1(aq) + OH-1(l) 2 H2O

    I 6.67x10-4 3.33x10-4C -3.33x10-4 -3.33x10-4E 3.33x10-4 0

    pH = -log (3.33x10-4

    ) = 3.47

    4. What is the pH when 40 mL of 0.01 M HI is mixed with 120 mL of 0.008 M CsOH?Acid = (40/160)(0.01 M) = 0.0025 M Base = (120/160)(0.008 M) = 0.006 M

    More base than acid so should be basic

    H3O+1(aq) + OH-1(l) 2 H2O

    I 0.0025 0.006C -0.0025 -0.0025

    E 0 0.0035pOH = -log (0.0035) = 2.46 pH = 11.54

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    5. What is the pH when 50 mL of 0.0025 M H2SO4is mixed with 75 mL of 0.0040 M LiOH?

    Acid = 2*(50/125)(0.0025 M) = 0.0020 M Base = (75/125)(0.0040 M) = 0.0024 M

    More base than acid so should be basic

    H3O+1

    (aq) + OH-1

    (l) 2 H2OI 0.0020 0.0024

    C -0.0020 -0.0020E 0 0.0004

    pOH = -log (0.0004) = 3.4 pH = 10.6

    6. What is the pH when 20 mL of 0.00050 M HClO4is mixed with 20 mL of 0.00040 M Ba(OH)2?Acid = (20/40)(0.00050 M) = 0.00025 M Base = 2* (20/40)(0.0004 M) = 0.0004 M

    More base than acid so should be basic

    H3O

    +1

    (aq) + OH

    -1

    (l)

    2 H2OI 0.00025 0.0004C -0.00025 -0.00025E 0 0.00015

    pOH = -log (0.00015) = 3.8 pH = 10.2

    Strong-Weak

    7. What is the pH when 25 mL of 0.25 M HCl is mixed with 25 mL of 0.25 M NH3? (Kb= 1.8x10-5

    )?Acid = (25/50)(0.25 M) = 0.125 M Base = (25/50)(0.25 M) = 0.125 M

    Equal amounts of strong acid/weak base so should be acidicShortcut hint- Equivalence point of a strong acid weak base gives a normal Kaproblem

    (you can skip the actual rxn and go straight to Kaof conj. acid of weak base)

    ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)

    H3O+1 (aq) + NH3(aq) H2O (l) + NH4

    +1(aq)I 0.125 0.125 0

    C -0.125 -0.125 +0.125E 0 0 0.125

    NOW FLIP RXN AND DO Ka (Ka= 1x10-14/1.8x10-5) = 5.56x10-10

    NH4+1(aq)+ H2O (l) H3O

    +1(aq) + NH3(aq)

    I 0.125 0 0C -x +x +xE 0.125 x x

    5.56x10-10= x2/0.125 x = 8.33x10-6 pH = -log (8.33x10-6) = 5.08

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    8. What is the pH when 20 mL of 0.020 M HNO3is mixed with 20 mL of 0.030 M C6H5NH2(Kb= 4x10-10)?

    Acid = (20/40)(0.02 M) = 0.01 M Base = (20/40)(0.030 M) = 0.015 M

    Not at equivalence point. Not at midpoint.

    ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)

    H3O+1

    (aq) + C6H5NH2(aq) H2O (l) + C6H5NH3+1

    (aq)I 0.01 0.015 0

    C -0.01 -0.01 +0.01E 0 0.005 0.01

    NOW FLIP RXN AND DO Ka (Ka= 1x10-14/4x10-10) = 2.5x10-5

    C6H5NH3+1(aq) + H2O (l) H3O

    +1(aq) + C6H5NH2(aq)

    I 0.01 0 0.005C -x +x +x

    E 0.01 x 0.0052.5x10

    -5= x(0.005/0.01) x = 5x10

    -5 pH = -log (5x10

    -5) = 4.30

    Shortcut hint- Simple buffer problem. Write Kaequation and plug in amounts after strong acid isneutralized (you can skip the actual rxn)

    Strong acid: 0.01 M neutralized is turned into a weak acidWeak base: 0.015 M -0.01 M neutralized by acid so 0.005 left

    Now plug straight into 2ndICE table above and solve

    9. What is the pH when 50 mL of 0.040 M NaOH is mixed with 50 mL of 0.040 M HN3(Ka= 1.9x10-5

    )?Base = (50/100)(0.04 M) = 0.02 M Acid = (50/100)(0.04 M) = 0.02 M

    Equal amounts of strong base/weak acid so should be basic

    ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)OH-1 (aq) + HN3(aq) H2O (l) + N3

