monoprotic acid-base equilibria

Monoprotic Acid-Base Equilibria

Review of Fundamentals

1.) Acids and Bases are essential to virtually every application of chemistry Analytical procedures such as chromatography and electrophoresis Protein purification, chemical reactions, environmental issues

Urban Stone DecayPollutants Contribute to Acid Rain

Forest Destruction

Yellowstone Air Pollution (same view)



2.) Strong Acids and Bases Completely dissociates

[H3O+] or [OH-] equals concentration of strong acid or base- What is the pH of a 0.1M solution of HCl?

- What is the pH of a 0.1M solution of KOH?

13.00101.0

M101.00.10

13

13

M)log(]log[HpH

101

][OH][H101]][OH[H

0.1M[KOH]][OH14-

14--

-

ww

KK

1.00M)(0.1log]log[HpH

M0.1[HCl]][H



2.) Strong Acids and Bases pH at other concentrations of a strong base

Relationship between pH and pOH:

[OH-] (M) [H+] (M) pH

1x10-1 1x10-13 13.0

1x10-2 1x10-12 12.0

1x10-3 1x10-11 11

1x10-4 1x10-10 10

1x10-5 1x10-9 9

C25 at 14.00KlogpOHpH ow

Acid Ka

HCl 103.9

HBr 105.8

HI 1010.4

HNO3 101.4



2.) Strong Acids and Bases Dilemma:

What is the pH of 1.0x10-8 M KOH?

6.00M)101.0log(]log[HpH

M101.010.01

101

][OH][H

M10.01[KOH]][OH

6

68-

14-

-8-

wK

How can a base produce an acidic solution?

M10.01[KOH]][OH -8-

Wrong Assumption!!



2.) Strong Acids and Bases Wrong Assumption!!

For large concentration of acid or base,

[H+] = [acid] or [OH-] = [base]

For small concentration, must account for water dissociation

In pure water [OH-] = 1.0x10-7M, which is greater than [KOH] = 1x10-8M

Must Use Systematic Treatment of Equilibrium



2.) Strong Acids and Bases Systematic Treatment of Equilibrium

Step 1: Pertinent reactions:

Kw

Step 2: Charge Balance:

][][][ -OHHK

Step 3: Mass Balance:

][][ 8100.1K All K+ comes from KOH

Completely dissociates, not pertinent




Step 4: Equilibrium constant expression (one for each reaction):

14w 100.1OHHK ]][[

Step 5: Count equations and unknowns:

Three equations:

Three unknowns:

][OH],[H],[K -

][][][ -OHHK

][][ 8100.1K

14w 100.1OHHK ]][[

(1)

(2)

(3)




Step 6: Solve (Seeking pH [H+]):

Set [H+] =x, and substitute mass balance equation into charge balance equation:

x100.1HKOH 8-- ][][][

][][ 8100.1K From mass balance

Substitute OH- equation into equilibrium equation:

x100.1OH 8-- ][14

w 100.1OHHK ]][[

148- 100.1)x100.1x )((

Monoprotic Acid-Base Equilibria Review of Fundamentals


Step 6: Solve (Seeking pH [H+]):

Solve the quadratic equation:

0100.1x100.1x 1482 )(

M101.1orM106.9x

12

100.1-14100.1100.1x

78

14-288

)(

))(()(

02.7106.9logHlogpH

M106.9H8-

8-

)(][

][

Negative number is physically meaningless

Use quadratic equation

pH slightly basic, consistent with low [KOH]


2.) Strong Acids and Bases Systematic Treatment of Equilibrium Three Regions depending on acid/base concentrations

High concentrations (≥10-6M), pH considered just from the added H+,OH-

low concentrations (≤10-8M), pH=7 not enough H+,OH- added to change pH

intermediate concentrations, (10-

6-10-8M), H2O ionization ≈ H+,OH- systematic equilibrium calculation necessary


3.) Water Almost Never Produces 10-7 M H+ and OH-

pH=7 only true for pure water

Any acid or base suppresses water ionization- Follows Le Châtelier’s principal

104 101OH4pH100.1HBr ][][][

In 10-4 M HBr solution, water dissociation produces only 10-10 M OH- and H+


4.) Weak Acids and Bases Weak acid/base do not completely dissociate Dissociation Ka for the acid HA:

