essay

Chapter9 Chapter 8 ( new edition)

Monoprotic Acid-Base Equilibria

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Review of Fundamentals

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 Decay

Pollutants Contribute to Acid Rain

Forest Destruction

Yellowstone Air Pollution (same view) 2

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Acids and Bases Reactions

Heat and gases

Review of Fundamentals

Knowledge of Acid-Base Equilibrium is Necessary to Understand:  Buffer preparation and pH control  Acid-Base Titrations  Complexation, etc.

 Acids and bases are essential to virtually every application of chemistry and for the intelligent use of analytical procedures such as chromatography and electrophoresis.

 It would be difficult to have a meaningful discussion of, say, protein purification or the weathering of rocks or acid reflux without understanding acids and bases.

Protein function and stability depends on pH, temperature and other conditions.

-

Protein activity is pH dependent

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Review of Fundamentals

- 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

M101.0 0.10

13

13

=×−=−=

×= ×

==⇒×==

==

−+

− −

++

M)log(]log[HpH

101 ][OH

][H101]][OH[H

0.1M[KOH]][OH 14-

14--

-

w w

K K

1.00M)(0.1log]log[HpH

M0.1[HCl]][H

=−=−=

== +

+

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Review of Fundamentals 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

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Strong Acids and Bases

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

6.00M)101.0log(]log[HpH

M101.0 10.01

101 ][OH

][H

M10.01[KOH]][OH

6

6 8-

14-

-8-

=×−=−=

×= ×

× ==

×==

−+

− −

+ wK

How can a base produce an acidic solution?

M10.01[KOH]][OH -8- ×≠≠  Wrong Assumption!!

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Strong Acids and Bases  Wrong Assumption!!  clearly, there is something wrong with our calculation. In particular, we have

not considered the contribution of OH− from the ionization of water. In pure water, [OH−] = 1.0 × 10−7 M, which is greater than the amount of KOH added to the solution.

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

 To handle this problem, we resort to the systematic treatment of equilibrium.

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Strong Acids and Bases  Systematic Treatment of Equilibrium

Step 1: Pertinent reactions: K w

Step 2: Charge Balance: ][][][ -OHHK =+ ++

Step 3: Mass Balance:

][][ 8100.1K −+ ×= All K+ comes from KOH

Completely dissociates, not pertinent

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

Step 4: Equilibrium constant expression (one for each reaction): 14

w 100.1OHHK −−+ ×== ]][[

Step 5: Count equations and unknowns: Three equations:

Three unknowns: ][OH],[H],[K -++

][][][ -OHHK =+ ++

][][ 8100.1K −+ ×=

14 w 100.1OHHK

−−+ ×== ]][[

(1)

(2)

(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 −×=+×)((

][][][ -OHHK =+ ++

][][ 8100.1K −+ ×=

14 w 100.1OHHK

−−+ ×== ]][[

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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

12 100.1-14100.1100.1

x

78

14-288

−−

−−

×−×=

×−×±×− =

)( ))(()(

02.7106.9logHlogpH

M106.9H 8-

8-

=×−=−=

×= +

+

)(][

][

Negative number is physically meaningless

Use quadratic equation

pH slightly basic, consistent with low [KOH]

solution of a quadratic equation:

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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

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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 AH

Ka −+

=

][ ]][[

B OHBH

Kb −+

=

)Klog(pK aa −= )Klog(pK bb −=

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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)

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Weak Acid Equilibria

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 given the formal

concentration of HA and the value of Ka.4 Let's call the formal concentration F

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):

][ ]][[

HA AH

Ka −+

= 14

w 100.1OHHK −−+ ×== ]][[

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Weak Acid Equilibria

Using General Systematic Treatment of Equilibrium  Find pH for a solution of a general weak acid HA given the formal concentration

of HA and the value of Ka. Let's call the formal concentration F ,

 After many steps of Systematic Treatment

xF )x)(x(

HA AH

Ka − ==

−+

][ ]][[

0)K)(F(x)K(x aa 2 =−+

]x[ )1(2

)K)(F)(1(4KK x a

2 aa −= +±−

=

Rearrange:

Solve quadratic equation:

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F-x x x

Ka=

Weak Acid Equilibria

Fraction of Dissociation  defined as the fraction of the acid(HA) in the form A−:

 Example:

 That is, the acid is 14% dissociated  at a formal concentration of 0.050 0 M

 Percent dissociation increases with dilution

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

%1414.0 M0500.0

M108.6 F x 3

== ×

== −

α Figure 9-2 Fraction of dissociation of a weak electrolyte increases as electrolyte is diluted. The stronger acid is more dissociated than the weaker acid at all concentrations o-Hydroxybenzoic acid is more dissociated than p-hydroxybenzoic acid at the same formal concentration because the ortho isomer is a stronger acid

ortho

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F-x x x

Example A Weak-Acid Problem

Find the pH of 0.050 M trimethylammonium chloride.

