BUFFERS AND TITRATIONSDr. HarrisCh 21 Suggested HW: Ch 21: 1, 13, 29

Buffers

• Solutions that contain both a weak acid and its conjugate base are called buffers.

• Buffers have the unique ability to resist sharp changes in pH• A buffer is able to do this because it contains both acid and base

components

• However, the acidic and basic components of the buffer themselves do not neutralize each other, a special property fulfilled by mixing conjugate pairs• If an external acid or base is added to the system, it is instantly

neutralized by one of the conjugate species

Creating A Buffer

• For example, an HF/F- buffer can be created by mixing 0.1M HF with 0.1 M NaF. Since NaF will fully dissociate, we ignore the neutral Na+. In solution, you’d have:

• Although HF is an acid, and F- is a base, the two CAN NOT neutralize each other because the reaction has no preferred direction.

• Thus, by mixing an acid with its own conjugate base, you create a buffered solution in which the components do not affect each other.

𝑯𝑭 (𝒂𝒒 )+𝑭−(𝒂𝒒)𝑭− (𝒂𝒒 )+𝑯𝑭 (𝒂𝒒 )𝑵𝑶 𝑹𝑿𝑵

How Buffers Work

HF F-HF F- HF F-

Initial Buffer solution

Base Added (NaOH)

pH slightly increases

pH slightly decreases

• The acidic component of the buffer neutralizes any base that is added to the system. The basic component of the buffer neutralizes any added acid. The resulting changes in pH are very small.

Acid Added (HCl)

C.I.R.L.: Blood as a Buffer

• Blood is the best example of a buffer in everyday life.

• The pH of the human body is 7.4. If the pH of the body becomes to acidic or basic, certain proteins and enzymes are chemically altered, rendering them inactive

• If the body’s pH drops below 6.5 or above 8.0, death may occur.

• Human blood acts as the body’s buffer to prevent pH changes

Blood as a Buffer

• O2 is carried throughout the body by binding with the hemoglobin protein found in red blood cells.

• Hemoglobin, Hb, binds both H+ and O2. These binding processes are represented below

• As expected by LeChatlier’s principle, when blood reaches oxygen-poor tissue, the reaction shifts left, releasing O2 in that region.

ΔH > 0

O

O

Blood As A Buffer (ex. exercising)

𝐶6𝐻12𝑂6→𝐶𝐻3𝐶𝑂𝐶𝑂𝑂−+𝐴𝑇𝑃+𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠

2𝐶𝐻3𝐶𝐻𝑂𝐻𝐶𝑂𝑂𝐻 (𝐿𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑)

anaerobic condition

𝐶𝑂2+𝐴𝑇𝑃+𝑁𝑎𝐷𝐻in O2

ExerciseNeed energy fast. Must burn more carbs.

Breathe faster.Bring in more O2

Heart rate increases to transport extra O2

Combustion of carbs yields CO2 and H+ in the muscles

Not enough O2? Anaerobic breakdown of glucose yields lactic acid

Muscle Blood

H+

CO2

O2

Concentration Gradients Shift Equilibria

Blood as a Buffer

• A sudden decrease in pH triggers the brain to increase the breathing rate, releasing CO2 faster and shifting the equilibrium right to consume H+.

• The body reacts O2 faster, which shifts the hemoglobin binding equilibrium left, removing more H+.

• Lactic acid is neutralized by bicarbonate, HCO3-. Blood pH remains intact.

• The primary buffer systems in human blood are a carbonic acid/bicarbonate buffer system and the HbH+/HbO2 system

𝐻+¿ (𝑎𝑞) +𝑯𝑪𝑶𝟑− (𝑎𝑔) 𝑯𝟐𝑪𝑶𝟑 (𝑎𝑞 ) 𝐻2𝑂 (𝐿 ) +𝐶𝑂 2(𝑔)¿

𝑯𝒃𝑯 +¿+𝑂 2𝑯𝒃𝑂 2+𝐻+¿¿ ¿

Calculating the pH of a Buffer Using the Henderson-Hasselbalch Equation

• The Henderson-Hasselbalch equation is used to determine the pH of a buffer system.