    -1(aq)

    I 0.02 0.02 0C -0.02 -0.02 +0.02

    E 0 0 0.02

    NOW FLIP RXN AND DO Kb (Kb= 1x10-14/1.9x10-5) = 5.26x10-10

    N3-1(aq)+ H2O (l) OH

    -1(aq) + HN3(aq)I 0.02 0 0

    C -x +x +x

    E 0.02 x x

    5.26x10-10= x2/0.02 x = 3.24x10-6pOH = -log (3.242x10-6) = 5.49 pH = 8.51

    Shortcut hint- Equivalence point of a strong base weak acid gives a normal Kbproblem(you can skip the actual rxn and go straight to Kbof conj. base of weak acid)

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    10. What is the pH when 120 mL of 0.0075 M KOH is mixed with 200 mL of 0.0050 M HOCl (Ka= 3.5x10-8)?

    Base = (120/320)(0.0075 M) = 0.00281 M Acid = (200/320)(0.005 M) = 0.00312 M

    Not at equivalence point. Not at midpoint.

    ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)

    OH-1

    (aq) + HOCl (aq) H2O (l) + OCl-1

    (aq)I 0.00281 0.00312 0

    C -0.00281 -0.00281 +0.00281E 0 0.00031 0.00281

    NOW FLIP RXN AND DO Kb (Kb= 1x10-14/3.5x10-8) = 2.89x10-7

    OCl-1(aq) + H2O (l) OH

    -1(aq) + HOCl(aq)

    I 0.00281 0 0.00031C -x +x +x

    E 0.00281 x 0.00031

    2.89x10-7= x(0.00031/0.00281) x = 2.59x10-6

    pOH = -log (2.59x10

    -6

    ) = 5.59 pH = 8.41

    Shortcut hint- Simple buffer problem. Write Kaequation and plug in amounts after strong base isneutralized (you can skip the actual rxn)

    Strong base = 0.00281 turns into weak baseWeak acid = 0.00312 neutralized and left with 0.00312-0.00281 = 0.00031

    So:HOCl (aq) + H2O (l) H3O

    +(aq) + OCl

    -1(aq)

    I 0.00031 0 0.00281C -x +x +xE 0.00031 x 0.00281

    3.5x10-8= x(0.00281/0.00031) x = 3.86x10-9 pH = 8.41

    Midpoint

    11. What is the pH when 25 mL of 0.050 M HCl is mixed with 50 mL of 0.050 C5H5N (Kb= 1.5x10-9)?

    acid = (25/75)(0.050 M) = 0.0166 M base = (50/75)(0.050 M) = 0.0333 M

    At midpoint! (Half as much strong acid as base)

    pH = pKa so Ka= (1x10-14/1.5x10-9) = 6.67x10-6 so pH = - log (6.67x10-6) = 5.18

    12. What is the pH when 40 mL of 0.80 M NaOH is mixed with 80 mL of 0.8 M HNO2(Ka= 4.5x10

    -4

    )?base = (40/120)(0.80 M) = 0.266 M acid = (80/120)(0.80 M) = 0.5333 M

    At midpoint! (Half as much strong base as acid)

    pH = pKa so pH = - log (4.5x10-4) = 3.35

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    13. What is the pH when 30 mL of 0.006 M HNO3is mixed with 30 mL of 0.012 M (CH3)3N (Kb= 7.4x10-5)?

    acid = (30/60)(0.006 M) = 0.003 M base = (30/60)(0.012 M) = 0.006 M

    At midpoint! (Half as much strong acid as base)

    pH = pKa so Ka= (1x10-14

    /7.4x10-5

    ) = 1.35x10-10

    so pH = - log (1.35x10-10

    ) = 9.87

    14. What is the pH when 100 mL of 0.0078 M KOH is mixed with 100 mL of 0.0156 M C6H5CO2H (Ka= 6.3x10-5)

    base = (100/200)(0.0078 M) = 0.0039 M acid = (100/200)(0.0156 M) = 0.0078 M

    At midpoint! (Half as much strong base as acid)

    pH = pKa so pH = - log (6.3x10-5) = 4.2

    Kspand pH1. What is the pH of a saturated solution of Mn(OH)2 Ksp= 1.9x10

    -13?