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Ka=

Ka=

]x[ )1(2

)K)(F)(1(4KK x a

2 aa −= +±−

=

Weak Base Equilibria 1.) Treatment of Weak Base is Very Similar to Weak Acid

-Using General Systematic Treatment of Equilibrium  Assume all OH- comes from base and not dissociation of water

][ ]][[

B OHBH

Kb −+

=

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):

14 w 100.1OHHK

−−+ ×== ]][[

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Weak Base Equilibria

1.) Treatment of Weak Base is Very Similar to Weak Acid

xF x

xF )x)(x(

B OHBH

K 2

b − =

− ==

−+

][ ]][[

0)K)(F(x)K(x bb 2 =−+

C

]OH[ )1(2

)K)(F)(1(4KK x b

2 bb −= +±−

=

Rearrange:

Solve quadratic equation:

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 After many steps of Systematic Treatment F-x x x

A Weak-Base Problem Example :

Find the pH of 0.10 M ammonia. Solution When ammonia is dissolved in water, its reaction is

Therefore, Kb for NH3 is

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Weak Base Equilibria Problem

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

46 2

101.3106.2 0372.0

−− ×==⇒×= −

= xOH x

x Kb

Kb=2.6x10-6

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Weak Base Equilibria Problem Example

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

11 4

14 102.3

101.3 100.1

][ ][ −−

−

− + ×=

× ×==

OH KH w

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

49.10)102.3log(]Hlog[pH 11 =×−=−= −+

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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)

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Fraction of Association  Fraction of Base B in BH+ form (α):

 Example:

F x

)xF(x x

BBH BH

= −+

= +

= +

+

][][ ][

α

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

%83.00083.0 M0372.0

M101.3 F x 4

== ×

== −

α 25

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xF x

xF )x)(x(

B OHBH

K 2

b − =

− ==

−+

][ ]][[

]OH[ )1(2

)K)(F)(1(4KK x b

2 bb −= +±−

=

F-x x x

][ ][ −

+ = OH

KH w

xF )x)(x(

HA AH

Ka − ==

−+

][ ]][[

][ )1(2

))()(1(4 ][

2 −=

+±− == x

KFKK xH aaa

F-x x x

Weak Base Equilibria Problem

Weak Acid Equilibria Problem

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 - A buffered solution resists changes in pH when acids or bases

are added or when dilution occurs. - The buffer is a mixture of an acid and its conjugate base.

There must be comparable amounts of the conjugate acid and base (say, within a factor of 10) to exert significant buffering.

Mixing a Weak Acid and Its Conjugate Base When you mix a weak acid with its conjugate base, you get what you mix!

If you mix A moles of a weak acid with B moles of its conjugate base, the moles of acid remain close to A and the moles of base remain close to B. Very little reaction occurs to change either concentration.

Buffers

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Consider an acid with pKa = 4.00 and its conjugate base with pKb = 10.00. Let's calculate the fraction of acid that dissociates in a 0.10 M solution of HA.

In a solution containing 0.10 mol of A− dissolved in 1.00 L, the extent of reaction of A− with water is even smaller.

The acid is only 3.1% dissociated under these conditions.

HA dissociates very little, and adding extra A− to the solution will make the HA dissociate even less. Similarly, A− does not react very much with water, and adding extra HA makes A− react even less.

Mixing a Weak Acid and Its Conjugated Base Very little reaction occurs Very little change in concentrations

Buffers

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Henderson-Hasselbalch Equation : for buffer  Rearranged form of Ka equilibrium equation:

][ ]][[

HA AH

Ka −+

=

][ ][

][ ][

]][[ HA A

logHlog HA

AH logKlog a

− +

−+ +==

][ ][

][ HA A

logKlogHlog a −

+ +−=−

Take log of both sides:

rearrange:

pH pKa

  

   

 +=

−

][ ][

HA A

logpKpH a

Buffers

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Henderson-Hasselbalch Equation  Determines pH of buffered solution

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

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

- Only one concentration of H+ in a solution

 Similar equation for weak base and conjugate acid

  

   

 +=

−

][ ][

HA A

logpKpH a

pKa is for this acid

Buffers

base

acid

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We use Pka for both acid and base

Example : Sodium hypochlorite (NaOCl, the active ingredient of almost all bleaches) was dissolved in a solution buffered to pH 6.20. Find the ratio [OCl−]/[HOCl] in this solution.

In Appendix G, we find that pKa = 7.53 for hypochlorous acid, HOCl. The pH is known, so the ratio [OCl−]/[HOCl] can be calculated from the Henderson-Hasselbalch equation.

Buffers Acid or base?

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Accidental mix of bleach and acid kills Buffalo Wild Wings employee Incidents like this, which create chlorine gas, occur more than 2,000 times per year in the US by Leigh Krietsch Boerner NOVEMBER 13, 2019 | APPEARED IN VOLUME 97, ISSUE 45

Henderson-Hasselbalch Equation  1-A strong acid and a weak base react “completely” to give the conjugate

acid:

 2- 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

Buffers

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 We can make buffer by :  1-mixing acid (or base) and its conjugate

 2- Or by adding a strong acid (or base) to the week base to produce the conjugate acid and make a buffer

BUFFER IN ACTION a widely used buffer called “tris,” which is short for tris(hydroxymethyl)aminomethane.

