Chem 222

#19 Ch 11, 12Nov 4, 2004

Announcement• