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Acid-Base Equilibria & Solubility Equilibria Base-part2 -  · PDF file Acid-Base Equilibria Buffer solutions ... Common Ion Effect The shift in equilibrium due to addition of a compound

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Text of Acid-Base Equilibria & Solubility Equilibria Base-part2 -  · PDF file Acid-Base...

  • 1

    Acid-Base Equilibria

    Buffer solutions Titrations

    And the beat goes on…

  • 2

    Common Ion Effect The shift in equilibrium due to addition of a compound having an ion in common with the dissolved substance.

  • 3

    Common Ion Effect

    HA(aq) + H2O H3O +(aq) + A-(aq)

    or HA H+ + A-

    Consider a mixture of: HA (weak acid) and NaA (conj. salt).

    from both HA & NaA

    reactant product

    A- is conj. base of HA

    Ka = [H+][A-]

    [HA]

  • 4

    Common Ion Effect Rearranging:

    [H+] = Ka[HA]

    [A-]

    -log[H+] = -logKa - log [HA] [A-]

    pH = pKa + log [A-] [HA]

    Ref. Tables

  • 5

    Henderson-Hasselbalch Approximation

    1.Neglect [A-] from ionization the of the weak acid (HA).

    2.The salt (NaA) is fully ionized.

    3.Thus [HA] = [HA]o & [A -] = [A-]o

    (from acid) (from salt)

    HA H+ + A-

    (where ‘o’ means original concentration)

  • 6

    (whenever both A- and HA are both “added” to a solution)

    Henderson-Hasselbalch Approximation

    pH = pKa + log [A-]o [HA]o

    Ka = [H+][A-]o

    [HA]o or:

  • 7

    H-H: Example

    What is the pH of a solution containing 0.20 M HAc (acid) and 0.30 M NaAc (conj. base) ?

    Compare it to the pH of just 0.20 M HAc.

    KHAc = 1.8 x 10 -5

  • 8

    H-H: Example

    Ka = [H+][A-]o

    [HA]o

    Can use either:

    pH = pKa + log [A-]o [HA]o

    or

  • 9

    H-H: Example

    = -log (1.8 x 10-5) + log (0.30/0.20)

    pH = 4.92

    = 4.74 + 0.18

    pH = pKa + log [A-]o [HA]o

    Note: no need for ICE table when both acid and its conj. base are added.

  • 10

    H-H: Example

    For just 0.20 M HAc (no NaAc) Must use ICE table.

    HAc H+ + Ac-

    Initial M 0.20 0 0 Change M -x +x +x Equil. M 0.20-x x x

  • 11

    H-H: Example

    1.8 x 10-5 = x2

    (0.20-x)

    x = 1.9 x 10-3 M

    pH = 2.72

    (vs. pH = 4.92 when both HAc and NaAc (conj. base) added together.

    HAc H+ + Ac-

  • 12

    H-H: You Try It !!

    What is the pH of a solution containing 0.30 M formic acid and 0.52 M potassium formate ?

    KHCOOH = 1.7 x 10 -4

  • 13

    H-H for Bases e.g. add both NH3 & NH4Cl to water

    base conj. acid

    NH3 + H2O NH4 + + OH-

    NH4 + NH3 + H

    + or equivalently:

    base on reactant side; use Kb

    conj. acid on reactant side;

    use Ka

  • 14

    H-H for Bases using Kb

    [OH-] = Kb[base]o [conj acid]o

    pOH = pKb + log [conj acid]o

    [base]o

    NH3 + H2O NH4 + + OH-

    Kb = [OH-] [conj acid]o

    [base]o

    See Reference table

  • H-H for Bases: Try It

    15

    What is the pH of a aqueous mixture that is 0.43 M NH3 and 0.35 M NH4Cl? (Window-side use Ka, hall-side use Kb)

    Compare this to a solution that is 0.43 M NH3 only.

    Kb (NH3) = 1.8 x 10 -5

  • 16

    Buffer Solutions  A weak acid or base and its

    conjugate salt. (H-H applies)

    Acid buffer: CH3COOH and CH3COONa

    Base buffer: NH3 and NH4Cl

    Buffers resist pH changes.

  • 17

    Buffer Solutions

    Each buffer has a specific pH, and is resistant to pH changes.

    Buffers are key to many biological functions.

  • 18

    Buffer Solutions CH3COO

    -/CH3COOH

    or simply Ac-/HAc

    If add acid:

    Ac- + H+  HAc

    If add base:

    HAc + OH-  H2O + Ac -

    “Buffer Capacity” ~ concentration

    Both rxns are one way

    arrows. WHY?

  • 19

    Which are Buffers?

    1. KH2PO4/H3PO4

    2. NaClO4/HClO4

    3. Na2CO3/NaHCO3

    Remember buffers are weak acid or weak base with its conjugate.

  • 20

    Proof of the Buffer Effect Calculate the pH of a buffer solution containing 1.0 M HAc and 1.0 M NaAc. What is the effect of adding 0.10 moles of HCl to 1.0 L of this buffer?

    Ka (HAc) = 1.8 x 10 -5

  • 21

    1.0 M HAc + 1.0 M NaAc

    Ka = [H+][Ac-]o

    [HAc]o

    [H+]= 1.8 x 10-5 pH = 4.74

    same as pKa!

    1.8x10-5 = [H+](1.0)

    1.0

    (H-H applies!) HAc H+ + Ac-

    remember lab?