    Ksp= [Mn+2][OH-1]2 1.9x10-13= 4x3

    x= 3.67x10-5 so [OH-1] = 7.32 x10-5 so pOH = 4.14 so pH = 9.86

    2. What is the pH of a saturated solution of Ca(OH)2Ksp= 5.5x10-5?

    Ksp= [Ca+2][OH-1]2 5.5x10-5= 4x3

    x= 0.0240 so [OH-1] = 0.0480 so pOH = 1.32 so pH = 12.68

    3. At what pH will a 0.0075 M solution of Mg(OH)2begin to form a precipitate (Ksp= 5.6x10-12)?

    Ksp= [Mg+2][OH-1]2 5.6x10-12= [0.0075][x]2

    x= 2.73x10-5

    so [OH-1

    ] = 2.73x10-5

    so pOH = 4.56 so pH = 9.44

    4. At what pH will a 0.00042 M solution of Pb(OH)2begin to form a precipitate (Ksp= 1.4x10-15)?

    Ksp= [Pb+2][OH-1]2 1.4x10-15= [0.00042][x]2

    x= 1.83x10-6

    so [OH-1

    ] = 1.83x10-6

    so pOH = 5.74 so pH = 8.26

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    Titration Curves1. Draw a titration curve of a strong base being titrated by a strong acid. Be sure to label a pH of 7 in the

    appropriate spot.

    2. Draw a titration curve of a strong acid being titrated by a strong base. Be sure to label a pH of 7 in theappropriate spot.

    3. Draw a titration curve of a weak base being titrated by a strong acid. Be sure to label a pH of 7 in theappropriate spot.

    pH

    mL acid added

    pH

    mL acid added

    pH

    mL acid added

    7

    7

    7

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    4. Draw a titration curve of a weak acid being titrated by a strong base. Be sure to label a pH of 7 in theappropriate spot.

    7. What is the significance of the midpoint of a titration? It is pH = pKa

    pH

    mL acid added

    7

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    8. If 40 mL of weak acid HA was titrated as the graph below shows, calculate:a. Show the pH at the equivalence point Eq. Pt @30 mL pH = 8.5

    b. Show the pH at the midpoint Mid Pt @15 mL pH = 5

    c. Calculate the original M of HA MaVa= MbVbMa(40 mL) = (0.10 M)(30 mL) Ma= 0.075 M

    d. Calculate Kaof HA pKa= pH at midpoint so pKa= 5 or Ka= 1x10-5

    Titration of 40 mL of ??? M weak acid HA

    with 0.10 M NaOH

    Eq. Pt @30 mL pH = 8.5

    Mid Pt @15 mL pH = 5

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    9. If 25 mL of weak base B was titrated as the graph below shows, calculate:a. Show the pH at the equivalence pointEq. Pt @60 mL pH = 4

    b. Show the pH at the midpointMid Pt @ 30 mL pH = 8

    c. Calculate the original M of B MaVa= MbVb (0.50 M)(60 mL) = (? M)(25 mL) Mb= 1.2 Md. Calculate Kbof B pKa= pH at midpoint so pKb= pOH at midpoint or Kb= 1x10

    -6

    Titration of 25 mL of ??? M weak base B

    with 0.50 M HCl

    Eq. Pt @60 mL pH = 4

    Mid Pt @ 30 mL pH = 8

    HCl

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    Buffers1. What is the definition of a buffer? A solution that resists changes in pH.2. What kind of an acid and base are present in a buffer? Weak acid and conj base or weak base

    and conj acid

    3. The Kaof carbonic acid is 4.3x10-7

    .

    a. What is the pH of a solution that is 0.25 M H2CO3and 0.25 M NaHCO3?-log (4.3x10

    -7) = 6.4

    b. What would the ratio of acid to base be if you wanted to buffer at exactly a pH of 6?4.3x10

    -7= 1x10

    -6(B/A) B:A = 0.43:1 or A:B = 2.32:1

    4. The Kaof hydrozoic acid is 1.9x10-5

    .

    a. What is the pH of a solution that is 0.25 M HN3and 0.75 M N3-1

    ?

    Ka= 1.9x10-5

    = [H3O+1

    ](0.25/0.75) = 5.7x10-5

    pH = -log(5.7x10-5

    ) = 4.24b. What would the ratio of acid to base be if you want to buffer exactly at a pH of 5?

