acid-base mixtures, buffers

Practical aspects of buffers

Chemistry 201

NC State University

Lecture 15

The everyday pH scale To review what pH means in

practice, we consider the pH

of everyday substances that

we know from experience.

Remember that [H+] = 10-pH.

pH + pOH = 14

Therefore that [OH-] = 10pH-14.

Two ways to make a buffer

Add the acid and conjugate base to the solution in a defined proportion.

Method 1 Method 2

Add a strong acid to the weak base (or vice versa) until the desired proportion [A-]/[HA] is obtained.

Buffer strength The ratio [A-]/[HA] should be as close as possible 1:1,

but the amounts may vary. To make a stronger buffer

you simply need to increase the amount of each

component. Let’s investigate.

Suppose we add 1 mL of 1 M HCl to 1 liter of solution.

The final concentration of HCl is 0.001 M.

Buffer strength The ratio [A-]/[HA] should be as close as possible 1:1,

but the amounts may vary. To make a stronger buffer

you simply need to increase the amount of each

component. Let’s investigate.

Suppose we add 1 mL of 1 M HCl to 1 liter of solution.

The final concentration of HCl is 0.001 M. pH = 3

Buffer strength

The ratio [A-]/[HA] should be as close as possible 1:1, but the

amounts may vary. To make a stronger buffer you simply

need to increase the amount of each component. Let’s

investigate.

Suppose we add 1 mL of 1 M HCl to 1 liter of solution. The

final concentration of HCl is 0.001 M. pH = 3

Suppose we add 1 mL of 1 M HCl to 1 liter of 10 mM optimal

phosphate buffer solution (pKa = 7.2) solution. The final

concentration of HCl is 0.001 M. pH = pKa + log10([A-]/[HA])

Buffer strength

Keeping in mind that the unbuffered solution in this example

([HCl] = 0.001 M) would be pH = 3

Suppose we add 1 mL of 1 M HCl to 1 liter of 10 mM

optimal phosphate buffer solution (pKa = 7.2) solution. The

final concentration of HCl is 0.001 M.

pH = pKa + log10([A-]/[HA])

[A-] = 0.005 – 0.001 = 0.004 [HA] = 0.005 + 0.001 = 0.006

pH = 7.2 + log10(0.004/0.006) = 7.02

If the target pH = 7.2 (i.e. pH = pKa) then this buffer is too

weak. An error of 0.2 pH units could be significant.

Buffer strength


([HCl] = 0.001 M) would be pH = 3





[A-] = 0.05 – 0.001 = 0.049 [HA] = 0.05 + 0.001 = 0.051

pH = 7.2 + log10(0.049/0.051) = 7.18

If the target pH = 7.2 (i.e. pH = pKa) then this buffer is

reasonable. The difference is only -0.02.

Buffer strength


([HCl] = 0.001 M) would be pH = 3





[A-] = 0.150 – 0.001 = 0.149 [HA] = 0.150 + 0.001 = 0.151

pH = 7.2 + log10(0.149/0.151) = 7.194

If the target pH = 7.2 (i.e. pH = pKa) then we would say that

this buffer is definitely strong enough, difference = -0.006

You can create a buffer either by adding the acid and

Its conjugate base to a solution or by titrating in strong

base to acid (or vice versa). Remember, regardless of

the method used to prepare it:

The buffering strength is maximum when [HA] = [A-]

The buffering range is considered to extend from

[HA] / [A-] = 0.1 to [HA] / [A-] = 10. This is subjective.

Wertz suggests 0.01 to 100 is an acceptable range.

Titrating to make a buffer

Understanding the titration curve

Starting point [HA] = [HA]0

Suppose we want to make a buffer by titrating [OH-].

We cannot use the H-H equation initially. We do not know

the concentration of [A-]. Instead at this initial point we will

use the other form of the

equilibrium constant and

make an ICE table.

Added [OH-] = 0


Starting point [HA] = [HA]0

We can calculate x = [H+]

and therefore the pH from

the equilibrium constant.

