# 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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Acid-Base Equilibria

Buffer solutions Titrations

And the beat goes on…

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

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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]

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Common Ion Effect Rearranging:

[H+] = Ka[HA]

[A-]

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

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

Ref. Tables

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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)

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(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:

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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

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H-H: Example

Ka = [H+][A-]o

[HA]o

Can use either:

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

or

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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.

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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

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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-

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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

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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

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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

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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.

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Buffer Solutions

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

Buffers are key to many biological functions.

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Buffer Solutions CH3COO

-/CH3COOH

or simply Ac-/HAc

Ac- + H+  HAc

HAc + OH-  H2O + Ac -

“Buffer Capacity” ~ concentration

Both rxns are one way

arrows. WHY?

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Which are Buffers?

1. KH2PO4/H3PO4

2. NaClO4/HClO4

3. Na2CO3/NaHCO3

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

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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

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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?

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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

H+ + Ac-  HAc

0.1 0.1 0.1 mol less more

(one way arrow)

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[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)

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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-

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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.

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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.

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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)

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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

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Strong Acid-Strong Base

Titration equation:

H+ + OH- H2O (one way arrow)

.10M .10M

strong strong

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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 *

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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)

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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:

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Strong Acid-Strong Base

ml NaOH

pH

14

7

0

MAVA = MBVB

excess OH-

excess H+

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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.

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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

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Weak Acid-Strong Base Titration equation:

HAc + OH-  Ac- + H2O

.10M .10M

.025L =.0025mol

(one way arrow due to addition of strong base)

weak strong

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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

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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!

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