• Example: Calculate the pH of a formic acid buffer (HCOOH) that is .25M HCOOH and .15 M of the conjugate base (HCOO-), given that the Ka of formic acid is 1.8 x 10-4

𝒑𝑯=𝒑 𝑲 𝒂+𝒍𝒐𝒈 ( [𝒃𝒂𝒔𝒆 ][𝒂𝒄𝒊𝒅 ] )

𝑝𝐻=− log (1.8 x10−4 )+ log( .15.25 )𝑝𝐻=3.53

Calculating the pH of a Buffer Solution After Exposure to an Acid/Base

• Let’s calculate the pH of a buffer solution following a response to a pH disturbance.

• It is important to know that a molecule of acid completely neutralizes a molecule of base, forming the conjugate base and conjugate acids

• So, to determine changes in concentration of either the acidic or basic component of a buffer, simply subtract the number of moles that are neutralized from the active component, and add those to the inactive one

acid baseneutralized!!

Example• Ex. You have a 100 mL of a buffer that is 0.10 M in CH3COOH and 0.10 M in

NaCH3COO at a pH of 4.74. If 10 mL of 0.10 M HCl is added to the buffer solution, what is the new pH? What is the change in pH?

𝑪𝑯𝟑𝑪𝑶𝑶𝑯 (𝑎𝑞 )+𝐻2𝑂 (𝐿)𝑪𝑯𝟑𝑪𝑶𝑶− (𝑎𝑞 )+𝐻3𝑂+¿ (𝑎𝑞 )𝐾𝑎=1.8𝑥 10

− 5¿

• Since HCl is an acid, it will only react with the basic component of the buffer, CH3COO-

• There are exactly .001 moles of HCl added to the buffer, so exactly .001 moles of CH3COO- will be consumed, forming exactly .001 moles of additional CH3COOH

Example, contd.

.001 mol HCl

Before addition of 10 mL of 0.10 M HCl

CH3COOH CH3COO-

.01 mol .01 mol

100 mL

CH3COOH CH3COO-

.011 mol .009 mol

110 mL

After

Example, contd.

• The new concentrations are:

[𝐶𝐻3𝐶𝑂𝑂𝐻 ]= .011𝑚𝑜𝑙.110𝐿 =.𝟏𝟎𝑴

[𝐶𝐻3𝐶𝑂𝑂−]= .009𝑚𝑜𝑙.110𝐿 = .𝟎𝟖𝟏𝟖𝑴

Buffer works well. This is a very small drop in pH considering that HCl is a strong acid.

ΔpH = -.09

* Since both the acid and base are in the same volume of solution, you may substitute moles for concentrations in the equation. The ratio is the same.

Group Example

• You have 100 mL of a buffer that is 0.3M NH3 and 0.45 M NH4Cl. The pH of the buffer is 9.07. Find the pKa. Calculate the change in pH after the addition of 5 mL of 4.0 M NaOH?

Buffer Capacity and pH Range• Two important characteristics of a buffer are its capacity and pH range.

• Buffer capacity is the amount of acid or base that a buffer can neutralize before the pH begins to change vastly.• ex. A buffer of 0.1M HF/ 0.1M NaF has the same pH as 1.0 M HF/ 1.0 M

NaF, but less buffer capacity

• pH range is the range over which a buffer works effectively. In real practice, the maximum buffer range is reached when one component is 10 times more concentrated than the other.

• • for B:A ratio of 10:1 , • for B:A ratio of 1: 10,

• When designing a buffer, the pKa of the acid should be as close as possible to the desired pH

max buffer range = pKa ± 1

Example: Selecting and Preparing a Buffer

• You work for a major pharmaceutical company that has developed a new drug to combat sickle cell anemia. You have a fresh batch of live cells to test this drug. The cells must be maintained in 100 mL of a buffered solution at pH=7.4. The buffering capacity should be high, so it is recommended that the concentration of the acid component be 0.5M. Assuming that you have access to any acid and base (non-ionic forms as well as sodium salts), what conjugate acid/base pair would you use to create your buffer? What masses of each would you dissolve into solution? Use the pKa table on the next slide.

Example contd.

• Potential options for a pH=7.4 buffer would be (acid/base):

• HClO/NaClO• pKa =7.40 (buffer range of 6.40 to 8.40)

• NaH2PO4/Na2HPO4• pKa = 7.21 (buffer range of 6.21 to 8.21)

Example contd.

