Titration Curve

Plateau RegionpH=pKa

Half-way PointpKa

HCOOH

0 0.5 1.0

pH

HCOO--2

4

6

8

10

What is the conjugateacid and conjugate base?

[HCOOH] = [HCOO-]

Smallest change in pH

Base equivalent added.

The titration curves of several weak acids.What 2 things do you notice?

Other important considerations:*In titrations of weak acids: When the acid is < 50% titrated,

pH < pKa; *At half titration, pH = pKa; *When acid is more than half titrated, pH > pK*Reacting acid with base generally yields salt and H2O*Stoichiometry of acid and base: need to consider equivalents:

Acid – dissociate into 2 H; base dissociates into 1 OH, so would need 2 base molecules to neutralize 1 acid molecule; would have a mol ratio of 1:2 acid:base or 2:1 base:acid

*Can a base be titrated?*Antilog of x = 10x pKa = -log Ka pH = -log[H+] So

10-pKa = Ka [H+] = 10-pH

Stoichiometry = Proportions

Polyprotic acid• When an acid has more than 1 dissociable H+,

dissociate in discrete steps; each proton of the weak acid has a distinct pKa

• Polyprotic acid – an acid that has more than one dissociable proton; for example phosphoric acid (H3PO4) has 3 pKa and 3 equivalents of OH- are required for neutralization

H3PO4 H+ + H2PO4- pKa = 2.15

H2PO4- H+ + HPO4

-2 pKa = 7.20**

HPO4-2 H+ + PO4

-3 pKa = 12.4

Figure 2.13 The titration curve for phosphoric acid.

Titration Curve of Phosphoric Acid

Polyprotic acidpK1 = 2.15pK2 = 7.2pK3 = 12.4

Sample ProblemIf the total cellular concentration of phosphate is 20 mM in plasma

at pH = 7.4. What is the distribution of the major phosphate species? pKa = pK2 = 7.2

pH = pKa + log [HPO4-2] or 7.4 = 7.2 + log [HPO4

-2] [H2PO4

-] [H2PO4-]

0.2 = log [HPO4-2] = 100.2 = [HPO4

-2] =1.58 What does this tell you? [H2PO4

-] [H2PO4-]

Tells you ratio is 1.58 or 1.58/1 so total of 2.58 partsKnow [HPO4

-2] + [H2PO4-] = 20 mM (or 0.02 M)

1.58 + 1 = 20 mMWhat portion of 0.02 M or 20 mM is [HPO4

-2] and [H2PO4-] ?

20 mM / 2.58 = 7.75 mMSo [HPO4

-2] = 12.25 mM and [H2PO4-] = 7.75 mM

Why pKa of 7.2?

What do you know?

What does A tell you?

What formula can be used with B?

What does C tell you? What can be determined?

What can be determined? What formula to use?

At pH=7.4pKa=2.15, 7.2, 12.4

Use pKa nearest pH; pKa =7.2

[A-]= HPO4-2

[HA]= H2PO4-

pH=7.4pKa = 7.2

7.4=7.2 +log [HPO4-2]

[H2PO4-]

0.2 = log [HPO4-2]

[H2PO4-]

Know 0.02 M and 2.58 parts;20 mM /2.58 =7.75 mM

[phosphate] = 20 mM

0.02 M phosphate

pH=pKa+log [A-] [HA]

Antilog x = 10x 1.58 (7.75)= 12.25 mM [HPO4

-2]

Use HH equation

100.2 = [HPO4-2] = 1.58

[H2PO4-]

So ratio = 1.58/1 or 2.58 parts

1(7.75)=7.75 mM [H2PO4

-]

What is the question being asked? How much -2 P compared to -3 P. A B C D E

Calculate the ratio of H2PO4- and HPO4

-2 at pH 3.7, 6.7, and 10.7.Recall: H3PO4 H+ + H2PO4- pK1 = 2.15

H2PO4- H+ + HPO4-2 pK2 = 7.20

HPO4-2 H+ + PO4

-3 pK3= 12.4

Which pKa should be used if the buffer is to be effective? Always use the pKa nearest the pH.Which of these buffers actually function in the cell?