Added [OH-] = 0


Maximum buffer capacity

[HA] = [A-].

When we have a buffer we can use the Hendersen-Hasselbach

equation. This is nice since it is the simplest treatment of the

acid-base equilibrium. In the case shown we have pH = pKa.

Added [OH-] = 1/2 [HA]0


Maximum buffer capacity

[HA] = [A-] when

[OH-] ~ 0.5 [HA]0

The buffer range is defined as approximately from:

pH = pKa – 1 R = [A-]/[HA] = 0.1 [OH-] ~ 0.091 [HA]0

to

pH = pKa + 1 R = [A-]/[HA] = 10 [OH-] ~ 0.91 [HA]0

Buffer region


Once the solution moves outside the buffer range the pH shoots

up. The equivalence point is reached when the added is equal

to the original acid concentration, i.e. [OH-] ~ [HA]0

At this point one can no longer

use the H-H equation. Instead,

we assume that [A-] ~ [HA]0

Then we use the base equil-

-ibrium:

Note that pKb = 14 - pKa

Buffer region

Types of buffers There are inorganic buffers, e.g. phosphate, but there are

many more organic buffers. In fact, the number of buffers

is staggering.

Organic buffers: Tris, HEPES, MOPS, MES…

“Biological” buffers: citrate, acetate, carbonate, malonate,

Proteins themselves are polyelectrolytes and therefore

tend to the buffer the solution they are in. This can have

important physiological impact (e.g. hemoglobin).

Tris buffer Tris(hydroxymethyl)aminomethane), is

an organic buffer with the formula

(HOCH2)3CNH2. Tris has a pKa = 8.07.

The buffer range is 7.07 – 9.07.

It is is widely used as a component for

solutions of nucleic acids, proteins and for

any application in which phosphate is not a

good choice.

For example, calcium phosphate has a low solubility product,

Which means that phosphate buffers are a poor choice in

any application where calcium is present.

Important point: Tris can react with aldehydes since it is a

primary amine. Choice of buffer should done by consulation

of the chemical interactions in your application.

Other organic buffers

HEPES: pKa = 7.5

MOPS: pKa = 7.2

Citrate buffer

Citric acid crystals

under polarized light

Citric acid is found in abundance

in citrus fruits. It is also part of the

citric acid cycle in biochemistry.

It is a triprotic acid, with three

carboxylic acid groups as seen

by its structure (on the right).

pKa1 = 3.14

pKa2 = 4.75

pKa3 = 6.40

Citrate buffer: species in solution

The middle H atom has the lowest

pKa. This is because the neighbor-

ing OH group has an electron with-

drawing effect that stabilizes the

negative charge created.

Amino acids are amphipathic

All amino acids contain the carboxylic acid and amino

group. These have very different pKa values so the amino

acids have a doubly charged form (zwitterion) at pH 7.

In proteins only the N- and C-terminus have these pKas.

Amino acids can be classified in part according to the pKa

of their side chains.

Amino acid side chain pKa

Proteins are made of up of a

number of “titratable” amino

acids. At pH 7 the terminal

carboxyl, aspartate and

glutamate have a negative

charge. Terminal amino,

lysine and arginine are

positively charged. Others

are neutral, but can be

charged due to interactions

within the protein. The pKa

of any of these groups may

be altered by the protein.

The isoelectric point of a protein pI

Proteins have many titratable groups on their surface. It is

not possible to define a single pH since all of the groups

have different pKa values. However, we can define the point

at which protein is neutral in charge: the isoelectric point.

At the isoelectric point the protein has 0 net charge, which

means that there as many positive as negative groups on

the surface.

The isoelectric point concept applies to polymers,

nanoparticles etc. Any macromolecule can be described in

terms of its overall charge.

IMPORTANT: When pH = pI a macromolecule has a neutral

surface. This is the minimum stability point. Macromolecules

tend to precipitate at this poin.