Base Hydrolysis constant Kb

pK is negative logarithm of equilibrium constant

- As Ka or Kb increase pKa or pKb decrease- Smaller pKa stronger acid

][

]][[

HA

AHKa

][

]][[

B

OHBHKb

)Klog(pK aa )Klog(pK bb


4.) Weak Acids and Bases Conjugate acid-base pair – related by the gain or loss of a proton

- Conjugate base of a weak acid is a weak base - Conjugate acid of a weak base is a weak acid- Conjugate base of a strong acid is a very weak base or salt

Acid-base pair

Formic acid (pKa 3.744) stronger acid than benzoic acid (pKa=4.202)

Monoprotic Acid-Base Equilibria Weak Acid Equilibria

1.) General Systematic Treatment of Equilibrium Unlike concentrated strong acid, need to account for water ionization Find pH for a solution of a general weak acid HA

Step 1: Pertinent reactions:KwKa


][][][ -- AOHH


][][ -AHAF F – formal concentration of acid


][

]][[

HA

AHKa

14

w 100.1OHHK ]][[


1.) General Systematic Treatment of Equilibrium Find pH for a solution of a general weak acid HA

Step 5: Count equations and unknowns:

Four Equations:

Four Unknowns: ][OH,][H[HA],],[A -

][][][ -- AOHH ][][ -AHAF ][

]][[

HA

AHKa

14w 100.1OHHK ]][[

(1)

(4)

(2) (3)

Step 6: Solve (Not easy to solve cubic equation results!):- Again, need to make assumptions to simplify equations- The goal is to determine [H+], so we can measure pH



Step 6: Solve (Not easy to solve cubic equation results!):

Make Some Initial Assumptions: For a typical weak acid, [H+] from HA will be much greater than [H+] from H2O If dissociation of HA is much greater than H2O, [H+] >> [OH-]

][][][][][ --- AHAOHH

Set [H+]=x:xAxH - ][][

xFAFHA - ][][

substitute



Step 6: Solve (Not easy to solve cubic equation results!):

Substitute into Equilibrium Equation:

xAH - ][][ xFHA ][

xF

)x)(x(

HA

AHKa

][

]][[

0)K)(F(x)K(x aa2

]x[)1(2

)K)(F)(1(4KKx a

2aa

Rearrange:

Solve quadratic equation:



Step 7: Verify Assumption:

Was the approximation [H+] ≈ [A-] justified ([H+] >>[OH-])?

Setting F = 0.050 M and Ka = 1.07x10-3 for o-hydroxybenzoic acid:

17.2xlogpH

]A[]H[M108.6x

)1(2

)1007.1)(0500.0)(1(4)1007.1(1007.1

)1(2

)K)(F)(1(4KKx

3

3233a

2aa

Determine [OH-] from water dissociation:

123

14w 105.1

108.6

101

]H[

K]OH[

[H+] >> [OH-]6.8x10-3M >> 1.5x10-12Massumption is justified!


2.) Fraction of Dissociation Fraction of acid HA in the form A-():

Example:

Percent dissociation increases with dilution

F

x

)xF(x

x

HAA

A

][][

][

What is the percent fraction dissociation for F = 0.050 M and Ka = 1.07x10-3 for o-hydroxybenzoic acid?

%1414.0M0500.0

M108.6

F

x 3

Monoprotic Acid-Base Equilibria Weak Base Equilibria

1.) Treatment of Weak Base is Very Similar to Weak Acid Assume all OH- comes from base and not dissociation of water

][

]][[

B

OHBHKb

Step 1: Pertinent reactions:KwKb


][][][ -OHBHH


][-F][][][ BHBBHBF F – formal concentration of base


14w 100.1OHHK ]][[


1.) Treatment of Weak Base is Very Similar to Weak Acid Assume all OH- comes from base and not dissociation of water

Step 6: Solve (Assume [BH+] >> [H+] [BH+] ≈ [OH-]):

Set [OH-]=x and substitute into Equilibrium Equation:

xOHBH - ][][ xFB ][

xF

x

xF

)x)(x(

B

OHBHK

2

b

][

]][[

0)K)(F(x)K(x bb2

]OH[)1(2

)K)(F)(1(4KKx b

2bb

Rearrange:

Solve quadratic equation:


2.) Example

What is the pH of cocaine dissolved in water? F = 0.0372 M and Kb = 2.6x10-6 for cocaine?

0.0372-x x x

462

101.3x106.2x0372.0

x

Kb=2.6x10-4


2.) Example

What is the pH of cocaine dissolved in water? F = 0.0372 M and Kb = 2.6x10-6 for cocaine?