    1.9x10-5

    = 1x10-5

    (B/A) B:A = 1.9:1 or A:B = 0.53:1

    5. The Kbof aniline is 4x10-10

    .

    a. What is the pH of a solution that is 0.050 M C6H5NH2and 0.050 M C6H5NH3+1

    ?

    pOH = -log (4x10-10

    ) = 9.4 so pH = 4.6or

    pH = -log (2.5x10-5

    ) = 4.6b. What would the ratio of acid to base be if you want to buffer exactly at a pH of 5?

    2.5x10-5

    = 1x10-5

    (B/A) B:A = 2.5:1 or A:B = 0.4:1

    6. The Kbof ethyleamine is 4.4x10-4

    .a. What is the pH of a solution that is 0.25 M C5H5N and 1.25 M C5H5NH

    +1?

    Ka= 1x10-14

    /4.4x10-4

    = 2.27x10-11

    2.27x10-11

    = [H3O+1

    ] (0.25/1.25) = 1.14x10-10

    pH = -log(1.14x10-10

    ) = 9.95b. What would the ratio of acid to base be if you want to buffer exactly at a pH of 10?

    2.5x10-11

    = 1x10-10

    (B/A) B:A = 0.25:1 or A:B = 4:1

    7. Which of the following buffer systems would you choose if you want to buffer at a pH of 2?a. HC3H5O2/ C3H5O2

    -1 Ka= 1.35x10

    -5

    b. HNO2/ NO2-1

    Ka= 7.2x10-4

    c. HIO / IO-1

    Ka= 2.3x10-11

    d. H3PO4/ H2PO4-1

    Ka= 7.6x10-3

    You would pick system d. The Kavalue is closest to the pH you want to buffer at.Describe how to make the buffer so it is exactly at a pH of 2.

    7.6x10-3

    = 1x10-2

    (B/A) so: B:A = 0.76:1 or A:B = 1.316

    8. Lactic acid has a Kaof 8.3x10-4

    . A buffer is made so that it is 0.50 M in HC3H5O3and 0.50M in C3H5O3

    -1.

    a. What is the pH of the buffer? pH = -log(8.3x10-4

    ) = 3.1

    b. What will the pH of the buffer be if 0.15 M of HCl are added?0.15 M HCl will react away 0.15 M of the weak base and add 0.15 M of the weak acid8.3x10

    -4= [H3O

    +1](0.35/0.65) = 1.54x10

    -3 pH = -log(1.54x10

    -3) = 2.8

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    9. Arsenous acid has a Kaof 6.6x10-10

    . A solution is made so that it is 50 mL of 1.2 M in

    HAsO2.

    a. What is the pH of the solution?6.6x10

    -10= x

    2/1.2 x = [H3O

    +1] = 2.8x10

    -5 pH = 4.55

    b. How many grams of NaAsO2must be added so the solution is a buffer at a pH of 10?Assume no change in volume.6.6x10

    -10 = 1x10

    -10([base]/1.2 M acid) so [base] = 7.92 M

    [base] = 7.92 M so 7.92 moles/L * 0.05 L = 0.396 moles NaAsO2

    0.396 moles NaAsO2* 129.9 g/mole = 51.4 g NaAsO2

    10. Methylamine has a Kbof 4.2x10-4

    . A buffer is made so that it is 0.75 M in CH3NH2and

    0.75 M in CH3NH3+1

    .

    a. What is the pH of the buffer?pOH = -log(4.2x10

    -4) = 3.4 so pH = 10.6

    or

    pH = -log(2.4x10-11) = 10.6b. What will the pH of the buffer be if 0.25 M of HNO3are added?

    0. 25 M HNO3will react away 0.25 M of the weak base and add 0.25 M of the weak acid

    2.4x10-11

    = [H3O+1

    ](0.50/1.0) = 4.8x10-11

    pH = -log(4.8x10-11

    ) = 10.3

    11. Quinoline has a Kbof 6.3x10-10

    . A solution is made so that it is 250 mL of 1.75 M C9H7N.

    a. What is the pH of the solution?

    6.3x10-10

    = x2/1.75 x = 3.32x10

    -5 pOH = 4.48 pH = 9.52

    b. How many grams of C9H7NHCl must be added so that the solution is a buffer exactly

    at a pH of 5? Assume no change in volume.