• Using the hypochloric acid/ hypochlorite buffer with [HClO]=0.5 M :

7 .4=7.4+ log ( [𝑁𝑎𝐶𝑙𝑂 ][𝐻𝐶𝑙𝑂 ] )

0= log( [𝑁𝑎𝐶𝑙𝑂 ][𝐻𝐶𝑙𝑂 ] )

100=[𝑁𝑎𝐶𝑙𝑂 ][𝐻𝐶𝑙𝑂 ]

1=[𝑁𝑎𝐶𝑙𝑂][𝐻𝐶𝑙𝑂 ]

Acid : 0.100 L x0.5mol H 2CO3

Lx 52.45 g𝐻𝐶𝑙𝑂mol𝐻𝐶𝑙𝑂 =𝟐 .𝟔𝟐𝐠

Base :0.100 Lx 0.5mol Na𝐶𝑙𝑂L

x74.45 g NaHCO3mol NaHCO3

=𝟑 .𝟕𝟐𝐠

Dissolve in 100 mL of solution.

Titration Curves

• In a titration, one substance of known concentration (e.g. base) is added to another (e.g. acid). A colored indicator or pH meter can be used to monitor the reaction progress.

• Using a pH meter, a titration curve can be generated, showing pH as a function of the volume of titrant added

Titration Curves

• The shape of a titration curve makes it possible to determine the equivalence point (point at which the moles of acid and base are stoichiometrically equivalent)

• Titration curves can also be used to determine values of Ka or Kb if weak acids/bases are being titrated

• To understand the characteristics of titration curves, we will examine two types of acid/base titrations:

1. Strong Base into Strong Acid2. Strong Base into Weak Acid3. Strong Base into Polyprotic acid

Strong Acid/Strong Base Titration Curves (e.g. 0.1M NaOH added to 0.1M HCl)

1. The general shape of this type of titration curve is shown to the right.

2. The initial pH is simply the pH of the HCl(aq) before addition of NaOH(aq)

3. At first, the pH rises slowly due to the presence of un-neutralized acid.

4. At the equivalence point, pH=7 and the moles of acid and base are stoichiometrically equal

5. Beyond equivalence, pH rises rapidly due to presence of free base

Group Example• Calculate the pH when 49 mL of 0.1M NaOH is added to 50 mL of

0.1M HCl?

• Calculate the pH when 51 mL of 0.1M NaOH is added to 50 mL of 0.1M HCl? (ans: pH = 11)

𝑁𝑎𝑂𝐻 (𝑎𝑞 )+𝐻𝐶𝑙 (𝑎𝑞 )→𝑁𝑎𝐶𝑙 (𝑎𝑞 )+𝐻 2𝑂 (𝑙)Initial

Final

𝑝𝐻=− log ( .0001𝑚𝑜𝑙.099𝐿 )=3

Weak Acid/Strong Base (e.g. 0.1M NaOH added to 0.1M CH3COOH with Indicator)

1. As the acid is neutralized, its conjugate base forms, resulting in the formation of a buffer when [B]/[A] > 1/10.

2. At the midpoint of the buffer region, pH = pKa; [CH3COOH] = [CH3COO-]

3. At the equivalence point, all of the acetic acid is now acetate, which is a weak base. Thus, the pH at the equivalence point is greater than 7.

Group Example

• Given that Ka of CH3COOH is 1.8 x 10-5 M, calculate the pH when 49 mL of 0.1 M NaOH is added to 50 mL of 0.1M CH3COOH.

• First, calculate the moles of acetic acid remaining

• Now, plug the concentrations into the Ka expression to find [H3O+], or use the Henderson-Hasselbach equation. The total volume is 99 mL.

I 0.005 mol 0.0049 mol 0C -.0049 mol -.0049 mol +.0049 molE .0001 mol 0 mol .0049 mol

1.8 𝑥10− 5=¿¿ MpH = 6.44

Strong Base/Polyprotic Acid Titration Curves (e.g. 0.1M NaOH added to 0.1M H3PO4)

From Last Image

• Initial point: • pH of 0.1M H3PO4

• First midpoint: • [H3PO4] = [H2PO4

-], pH at 1st midpoint = pKa # 1• First equivalence point:

• No H3PO4 remains. • Second midpoint:

• [H2PO4-] = [HPO4

2-], pH at 1st midpoint = pKa # 2• Second equivalence point:

• No H2PO4- remains

• Third midpoint: • [HPO4

2-] = [PO43-] , pH at 1st midpoint = pKa # 3

• Third equivalence point:• Only phosphate remains