114

14w 102.3

101.3100.1

]OH[K]H[

Because x=[OH-], we need to solve for [H+]

49.10)102.3log(]Hlog[pH 11


4.) Fraction of Association Fraction of Base B in BH+ form ():

Example:

F

x

)xF(x

x

BBH

BH

][][

][

What is the percent fraction dissociation of cocaine reacted with water? F = 0.0372 M and Kb = 2.6x10-6 for cocaine?

%83.00083.0M0372.0

M101.3

F

x 4

Monoprotic Acid-Base Equilibria Weak Acid Base Equilibria

5.) Example

A 0.0450 M solution of benzoic acid has a pH of 2.78. Calculate pKa for this acid. What is the percent fraction dissociation?

Monoprotic Acid-Base Equilibria Buffers

1.) A buffered solution resists changes in pH when acids or bases are added Buffer: is a mixture of a weak acid and its conjugate base

- Must be comparable amounts of acid & base For an organism to survive, it must control the pH of each subcellular

compartments- Enzyme-catalyzed reactions are pH dependent

Thermophilic archaea Picrophilus oshimae and Picrophilus torridus grow at pH =0.7

(Stomach acid 1-3 pH)

Nature (London) (1995), 375(6534), 741-2.

Bacteria growing in hot springs (acidic pH)

Bacteria growing in lung tissues (neutral pH)


2.) Mixing a Weak Acid and Its Conjugated Base Very little reaction occurs Very little change in concentrations Example:

Consider a 0.10 M of acid with pKa of 4.00

0.10-x x x

%1.3031.0M10.0

M101.3

F

x 3

3a

2101.3xK

xF

x


2.) Mixing a Weak Acid and Its Conjugated Base Example:

Consider adding 0.10 M of conjugate base with pKb of 10.00

0.10-x x x

6b

2102.3xK

xF

x

56

102.3M10.0

M102.3

F

x

HA dissociates very little and A- reacts very little with water


3.) Henderson-Hasselbalch Equation Rearranged form of Ka equilibrium equation:

][

]][[

HA

AHKa

][

][][

][

]][[

HA

AlogHlog

HA

AHlogKlog a

][

][][

HA

AlogKlogHlog a

Take log of both sides:

rearrange:

pH pKa

][

][

HA

AlogpKpH a


3.) Henderson-Hasselbalch Equation Determines pH of buffered solution

- Need to know ratio of conjugate [acid] and [base]

- If [A-] = [HA], pH = pKa

- All equilibria must be satisfied simultaneously in any solution at equilibrium- Only one concentration of H+ in a solution

Similar equation for weak base and conjugate acid

][

][

HA

AlogpKpH a

][

][

BH

BlogpKpH a

pKa is for this acid

[A-]/[HA] pH

100:1 pKa + 2

10:1 pKa + 1

1:1 pKa

1:10 pKa - 1

1:100 pKa - 2

Buffers

3.) Henderson-Hasselbalch Equation A strong acid and a weak base react “completely” to give the conjugate acid:

Also, a strong base and a weak acid react “completely” to give the conjugate base:


Weak base

Strong acid

conjugate acid

Weak acid

Strong base

conjugate base


3.) Henderson-Hasselbalch Equation Example:

Calculate how many milliters of 0.626 M KOH should be added to 5.00 g of MOBS to give a pH of 7.40?

What is the pH if an additional 5 mL of the KOH solution is added?

FW = 223.29

pKa = 7.48


4.) Why Does a Buffer Resist Changes in pH? Strong acid or base is consumed by B or BH+

Maximum capacity to resist pH change occurs at pH=pKa

Buffer Capacity (): measure of a solutions resistance to pH change

5.) Choosing a Buffer Choose a buffer with pKa as close as possible to

desired pH Useful buffer range is pKa ± 1 pH units Buffer pH depends on temperature

and ionic strength activity coefficients

pH

C

pH

C ab

d

d

d

d

where Ca and Cb are the number of moles of strong acid and strong base per liter needed to produce a unit change in pH

When preparing a buffer, you need to monitor the pH.

Can not assume the added HA and A- will yield the desired pH.

pH dependent on: - activity - temperature - ionic strength

Wide number of buffers available that cover an essential complete range of pHs.

Choose a buffer with a pKa as close as possible to the desired pH.

monoprotic acid-base equilibria

Documents

strong acids

concentration of strong

large concentration

pure waterany acid

seeking ph h

strong baserelationship

weak acids

added h