    Ka= 1x10-14

    /6.3x10-10

    = 1.59x10-5

    1.59x10-5= 1x10-5(1.75 M/[acid]) so [acid] = 1.1 M[acid] = 1.1 M so 1.1 moles/L * 0.25 L = 0.275 moles acid

    0.275 moles acid * 165.5 g/mole = 45.5 grams C9H7NHCl

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    Lewis Acids & Bases1. What is the definition of a Lewis Acid? Electron pair acceptor2. What is the definition of a Lewis Base? Electron pair donor

    3. What is an amphoteric substance? A substance that can be a Bronsted base or a Lewis acid

    4. Give 4 examples of amphoteric substances. Al(OH)3 Zn(OH)2 Cr(OH)3

    Sn(OH)45. Aluminum hydroxide, Al(OH)3, is a famous amphoteric substance.

    a. Write an equation showing Al(OH)3acting as a Bronsted baseAl(OH)3 + 3 H3O

    +1 6 H2O + Al

    +3

    b. Write an equation showing Al(OH)3acting as a Lewis acid

    Al(OH)3 + OH-1

    [Al(OH)4]-1

    6. In the following equations, which species is acting as the Lewis acid and the Lewis base?a. NH3 + BF3 BF3NH3

    LB LA

    b. SnCl4 + 2 Cl-1

    [SnCl6]-2

    LA LB

    c. H++ OH

    -1 H2O

    LA LB

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    Combination Problems1.

    a) A 0.10 M solution of it has a starting pH of 3. If it was a strong acid it should have a pH of 1.

    Also, the equivalence point is at about a pH of 9. It it was a strong acid, the equivalence point

    should be at 7 through this titration with a strong base NaOH.

    b) Find the pH at the midpoint of the titration. The equivalence point is at 25 so the midpoint is

    at 12.5 mL. The pH at this point is 5 so pKa= 5 so Ka= 1x10-5

    c) See above line in red

    d) i) MaVa= MbVb (0.10)(25 mL) = (0.20 M)(x mL Vb= 12.5 mL

    ii) Less NaOH will be needed so the concentration will be higher resulting in a more

    concentrated base and so a higher equivalence point

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    2.

    a) NH3 + H3O+1

    NH4+1

    + H2O

    b) Starting pH: 1.8x10-5

    = x2/0.10 x = 0.00134 pOH = 2.87 pH = 11.13

    Equivalence point: (30 mL) (0.1 M) = (x mL)(0.2 M) x mL = 15 mLNH3at equivalence point: (0.10 M)(30/45) = 0.0667 MHCl at equivalence point: (0.20 M)(15/45) = 0.0667 M

    pH at equivalence point: 1x10-14

    /1.8x10-5

    = x2/0.0667 x = 6.1x10

    -6pH = 5.2

    c) Since the equivalence point is at 4.71, methyl red will be the closest to that as it changes colorat a pH of 5.5

    d) Since Kbfor NH3is 1.8x10-5

    that means that Kafor NH4+1

    is 1x10-14

    /1.8x10-5

    = 5.6x10-10

    .

    Since Kbis larger than Ka, the solution will be basic

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    3. For the reaction:

    NH3(aq) + H2O (l) NH4+1

    (aq) + OH-1

    (aq)

    In 0.0180 M NH3at 25oC, the [OH

    -1] is 5.6x10

    -4M.

    a) Write the equilibrium-constant expression for the reaction above.

    Kb= [NH4+1][OH-1]/[NH3]b) Determine the pH of 0.0180 M NH3

    pOH = -log (5.6x10-4

    ) = 3.25 pH = 10.75

    c) Determine the value of Kbfor NH3Kb= (5.6x10

    -4)2/(0.0180) = 1.74x10

    -5

    d) Determine the % ionization of NH3

    % Ionization = (5.6x10-4

    )/(0.0180) *100% = 3.11%e) In an experiment, a 20.0 mL sample of 0.0180 M NH3was placed in a flask and

    titrated to the equivalence point and beyond using 0.0120 M HCl

    i) Determine the volume of 0.0120 M HCl that was added to reach the equivalence pt

    MaVa= MbVb (0.0120)(Va) = (0.0180)(20 mL) Va= 30 mL

    ii) Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120 MHCl was added

    This is the midpoint of the titration so the midpoint pH = pKa

    Kb= 1.74x10-5

    so Ka= 1x10-14

    /1.74x10-5

    = 5.747x10-10

    pH = -log (5.747x10

    -10) = 9.24

    iii) Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M

    was added.HCl: 0.0120 M (40/60) = 0.008 M

    NH3: 0.0180 M (20/60) = 0.006 M

    H3O+1

    + NH3 NH4+1

    + H2O

    0.008 0.006

    0.002 0

    pH = log (0.002) = 2.70

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    4. For the reaction:C6H5NH2 (aq) + H2O (l) C6H5NH3

    +1 (aq) + OH

    -1(aq)

    a) Write the equilibrium constant expression, Kb, for the reaction above.

