monoprotic acid-base equilibria
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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. Forest Destruction. - PowerPoint PPT PresentationTRANSCRIPT
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)
Monoprotic Acid-Base Equilibria Review of Fundamentals
2.) Knowledge of Acid-Base Equilibrium is Necessary to Understand: Buffer preparation and pH control Acid-Base Titrations Complexation, etc. Illustration:
Blood clot formation on heart valve
Milk precipitation on heart valve
Protein function and stability depends on pH, temperature and other conditions.- Blood and milk have high protein concentrations- Common problem in mechanical heart valve is clot formation- Protein precipitation from milk provides a cheap and easy mimic of blood clotting for testing new heart valves avoids expensive and long animal studies
Protein activity is pH dependent
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) 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
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) Strong Acids and Bases pH at other concentrations of a strong base
Relationship between pH and pOH:
[OH-] (M) [H+] (M) pH1x10-1 1x10-13 13.01x10-2 1x10-12 12.01x10-3 1x10-11 111x10-4 1x10-10 101x10-5 1x10-9 9
C25 at 14.00KlogpOHpH ow
Acid Ka
HCl 103.9
HBr 105.8
HI 1010.4
HNO3 101.4
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) 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!!
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) 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
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) 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
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) Strong Acids and Bases Systematic Treatment of Equilibrium
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)
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) Strong Acids and Bases Systematic Treatment of Equilibrium
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
3.) Strong Acids and Bases Systematic Treatment of Equilibrium
Step 6: Solve (Seeking pH [H+]):
Solve the quadratic equation:0100.1x100.1x 1482 )(
M101.1orM106.9x
12100.1-14100.1100.1x
78
14-288
)())(()(
02.7106.9logHlogpH
M106.9H8-
8-
)(][
][
Negative number is physically meaningless
Use quadratic equation
pH slightly basic, consistent with low [KOH]
Monoprotic Acid-Base Equilibria Review of Fundamentals
3.) 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
Monoprotic Acid-Base Equilibria Review of Fundamentals
4.) 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+
Monoprotic Acid-Base Equilibria Review of Fundamentals
5.) 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
][]][[
HAAHKa
][]][[
BOHBHKb
)Klog(pK aa )Klog(pK bb
Monoprotic Acid-Base Equilibria Review of Fundamentals
5.) 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.) Any Effect that Increases the Stability of the Product Drives the Reaction Forward Formation of internal hydrogen bond for an acid/base
Ortho isomer is stronger acid because product forms internal hydrogen bond
Can’t form internal hydrogen bond
Monoprotic Acid-Base Equilibria Weak Acid Equilibria
2.) 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
Step 2: Charge Balance:
][][][ -- AOHH
Step 3: Mass Balance:
][][ -AHAF F – formal concentration of acid
Step 4: Equilibrium constant expression (one for each reaction):
][]][[
HAAHKa
14
w 100.1OHHK ]][[
Monoprotic Acid-Base Equilibria Weak Acid Equilibria
2.) 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 ][]][[
HAAHKa
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
Monoprotic Acid-Base Equilibria Weak Acid Equilibria
2.) General Systematic Treatment of Equilibrium Find pH for a solution of a general weak acid HA
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
Monoprotic Acid-Base Equilibria Weak Acid Equilibria
2.) General Systematic Treatment of Equilibrium Find pH for a solution of a general weak acid HA
Step 6: Solve (Not easy to solve cubic equation results!):
Substitute into Equilibrium Equation:xAH - ][][ xFHA ][
xF)x)(x(
HAAHKa
][]][[
0)K)(F(x)K(x aa2
]x[)1(2
)K)(F)(1(4KKx a
2aa
Rearrange:
Solve quadratic equation:
Monoprotic Acid-Base Equilibria Weak Acid Equilibria
2.) General Systematic Treatment of Equilibrium Find pH for a solution of a general weak acid HA
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(4KK
x
3
3233a
2aa
Determine [OH-] from water dissociation:
123
14w 105.1
108.6101
]H[K]OH[
[H+] >> [OH-]6.8x10-3M >> 1.5x10-12Massumption is justified!
Monoprotic Acid-Base Equilibria Weak Acid Equilibria
3.) Fraction of Dissociation Fraction of acid HA in the form A-():
Example:
Percent dissociation increases with dilution
Fx
)xF(xx
HAAA
][][][
What is the percent fraction dissociation for F = 0.050 M and Ka = 1.07x10-3 for o-hydroxybenzoic acid?
%1414.0M0500.0
M108.6Fx 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
][]][[
BOHBHKb
Step 1: Pertinent reactions:KwKb
Step 2: Charge Balance:
][][][ -OHBHH
Step 3: Mass Balance:
][-F][][][ BHBBHBF F – formal concentration of base
Step 4: Equilibrium constant expression (one for each reaction):
14w 100.1OHHK ]][[
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
Step 6: Solve (Assume [BH+] >> [H+] [BH+] ≈ [OH-]):
Set [OH-]=x and substitute into Equilibrium Equation:xOHBH - ][][ xFB ][
xFx
xF)x)(x(
BOHBHK
2b
][]][[
0)K)(F(x)K(x bb2
]OH[)1(2
)K)(F)(1(4KKx b
2bb
Rearrange:
Solve quadratic equation:
Monoprotic Acid-Base Equilibria Weak Base Equilibria
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
Monoprotic Acid-Base Equilibria Weak Base Equilibria
2.) Example
What is the pH of cocaine dissolved in water? F = 0.0372 M and Kb = 2.6x10-6 for cocaine?
114
14w 102.3101.3
100.1]OH[
K]H[
Because x=[OH-], we need to solve for [H+]
49.10)102.3log(]Hlog[pH 11
Monoprotic Acid-Base Equilibria Weak Base Equilibria
3.) Illustration
http://www.stopmethaddiction.com/meth-lab-photos.htm
Consider an Illegal Drug Lab (pictured)
Criminal Investigation of the illegal lab will require identifying:
What drugs are present? How much of each drug is present? What is the purity of the drugs?
pH change or titration indicator may help identify/quantify drug.(positive field test for cocaine)
Monoprotic Acid-Base Equilibria Weak Base Equilibria
4.) Fraction of Association Fraction of Base B in BH+ form ():
Example:
Fx
)xF(xx
BBHBH
][][][
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
M101.3Fx 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)
Monoprotic Acid-Base Equilibria Buffers
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
M101.3Fx 3
3a
2101.3xK
xFx
Monoprotic Acid-Base Equilibria Buffers
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
xFx
56
102.3M10.0
M102.3Fx
HA dissociates very little and A- reacts very little with water
Monoprotic Acid-Base Equilibria Buffers
3.) Henderson-Hasselbalch Equation Rearranged form of Ka equilibrium equation:
][]][[
HAAHKa
][][
][][]][[
HAAlogHlog
HAAHlogKlog a
][][
][HAAlogKlogHlog a
Take log of both sides:
rearrange:
pH pKa
][][
HAAlogpKpH a
Monoprotic Acid-Base Equilibria Buffers
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
][][
HAAlogpKpH a
][
][BH
BlogpKpH a pKa is for this acid
[A-]/[HA] pH100:1 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:
Monoprotic Acid-Base Equilibria
Weak base
Strong acid
conjugate acid
Weak acid
Strong base
conjugate base
Monoprotic Acid-Base Equilibria Buffers
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
Monoprotic Acid-Base Equilibria Buffers
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
pHC
pHC ab
dd
dd
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.