In Appendix G, pKa for the conjugate acid of tris listed as 8.072.

An example of a salt containing the BH+ cation is tris hydrochloride, which is BH+Cl−.

When BH+Cl− is dissolved in water, it dissociates to BH+ and Cl−.

BH+Cl− conjugate acid

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Example : Find the PH of a solution prepared by dissolving 12.43 g of tris ( FM 121.135) plus 4.67 g of tris hydrochloride (FM 157.596) in 1.00 L of water.

The concentrations of B and BH+ added to the solution are

Assuming that what we mixed stays in the same form, we plug these concentrations into the Henderson-Hasselbalch equation to find the PH:

Notice that the volume of solution is irrelevant, because volume cancels in the numerator and denominator of the log term:

BUFFER------ BUFFER !????

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 We can make buffer by :  mixing acid (or base) and its conjugate

If we add 12.0 mL of 1.00 M HCl to the solution used in the previous example, what will be the new pH? The key to this problem is to realize that, when a strong Acid is added to a weak base, they react completely to give BH+ We are adding 12.0 mL of 1.00 M HCl, which contains

(0.012 0 L)(1.00 mol/L) = 0.012 0 mol of H+. This much H+ will consume 0.012 0 mol of B to create 0.012 0 mol of BH+,

Effect of Adding Acid to a Buffer

We see that the pH of a buffer does not change very much when a limited amount of strong acid or base is added. Addition of 12.0 mL of 1.00 M HCl changed the pH from 8.61 to 8.41. Addition of 12.0 mL of 1.00 M HCl to 1.00 L of unbuffered solution would have lowered the pH to 1.93. 36

How Buffers Work A Buffer resists changes in pH because the added acid or base is consumed by the Buffer. As the Buffer is used up, it becomes less resistant to changes in pH.

a mixture containing approximately a 10:1 mole ratio of is prepared. Because pKa for is 7.2, the pH should be approximately

When formaldehyde is added, the net reaction is the consumption of , but not of :

[A-]/[HA] pH

100:1 pKa + 2

10:1 pKa + 1

1:1 pKa 1:10 pKa - 1

1:100 pKa - 2 37

Calculating How to Prepare a Buffer Solution

How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.60 in a final volume of 250 mL? Solution The moles of tris hydrochloride in 10.0 g are (10.0 g)/(157.596 g/mol) = 0.063 5. We can make a table to help solve the problem:

This many moles of NaOH is contained in 38

 We can make buffer by :  Or by adding a strong acid (or base) to the week base to produce the conjugate acid and make a buffer

Buffers

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

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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: 1 - activity 2 – temperature

Buffer pH and pKa depends on temperature . Tris has an exceptionally large dependence, −0.028 pKa units per degree, near room temperature. A solution of tris with pH 8.07 at 25°C will have pH ≈ 8.7 at 4°C and pH ≈ 7.7 at 37°C.

3- ionic strength  activity coefficients 40

Preparing a Buffer in Real Life!

If you really wanted to prepare a tris buffer of pH 7.60, you would not do it by calculating what to mix. Suppose that you wish to prepare 1.00 L of buffer containing 0.100 M tris at a pH of 7.60. You have available solid tris hydrochloride and approximately 1 M NaOH. Here's how you would do it: 1. Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker containing about 800 mL of water. 2. Place a calibrated pH electrode in the solution and monitor the pH. 3. Add NaOH solution until the pH is exactly 7.60. 4. Transfer the solution to a volumetric flask and wash the beaker a few times. Add the washings to the volumetric flask. 5. Dilute to the mark and mix.

You do not simply add the calculated quantity of NaOH, because it would not give exactly the desired pH. The reason for using 800 mL of water in the first step is so that the volume will be reasonably close to the final volume during pH adjustment. Otherwise, the pH will change slightly when the sample is diluted to its final volume and the ionic strength changes

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Buffers

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 (β): measure of a solutions resistance to pH

change

 Choosing a Buffer  Choose a buffer with pKa as close as possible to

desired pH  Useful buffer range is pKa ± 1 pH units

Outside this range, there is not enough of either the weak acid or the weak base to react with added base or acid.

Buffer capacity can be increased by increasing the concentration of the buffer.

pH C

pH C ab

d d

d d

−==β

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

a) Cb versus pH for a solution containing 0.100 F HA with pKa = 5.00. (b) Buffer capacity versus pH for the same system reaches a maximum when pH = pKa. The lower curve is the derivative of the upper curve.

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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.

43

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?

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In summary,

•Abuffer consists of a mixture of a weak acid and its conjugate base. The buffer is most useful when pH ≈ pKa.

•Over a reasonable range of concentration, the pH of a buffer is nearly independent of concentration.

•A buffer resists changes in pH because it reacts with added acids or bases. If too much acid or base is added, the buffer will be consumed and will no longer resist changes in pH.

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