acid-base equilibria & solubility equilibria base-part2 -student.pdf · acid-base equilibria...

1

Acid-Base Equilibria

Buffer solutionsTitrations

And the beat goes on…

2

Common Ion EffectThe 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 EffectRearranging:

[H+] =Ka[HA]

[A-]

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

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

Ref. Tables

5

Henderson-HasselbalchApproximation

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]oor:

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

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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.

10

H-H: Example

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

HAc H+ + Ac-

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

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

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H-H for Basese.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


Kb =[OH-] [conj acid]o

[base]o

See Reference table

H-H for Bases: Try It

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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 SolutionsCH3COO-/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?

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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 EffectCalculate the pH of a buffer solution containing 1.0 M HAcand 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 10-5pH = 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

Due to addition of HCl

H+ + Ac- HAc

0.1 0.1 0.1 molless 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’sHow 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 HPO42- 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

andKa2 = 4.8 x 10-11

Write the reactions that occur if HClor 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

pH14

7

0

NaOH0.10 M“titrant”

HCl25mL, 0.10 M

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Titration equation:

H+ + OH- H2O (one way arrow)

.10M .10M

.025L vol changesflask buret

strong strong

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Strong Acid-Strong BaseEquivalence 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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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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.025L x .10M .035L x .10M=.0025 = .0035

~0 .0010 (excess)

-.0025 .0025

H+ + OH- H2Oinit.

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


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pOH = 1.77 and pH = 12.23

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

Calculate pH from the concentration of excess reagent:

34


ml NaOH

pH

14

7

0

MAVA = MBVB

excess OH-

excess H+

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Go For ItCalculate 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

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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

NaOH0.10 M

HAc0.10 M25mL

gradual rise

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Weak Acid-Strong BaseTitration equation:

HAc + OH- Ac- + H2O

.10M .10M

.025L=.0025mol

flask buret

(one way arrowdue 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=35mLmol 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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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 ICE in moles (volume changes)

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The buffer thus contains:0.0015 mol/0.035 L = .0428 M HAc0.0010 mol/0.035 L = .0286 M Ac-

Ka =[H+][Ac-]o

[HAc]o

Use H-H

1.8x10-5 = [H+](.0286)(0.0428)

[H+] = 2.7 x 10-5 & pH = 4.57

Weak Acid-Strong Base: Before Equiv. Point

42ml NaOH

pH

14

7

0gradual rise since buffer

HAc + OH- Ac- + H2O

excess limiting

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Weak Acid-Strong BaseAt equivalence point:

HAc + OH- Ac- + H2O

But Ac- (conj.base of a weak acid) undergoes hydrolysis (equilibrium):

Ac- + H2O HAc + OH-

(one way arrow)

So the pH will be > 7.0

~100%

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At equivalence point:

mol HAc = mol NaOH

So total volume is 0.050 L

(0.10 M)(0.025 L) = (.10 M) (VB)

VB = 0.025 L

MA VA = MB VB

Need to calculate total volume.

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At equivalence, the moles of Ac-

formed is the same as starting HAc.

= 0.050MThus [Ac-] =.0025 mol

.050 L

HAc + OH- Ac- + H2O

.0025mol

.0025mol

+ .0025mol

46


At equivalence: 0.050 M Ac-

Set up an ICE table using Kb with Ac- as reactant.

I 0.050 0 0

C -x x x

E 0.050-x x x

Now plug in to Kb = 5.6 x 10-10

Ac- + H2O HAc + OH-

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5.6 x 10-10 =x2

0.050-x

x = [OH-] = 5.29 x 10-6M

pOH = 5.28

pH = 8.72

(at equivalence point)

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Weak Acid-Strong BaseWhat about pH after equivalence point, e.g. after 35 mL NaOH?

Solution is basic due to OH- and Ac-, but since OH- is so much stronger, it alone determines pH.