Hint to work the homework problem 1:pH = pKa + log [conjugate base] pH-pKa = log [conjugate base]

[acid] [acid]

10(pH-pKa) = [conjugate base] Not graded [acid]

Homework: Practice Problem 1.

Before next class: Read pages 41-43 in text.

Pay special attention to page 43.

Good practice problems: end of chapter 2: 1a and 1b, 2a-d , and 3;

Later 4-7 as we get to them

More on Acids and Bases• Acids donate a proton (H+) and bases accept a proton • Needs to be expanded. Lewis: acids accept an electron pair

and bases donates an electron pair. Brǿnsted: brought in the idea of conjugate acids and bases.

Eliminates the need to only use weak and strong acids in water. For example, ammonia (NH3) and the ammonium ion. Important in drug design.

For our purposes, we will work almost always in water.

HA + H2O H3O+ + A- or can infer the water –

HA H+ + A- which infers a weak acid in water.

Situations where you add a strong acid/base to a weak acid/base OR add a salt solution to a weak acid/base. What then? Take it one step at a time!1. Recall strong acids, bases, salts completely dissociate,

so if have acetic acid (weak) and add sodium acetate (salt): then have HA H+ + A- (don’t know how much without the pKa). Also have NaA Na+ + A-. So have 2 sources of A-. Must consider both: Must add [A-] from HA to the [A-] from NaAc.

2. Sometimes will have 2 sources of H+ (mixing a strong and weak acid). Again, must consider BOTH sources of H+ in your calculations: add [H+] from HA to the [H+] from strong acid.

HA H+ + A- must consider Ka and pKa

HCl H+ + Cl- 100% dissociation

3. If adding a strong base – requires special consideration. Must consider [OH- or A-] from the weak base and the strong base. The strong base will neutralize an equivalent number of mol of HA – gives you a new starting [HA] AND yields an equivalent amount of A-. Why?

HA H+ + A-

NaOH Na+ + OH-

4. Remember to take one step at a time. Sometimes you will have to calculate several intermediate values before you can calculate what the problem is asking for.

Cont’d

5. NOTE: At any point during the titration of HA, the pH can be calculated using the Henderson-Hasselbalch Equation:pH = pKa + log [A-] / [HA]

Example: After adding 100 ml of 0.1 M KOH to 500 ml of 0.1 M of weak acid (Ka = 1 x 10-5, pKa = 5), what is the pH?

What we know: 0.1 L of 0.1 M KOH has been added to a weak acid.0.1 M = 0.1 mol/L (0.1 L) = 0.01 mol, So 0.01 mol of OH- neutralize 0.01 mol of HA and forms 0.01 M of A-. How many mol HA remains? How much did we start with?

Started with 0.1 M = 0.1 mol/L (0.5 L) = 0.05 mol of HA initially. Added 0.01 mol of OH- (neutralized 0.01 mol of HA). So have 0.05 – 0.01 = 0.04 mol HA left. Generated 0.01 mol A-

Cont’d

Continued: The volume also has changed but NOT the ratio of mol A-/mol HA (mol is volume independent); So mol of HA remains = 0.04 mol

Use HH equation: pH= pKa + log [A-]/[HA]pH = 5.0 + log 0.01/0.04 = 5.0 + log 0.25 A- from HA can be ignored. Why?

Then have a new problem:The log of 0.25 is a negative number!! pH can NOT be a negative #.

**To avoid dealing with the log of a negative number – Henderson-Hasselbalch can be written as follows:

Change pH= pKa + log [A-]/[HA] to pH = pKa – log [HA]/[A-] in the same way –log x = log 1/x.

Gone From: pH= pKa + log [A-]/[HA] To: pH = pKa – log [HA]/[A-]

So pH = 5 – log 0.04/0.01 = 5 - log 4 = 5 - 0.602= 4.40Can be seen in the Henderson-Hasselbach derivation (next slide).

This is with 100 ml of 0.1 M KOH.

If the problem reads: When 250 ml of 0.1 M KOH is added, then HA is 50% titrated. Why? What does this tell you?[HA] = 0.1 M = 0.1 mol/L (0.5 L) = 0.05 mol[KOH] = 0.1 mol/L (0.25 L) = 0.025 mol of OH- (neutralizes 50% of HA)

Tells you [A-] = [HA] and log [A-]/[HA] = log 1 = 0So pH = pKa.