    Kb= [C6H5NH3+1

    ][OH-1

    ]/[C6H5NH2]

    b) A sample of C6H5NH2is dissolved in water to produce 25.0 mL of a 0.010 M solution.

    The pH of the solution is 8.82. Calculate Kb.b) pH = 8.82 pOH = 5.18 [OH-1] = 10-5.18= 6.61 x10-6

    C6H5NH2 + H2O C6H5NH3+ + OH-

    []o 0.10 0 0

    -6.61 x10-6

    +6.61 x10-6

    +6.61 x10-6

    []eq 0.1 6.61 x10-6

    6.61 x10-6

    Kb= (6.61 x10-6)2/0.1 = 4.37 x10-10

    c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calclulate the pH of thesolution when 5.0 mL of the acid has been added.

    c) Base = (0.1) (25/30) = 0.0833Acid = (0.1) (5/30) = 0.0167

    C6H5NH2 + H3O+ C6H5NH3

    + + H2O

    []o 0.0833 0.0167 0

    -0.0167 -0.0167 +0.0167

    []eq 0.0666 0 0.0167

    C6H5NH2 + H2O C6H5NH3+ + OH-

    []o 0.0666 0.0167 0

    -x +x +x

    []eq 0.0666 0.0167

    x

    4.37 x10-10= [C6H5NH3+][OH-1]/[ C6H5NH2]

    4.37 x10-10= [0.0167][x]/[ 0.0666]

    X = 1.74 x10-9

    pOH = -log 1.74 x10-9

    = 8.76 pH = 5.24

    d) Calculate the pH at the equivalence point of the titration in (c)d) C6H5NH2 + H3O

    + C6H5NH3+ + H2O

    []o 0.05 0.05 0

    -0.05 -0.05 +0.05

    []eq 0 0 0.05

    C6H5NH3+ + H2O C6H5NH2 + H3O

    +

    []o 0.05 0 0

    -x +x +x

    []eq 0.05 x x

    Ka= 1x10-14

    /4.37x10-10

    = 2.29 x10-5

    x2/0.05 = 2.29 x10

    -5= Ka

    X = 0.00107 pH = -log 0.00107 = 2.97

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    e) The pKavalues for several indicators are given below. Which is most suitable for thistitration? Justify.

    Erythrosine pKa= 3

    Litmus pKa= 7

    Thymolphthalein pKa= 10

    e) As the pH changes at about 3, erythrosine is the best as it changes at the same pH value.

    5. Hypochlorous acid, HOCl, is a weak acid. The Kafor HOCl is:

    Ka= [H3O+1

    ][OCl-1

    ] = 3.2x10-8

    [HOCl]

    a) Write a chemical equation showing how HOCl behaves as an acid in water.a) HOCl (aq) + H2O (l) H3O

    + (aq) + OCl-1(aq)

    b) Calculate the pH of a 0.175 M solution of HOCl

    3.2 x10-8= x2/0.175 x = 7.48 x10-5 pH = -log x = log 7.48 x10-5= 4.13c) Write the net ionic equation for the reaction between HOCl and NaOHc) HOCl (aq) + OH

    -1(aq) H2O (l) + OCl

    -1(aq)

    d) In an experiment, 20.00 mL of 0.175 M HOCl is titrated with 6.55 mL of 0.435 M NaOH

    i) Calculate the number of moles of NaOH addeddi) 0.0655 L x 0.435 moles/L = 0.0285 moles OH-added

    ii) Calculate the [H3O+1

    ] in the flask after the NaOH has been addeddii) acid = 0.175 (20/26.55) = 0.132 M

    base = 0.435 (6.55/26.55) = 0.107 M

    HOCl (aq) + OH-1

    (aq) H2O (l) + OCl-1

    (aq)

    []o 0.132 0.107 0

    -0.107 -0.107 +0.107[]eq 0.025 0 0.107

    HOCl (aq) + H2O (l) H3O+ (aq) + OCl

    -1(aq)

    []o 0.025 0 0.107

    -x +x +x

    []eq 0.025 x 0.107

    3.2x10-8

    = x(0.107/0.025) x = 7.48x10-9

    pH = 8.13

    ii) Calculate the [OH-1

    ] in the flask after the NaOH has been added

    1x10-14

    /7.48x10-9

    = 1.34x10-6

    M OH-1