(Region of excess OH-, just like strong acid-strong base case)

Solution vol. = 60 mL = 0.060 L

49

Weak Acid-Strong BaseExcess mol OH- = mol OH- added –

init. mol HAc

0.0010 mol0.060L

= 0.017 M[OH-] =

pOH = 1.77 and pH = 12.23

= MBVB - MAVA

= .010M x.035L - .010M x.025L

= 0.0010 mol

50

pH

14

7

0ml NaOH


excess [OH-] pH

Conj base of weak acid (ICE)

buffer (via titration) (H-H)

Half-Equivalence Point

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The point in the titration where the moles of base added is one-half the moles of weak acid.

HAc + OH- Ac- + H2O

init mol .050 .025 0

D mol -.025 -.025 +.025

equil mol .025 ~0 .025

equal

Half-Equivalence Point

52

Plugging into H-H:

equal

Ka = [H+]

pKa = pH

ml NaOH

pH

14

7

0

pKa

Ka =[H+][Ac-]o

[HAc]o

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Titration Calculation Step-by- Step

Example: Weak acid - strong base.

Distinguish between the equilibrium equation and titration equation.

HA H+ + A-Equilibrium, governed by Ka

one way

HA + OH- A- + H2OTitration


54

Example: Weak acid - strong base.

Write the titration equation:one way

HA + OH- A- + H2O

1. Calculate:• moles of acid added (MAVA)• moles base added (MBVB)• total volume (VT = VA + VB)

Example.032 mol.018 mol42 mL

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HA + OH- A- + H2O

Three possibilities:Either mol HA or mol OH- is limiting, or they are equal.

I .032 mol .018 mol

C

E


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2. Determine the net moles of HA, OH- and A- remaining as a result of the titration equation.

HA + OH- A- + H2O

I .032 mol .018 mol

C -.018 mol -.018 mol +.018 mol

E .014 mol ~0 .018 mol


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3. Convert net moles to concentration.

HA + OH- A- + H2O

I .032 mol .018 mol

C -.018 mol -.018 mol +.018 mol

E .014 mol ~0 .018 mol

If VT = 42 mL

[HA] = = .33 M & [A-] = .43 M.014 mol.042 L


[M] .33M .43M

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4. There are 3 possibilities.

Possibility #1: OH- was limiting.• Thus only HA and A- remain. • This is the buffer region. • Use Ka and the H-H approx. to

solve for [H+] and pH.• No ICE table is needed.


59

Possibility #1 cont.• There is a special case when

mol OH- added was one-half the mol of HA originally present.

• Then [HA] = [A-] and pH = pKa. This is the half-equiv. point.

pH =pKa


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Possibility #2: mol HA added was equal to mol OH- added. • This is the equivalence point and

only A- remains. • Use Kb and set up an ICE table

with A- as the reactant. • Solve for [OH-] and pH.


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Possibility #3: mol OH- was excess. • This is past the equiv. point. • The excess [OH-] alone

determines pOH and pH.• No ICE table is needed.


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Grind It Out !!!

Initial pH in the flaskpH after 15 ml KOH addedpH after 24 mL KOH addedpH at equivalence pointpH after 58 mL KOH added

80. mL of 0.12 M benzoic acid is titrated with 0.20 M KOH.

Ka(BA) = 6.5 x 10-5. Calculate:

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http://www.google.com/url?sa=i&rct=j&q=benzoic+acid&source=images&cd=&cad=rja&docid=yq77uoXjaPfdAM&tbnid=8FZHRhCfl8oxRM:&ved=0CAUQjRw&url=http://www.premierindiaa.com/organic-chemicals.html&ei=DPcQUuzVDdGn4AO_xICQDg&bvm=bv.50768961,d.aWc&psig=AFQjCNFL4n3F9y2DgfBKpWuDiBRQcthd_A&ust=1376929910017118

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Strong Acid-Weak Base

Map the titration of 25 mL of 0.10 M NH3 with 0.10 M HCl.

ml HCl

pH14

7

0

HCl0.10 M

NH30.10 M25 mL

gradual slope


64

Keep two equations straight!