Hemoglobin is the most abundant protein inside of red blood cells and accounts for one-third of the mass of the cell. During the conversion of CO2 into HCO3

-, H+ liberated in the reaction are buffered by hemoglobin, which is reduced by the dissociation of oxygen. This buffering helps maintain normal pH. The process is reversed in the pulmonary capillaries to re-form CO2, which then can diffuse into the air sacs to be exhaled into the atmosphere.

Hemoglobin as a buffer

As with the phosphate buffer, a weak acid or weak base captures the free ions, and a significant change in pH is prevented. Bicarbonate ions and carbonic acid are present in the blood in a 20:1 ratio if the blood pH is within the normal range. With 20 times more bicarbonate than carbonic acid, this capture system is most efficient at buffering changes that would make the blood more acidic. This is useful because most of the body’s metabolic wastes, such as lactic acid and ketones, are acids. Carbonic acid levels in the blood are controlled by the expiration of CO2 through the lungs.

Bicarbonate (hydrogen carbonate): an important regulator in the body

The role of carbonic anhydrase

The enzyme does not change the equilibrium, but it accelerates the Rate of reaching the equilibrium on each side of a membrane.

Acid/Base Mixtures : Reactions

• How do you calculate pH after an acid/base reaction occurs?

Text : Section 7.3

What is the pH when:

a) 25 mL of 0.30 M HCl are added

to 35 mL of 0.20 M NaOH?

b) 15 mL of 0.25 M HClO4 are added

to 25 mL of 0.20 M NaOH?

Examples: Strong acids and bases

What is the pH when 25 mL of 0.30 M HCl are

added to 35 mL of 0.20 M NaOH?





Step 1. Calculate dilutions. First add the volumes

Total volume = 25 mL + 35 mL = 60 mL

Calculate concentrations in the solution




Step 2. Write a balanced chemical reaction for the limiting

reaction and the excess reaction.

Limiting reaction

Species HCl NaOH Na+ Cl-

Initial 0.125 0.117 0.0 0.0

Difference -x -x x x

Final 0.125-x 0.117-x x x




Step 2. Write a balanced chemical reaction for the limiting

reaction and the excess reaction.

Limiting reaction

Excess reaction

Species HCl NaOH Na+ Cl-

Initial 0.125 0.117 0.0 0.0

Difference -0.117 -0.117 0.117 0.117

Final 0.008 0.0 0.117 0.117

Species HCl H+ Cl-

Initial 0.008 0.0 0.0

Final 0.0 0.008 0.008




Recognize that both HCl and NaOH are strong acid/base,

respectively. Therefore, rather than find the equilibrium

constant, we assume that the reaction goes to completion.

In this case we find the limiting reagent which is NaOH.

In the general case we could include both H+ and OH- on

the right hand side. We may not know initially which one

is going to dominate, since we must first calculate the

limiting reagent.




Short cut method:

Step 1. calculate number of moles of each reagent




Short cut method:

Step 1. calculate number of moles of each reagent

Step 2. calculate the total volume (0.025 + 0.035 = 0.060 L)

Step 3. make a table considering only H+ and OH-

Species H+ OH- H2O

Initial 7.5 7.0 0.0

Difference -7.0 -7.0 +7.0

Final 0.5 0.0 7.0




Short cut method:

Step 4. calculate the final concentration of [H+ ]

Step 5. calculate the pH

15 mL of 0.25 M HClO4 are added to 25 mL of

0.20 M NaOH?



0.20 M NaOH?






0.20 M NaOH?


Step 2. Write a balanced chemical reaction for the limiting rxn.

Excess rxn is

Species HClO4 NaOH Na+ ClO4-

Initial 0.0937 0.125 0.0 0.0

Difference -0.0937 -0.0937 0.0937 0.0937

Final 0.0 0.0313 0.0937 0.0937

Species NaOH Na+ OH-

Initial 0.0313 0.0 0.0

Final 0.0 0.0313 0.0313

Strong Strong

Acid Base

Weak Weak

Acid Base

use rxn table

goes 100%

calc pH or pOH

What is the pH when 50. mL of 0.25 M NaOH are

added to 40. mL of 0.20 M HF?