Derivation of Henderson-HasselbalchStart with Ka = [H+][A-] / [HA]

Rearrange the expression by (a) inverting each side and (b) multiply by Ka and [H+] :1/Ka = [HA]/ [H+] [A-] Ka[H+] (1/Ka) = Ka [H+] ([HA]/ [H+][A-])=(Ka [H+]) [HA]/ [H+] [A-])

Gives you: [H+] = Ka ([HA]/[A-])Take the -log of each side: -log [H+] = -log Ka –log[HA]/[A-] Which is –pH = pKa – log[HA]/[A-]Or pH = pKa + log [A-]/[HA]

Buffers – Why important?• Maintenance of pH is vital to all cells, hence have a variety of

mechanisms to keep intracellular and extracellular pH stable. When altered – disease.

• Primary mechanism of pH maintenance: buffer systems• Buffer = solutions that tend to resist a change in pHas acid or

base is added; composed of weak acid + conj. base• Properties of cellular buffer systems: pKa near pH 7

chemically compatible with the cell

• 2 primary buffer systems that maintain intracellular pH:(1) phosphate (HPO4

-2 / H2PO4-) system and (2)

histidine (an amino acid) buffer system • Primary buffer system that maintains extracellular (blood) pH

is the bicarbonate / carbonic acid (HCO3- / H2CO3) system

Buffers – What do they do?1. Buffers stabilize the pH. 2. Buffer needs both – weak acid and its conjugate base.

Why? Together can partially absorb additions of H+ or OH- to the system.

3. When pH = pKa, [conjugate base] and [weak acid] are equal – greatest buffering capacity.

4. Note that the titration curve at and near pKa is flat, ie, the pH remains relatively constant. Basis of buffers: addition of more acid or base is absorbed by H+ and/or A- of the buffer.

• NOTE: the pH does change in a buffered system – the change is much less – dependent on strength of the buffer and [conjugate base]/[acid] ratio

5. Buffers can only be used reliably within one pH unit of their pKa. Why? At ph>1 from the pKa, the concentration of one of the buffer components is too low to absorb the influx of H+ and OH-6. Molarity of a buffer is defined as the SUM of the concentrations of the acid and conjugate base forms.7. Cellular buffers – must have a pKa near 7 AND compatible with the metabolism of the cell; intracellular and extracellular fluids must maintain a relatively constant pH.

Intracellular buffers: *phosphate system (HPO4-2 / H2PO4

-)

histidine system; pH range 6.9-7.4Extracellular fluid: bicarbonate/carbonic acid system

(HCO3- / H2CO3) Blood is maintained at pH

7.4; Closely connected to metabolism, lung and kidney function.

Buffers - Properties

Buffer Problem. What are the concentrations of HA and A- in a 0.2 M acetate buffer, pH=5 ? (acetic acid pKa = 4.77, Ka = 1.7 x 10-5) Buffer means has acid and conjugate base.NOTE: Can calculate 2 ways: using Ka or Henderson Hasselbalch.

Using Ka: Ka = [H+][A-]/[HA] Using HH: still x = [A-]; [HA] = 0.2 – x 1.7 X 10-5 = [H+][A-] /[HA] 5 = 4.77 + log (x) / 0.2-xx = [A-]; [H+]=10-5; [HA] = 0.2 – x; 0.23 = log (x) / 0.2 - x1.7 X 10-5 = (10-5)(x) / 0.2 – x antilog 0.23 = x/0.2-x 1.7 X 10-5 (0.2-x) = 1 X 10-5x 1.7 = x / 0.2-x3.4 X 10-6 – 1.7 X 10-5x = 1 X10-5x 1.7 (0.2 – x) = x3.4 X 10-6 = 2.7 X 10-5 x 0.34 – 1.7x =xx = 3.4 X 10-6 / 2.7 X 10-5 0.34 = 2.7 xx = 0.126 M = [A-] x = 0.34/2.7 or x = 0.126 M A-0.2 -0.126 = 0.074 M HA 0.2 – 0.126 = 0.074 M HA