  • 22

    1.0 M HAc + 1.0 M NaAc + 0.1 mol HCl

    Thus:

    HAc = 1.0 + 0.1 = 1.1 mol

    Ac- = 1.0 – 0.1 = 0.9 mol

    Due to addition of HCl

    H+ + Ac-  HAc

    0.1 0.1 0.1 mol less more

    (one way arrow)

  • 23

    [H+](0.9) 1.1

    1.8x10-5 =

    1.0 M HAc + 1.0 M NaAc + 0.1 mol HCl

    Ka = [H+][Ac-]o

    [HAc]o

    [H+] = 2.2 x 10-5 M

    pH = 4.66 (vs. 4.74) (very small change)

  • 24

    Buffers with Specific pH’s How do you prepare a “phosphate buffer” at pH 7.4 ?

    Of the 3 ionizations for H3PO4, the following is best because its pKa2 is closest to the target pH.

    Ka2 = 6.2 x 10 -8 or pKa2 = 7.21

    H2PO4 - H+ + HPO4

    2-

  • 25

    Buffers with Specific pH

    pH = pKa + log [conj. base]

    [acid]

    7.4 = 7.21 + log [HPO4

    2-] [H2PO4

    -] [HPO4

    2-] [H2PO4

    -] = 1.5

    e.g. add 1.5 mol HPO4 2- and

    1.0 mol H2PO4 - to water.

  • 26

    Buffers: Try It !!! How would you prepare a carbonate buffer at a pH of 10.10?

    For carbonic acid: Ka1 = 4.2 x 10

    -7

    and Ka2 = 4.8 x 10

    -11

    Write the reactions that occur if HCl or if NaOH is added to this buffer.

  • 27

    Acid-Base Titrations

    3 types are common:

    •Strong acid – strong base (no hydrolysis of the salt formed)

    •Weak acid – strong base (hydrolysis of the salt anion)

    •Strong acid – weak base (hydrolysis of the salt cation)

  • 28

    Strong Acid-Strong Base

    Map the titration of 25 mL of 0.10 M HCl with 0.10 M NaOH.

    ml NaOH

    pH 14

    7

    0

    NaOH 0.10 M “titrant”

    HCl 25mL, 0.10 M

  • 29

    Strong Acid-Strong Base

    Titration equation:

    H+ + OH- H2O (one way arrow)

    .10M .10M

    .025L vol changes flask buret

    strong strong

  • 30

    Strong Acid-Strong Base Equivalence point is easy to find.

    mol H+ = mol OH-

    (0.10 M)(25 mL) = (.10 M) (VB)

    VB = 25 mL

    Since [H+] = [OH-], pH = 7.0 (if temperature is 25oC)

    MA VA = MB VB *

  • 31

    Strong Acid-Strong Base

    The pH at any point in the titration can be calculated by determining the concentration of the excess reagent (H+ or OH-):

    moles excess H+ or OH-

    total liters of solution

    H+ + OH-  H2O (one way arrow)

  • 32

    Strong Acid-Strong Base

    .025L x .10M .035L x .10M =.0025 = .0035

    ~0 .0010 (excess)

    -.0025 .0025

    H+ + OH- H2O init.

    moles

    equil. moles

    D moles

    e.g. after 35 mL of 0.10 M NaOH added

    Total volume = 25 mL + 35 mL = 60 mL = 0.060 L

  • Strong Acid-Strong Base

    33

    pOH = 1.77 and pH = 12.23

    = .017M[OH-] = .0010 mol .060 L

    Calculate pH from the concentration of excess reagent:

  • 34

    Strong Acid-Strong Base

    ml NaOH

    pH

    14

    7

    0

    MAVA = MBVB

    excess OH-

    excess H+

  • 35

    Go For It Calculate the following when 32 mL of 0.15 M NaOH is titrated with 0.18 M HCl :

    •pH after 17 mL HCl is added. •Volume of HCl to reach eq. pt. •pH after 30 mL HCl is added

  • Titrations Involving Weak Acids or Bases

    36

    Two types are most common: • Weak acid – strong base • Strong acid – weak base

    These are complicated by the fact that the weak acid or weak base will hydrolyze.

  • 37

    Weak Acid-Strong Base

    Map the titration of 25 mL of 0.10 M HAc with 0.10 M NaOH.

    ml NaOH

    pH

    14

    7

    0

    NaOH 0.10 M

    HAc 0.10 M 25mL

    gradual rise

  • 38

    Weak Acid-Strong Base Titration equation:

    HAc + OH-  Ac- + H2O

    .10M .10M

    .025L =.0025mol

    flask buret

    (one way arrow due to addition of strong base)

    weak strong

  • 39

    The pH at any point in the titration can be calculated. e.g. after 10 mL of NaOH added. This is before the equiv. point.

    Total volume = 25mL+10mL=35mL mol OH- added is: 0.010 L x 0.10 M = 0.0010 mol

    This is also the mol of Ac- formed.

    25mL, .10M .10M

    HAc + OH-  Ac- + H2O

  • 40

    Weak Acid-Strong Base

    HAc + OH-  Ac- + H2O

    I (mol) .0025 .0010

    C (mol)-.0010 -.0010 .0010

    E (mol) .0015 ~0 .0010

    Before the equivalence point,

    the titration created a buffer!

    Do

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