Examples: One strong and one weak

What is the pH when 50.0 mL of 0.25 M NaOH are

added to 40.0 mL of 0.20 M HF?








Step 2. Write a balanced chemical reaction and determine

the form of the equilibrium constant. Make a reaction table.




Step 2. contd. We calculate 1/Kb from the data in the tables

Ka = 6.6 x 10-4.

Step 3. Make a reaction table.

Species HF OH- F-

Initial 0.0888 0.139 0.0

Difference -x -x x

Final 0.0888-x 0.139-x x




Step 4. Solve for x




Step 5. Calculate OH- and pOH.

We see from the table that [OH-] = 0.139 – x = 0.0511

BIG PICTURE: This example is very high on the titration

Curve. We can get [OH-] approximately from

[OH-] = [OH-]0 – [HF] 0 = 0.139 – 0.888 = 0.051

The additional information from the ICE is in the third

decimal place.

Strong base exceeds weak acid

The key point of the previous problem is that we are no longer

in the buffer range. We cannot use H-H in this case. Since:

[OH-]0 > [HA]0

While Kb still applies it is often

unnecessary since [OH-] is in

excess.

If you need to use Kb

then use:

Buffer region

Strong Strong

Acid Base

Weak Weak

Acid Base

use rxn table

goes 100%

calc pH or pOH

use rxn table

goes 100%

buffer : H-H

base : pOH

What is the pH when 25.0 mL of 0.40 M HCl are

added to 40.0 mL of 0.30 M NH3?


What is the pH when 25.0 mL of 0.40 M HCl are

added to 40.0 mL of 0.30 M NH3?


The total volume is 65 mL so the final concentrations

are:

[HCl] = 25/65(0.40 M) = 0.154 M

[NH3] = 40/65(0.30 M) = 0.184 M

In this case the [HCl] < [NH3] so this will make a buffer.

Assume that the strong acid reacts completely then at

equilibrium we have:

[NH3] = 0.184 – 0.154 M = 0.03 M and [NH4+] = 0.154 M

Strong Strong

Acid Base

Weak Weak

Acid Base

use rxn table

goes 100%

calc pH or pOH

use rxn table

goes 100%

buffer : H-H

base : pOH

use rxn table

goes 100%

buffer : H-H

acid : pH

For a reaction of a weak acid and a weak base

we need to calculate the equilibrium constant

from the known Ka’s. We take the example of

ammonium acetate.

We see that the overal reaction is composed

of two acid-base equilibria

Examples: Weak acid and weak base

for acetate

for ammonia

Therefore, the overall equilibrium constant for

the reaction is

Now, that we can see how to calculate the

Equilibrium constant, we can solve any acid-base

Reaction problem using the standard methods

That we have used.

Step. 1. determine dilutions

Step. 2. set up the reaction table

Step. 3. solve for the unknown and then calculate pH


20.0 mL of 0.30 M NaH2PO4 was added to

20.0 mL of 0.30 M NaHS. What are the concs.

of all species at equil.?


20.0 mL of 0.30 M NaH2PO4 was added to

20.0 mL of 0.30 M NaHS. What are the concs.

of all species at equil.?


Solution: Look up the Ka for each reaction involved

in this acid-base equilibrium.


Species

Initial 0.15 0.15 0.0 0.0

Difference -x -x x x

Final 0.15-x 0.15-x x x

Strong Strong

Acid Base

Weak Weak

Acid Base

if not conjugates

calc. K

calc concs.

if conjugates

use H-H

calc pH

Goals • Calculate the pH of acid/base mixtures

• Calculate the pH at any point in a titration

• Calculate the concentration of all species in acid/base mixtures at equilibrium

acid-base mixtures, buffers

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