1. Equilibrium equation for NH3.


2. Titration equation:

H+ + NH3 NH4+ (one way)

Governed by: Kb = [NH4+][OH-]

[NH3]

65


H+ + NH3 NH4+

Titration equation:

.10M

.025L

.0025 mol

.10M

Before equivalence point, this reaction creates a buffer, so use H-H.


66

All calculations are analogous to the strong base-weak acid example.

ml HCl

pH14

7

0

buffer (H-H)

excess [H+]

equiv pt.


67

H+ + NH3 NH4+

What happens at the half-equiv. point?

.0025 .00125 Start mol

-.00125 +.00125 -.00125 D mol-.00125 +.00125 ~0 D mol

equal

Kb = [NH4+][OH-]

[NH3]and pOH = pKb

68

I’ll do the equivalence point pH:

H+ + NH3 NH4+

But NH4+ undergoes hydrolysis:

NH4+ NH3 + H+

So the pH will be < 7.0 (acidic)


100%

69

At equivalence point:mol NH3 = mol HCl

So total volume is 0.050 L

(0.10 M)(0.025 L) = (.10 M) (VA)

VA = 0.025 L


MB VB = MA VA

70

At equivalence, the amount of NH4

+ formed is:


H+ + NH3 NH4+

.0025mol

.0025mol

+

= 0.050MThus [NH4+] = .0025 mol

.050 L

.0025mol

71

At equivalence: 0.050 M NH4+

which then hydrolyzes:

I 0.050 0 0

C -x x x

E 0.050-x +x +x

Now plug in to Ka = 5.6 x 10-10


NH4+ NH3 + H+

72

5.6 x 10-10 =x2

0.050-x

x = [H+] = 5.29 x 10-6M

pH = 5.28

(At equivalence point)


73ml HCl

pH

14

7

0 excess [H+]

conj acid of weak base (ICE)

buffer via titration (H-H)


pOH= pKb

half titr.

74

One More !!!

60. mL of 0.15 M ammonia is titrated with 0.18 M HCl

Initial pH in the flaskpH after 25 mL HCl addedpH after 35 mL HCl addedpH at equivalence pointpH after 60 mL HCl added

Kb = 1.8 x 10-5. Calculate:

75

Titration Hints(also see summary in packet)

1. Often must work in moles and then converted to concentration since volume changes.

2. Calculate moles of acid & base to see where in titration you are.(MAVA and MBVB)

3. Buffer region: excess weak acid (or weak base). Use H-H.

Titration Hints cont.

76

4. Be on the lookout for the half titration point!

5. At equiv. point, moles acid and moles base are equal resulting in 100% conjugate. Use ICE. This is the most involved calculation.

6. If excess strong acid or strong base, pH determined by its excessconcentration alone.

Diprotic Acid Titrations

77

H2A 2H+ + A-2 Ka1 x Ka2

H2A H+ + HA- Ka1

HA- H+ + A-2 Ka2

pKa2

pKa1

B=1st end pointC=1st end point

Volume of NaOH

What are the major species at D, C, B, A?

78

Acid Base Indicators

Weak acids that have different acid and conjugate base colors.

HIn H+ + In-HIn In-

Le Chatelier’s principle: if excess H+, color is yellow. If excess OH-, it is blue.

Acid Base Indicators

79

pH

14

7

0ml NaOH

Want pKa of indicator dye to be near equivalence point pH.

80

Acid Base Indicatorse.g. phenolphthalein (HPh) has a pKaof 9.7. It is colorless in acid, but starts to turn pink at pH > 8.3.

It is a good indicator for a strong acid-strong base titration since pH 8.3 lies on the steep part of the titration curve.

HIn H+ + In-HPh Ph-

acid-base equilibria & solubility equilibria base-part2 -student.pdf · acid-base equilibria...

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