Acid-Base Mixtures, Buffers - NC State: WWW4 Server

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Chemistry 201 Lecture 15

Practical aspects of buffers

NC State University

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 Method 1

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

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 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 100 mM optimal phosphate buffer solution (pKa = 7.2) solution. The final concentration of HCl is 0.001 M. pH = pKa + log10([A-]/[HA]) [A-] = 0.05 – 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 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 300 mM optimal phosphate buffer solution (pKa = 7.2) solution. The final concentration of HCl is 0.001 M. pH = pKa + log10([A-]/[HA]) [A-] = 0.150 – 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

Titrating to make a buffer 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.

Understanding the titration curve 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.

Starting point [HA] = [HA]0 Added [OH-] = 0

Understanding the titration curve 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.

Starting point [HA] = [HA]0 Added [OH-] = 0

Understanding the titration curve

We can calculate x = [H+] and therefore the pH from the equilibrium constant.

Starting point [HA] = [HA]0 Added [OH-] = 0

Understanding the titration curve 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.

Maximum buffer capacity [HA] = [A-]. Added [OH-] = 1/2 [HA]0

Understanding the titration curve 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

Maximum buffer capacity [HA] = [A-] when [OH-] ~ 0.5 [HA]0

Buffer region

Understanding the titration curve 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:

Buffer region

Note that pKb = 14 - pKa

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 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

Citric acid crystals under polarized light

Citrate buffer: species in solution

The middle H atom has the lowest pKa. This is because the neighboring OH group has an electron withdrawing 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 as a buffer 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.

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

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.

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

Examples: Strong acids and bases 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?

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

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 2. Write a balanced chemical reaction for the limiting reaction and the excess reaction. Limiting reaction Species Initial Difference Final

HCl 0.125 -x 0.125-x

NaOH 0.117 -x 0.117-x

Na+ 0.0 x x

Cl0.0 x x

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 2. Write a balanced chemical reaction for the limiting reaction and the excess reaction. Limiting reaction Species Initial Difference Final

HCl 0.125 -0.117 0.008

NaOH 0.117 -0.117 0.0

Na+ 0.0 0.117 0.117

Cl0.0 0.117 0.117

Excess reaction Species Initial Final

HCl 0.008 0.0

H+ 0.0 0.008

Cl0.0 0.008

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? 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.

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?

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? Short cut method: Step 1. calculate number of moles of each reagent

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? 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 OHSpecies Initial Difference Final

H+ 7.5 -7.0 0.5

OH7.0 -7.0 0.0

H2O 0.0 +7.0 7.0

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? Short cut method: Step 4. calculate the final concentration of [H+ ]

Step 5. calculate the pH

Examples: Strong acids and bases 15 mL of 0.25 M HClO4 are added to 25 mL of 0.20 M NaOH?

Examples: Strong acids and bases 15 mL of 0.25 M HClO4 are added to 25 mL of 0.20 M NaOH? Step 1. Calculate dilutions. First add the volumes Total volume = 25 mL + 15 mL = 40 mL

Calculate concentrations in the solution

Examples: Strong acids and bases 15 mL of 0.25 M HClO4 are added to 25 mL of 0.20 M NaOH? Step 2. Write a balanced chemical reaction for the limiting rxn. Species Initial Difference Final

HClO4 0.0937 -0.0937 0.0

NaOH 0.125 -0.0937 0.0313

Na+ 0.0 0.0937 0.0937

ClO40.0 0.0937 0.0937

Excess rxn is Species Initial Final

NaOH 0.0313 0.0

Na+ 0.0 0.0313

OH0.0 0.0313

use rxn table goes 100% calc pH or pOH

Strong Acid

Strong Base

Weak Acid

Weak Base

Examples: One strong and one weak 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 1. Calculate dilutions. First add the volumes Total volume = 50 mL + 40 mL = 90 mL Calculate concentrations in the solution

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.

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. contd. We calculate 1/Kb from the data in the tables Ka = 6.6 x 10-4.

Step 3. Make a reaction table. Species Initial Difference Final

HF 0.0888 -x 0.0888-x

OH0.139 -x 0.139-x

F0.0 x x

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 4. Solve for x

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 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

use rxn table goes 100% calc pH or pOH Strong Acid

use rxn table goes 100% Weak buffer : H-H Acid base : pOH

Strong Base Weak Base

Examples: One strong and one weak What is the pH when 25.0 mL of 0.40 M HCl are added to 40.0 mL of 0.30 M NH3?

Examples: One strong and one weak 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

Examples: One strong and one weak

use rxn table goes 100% calc pH or pOH

use rxn table goes 100% buffer : H-H base : pOH

Strong Acid Weak Acid

Strong Base Weak Base

use rxn table goes 100% buffer : H-H acid : pH

Examples: Weak acid and weak base 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 for acetate

for ammonia

Examples: Weak acid and weak base 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

Examples: Weak acid and weak base 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.?

Examples: Weak acid and weak base 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.

Examples: Weak acid and weak base Species Initial Difference Final

0.15 -x 0.15-x

0.15 -x 0.15-x

0.0 x x

0.0 x x

Strong Acid

Strong Base

Weak Acid

Weak Base

if conjugates use H-H calc pH

if not conjugates calc. K calc concs.

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

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Acid-Base Mixtures, Buffers - NC State: WWW4 Server

Chemistry 201 Lecture 15 Practical aspects of buffers NC State University The everyday pH scale To review what pH means in practice, we consider t...

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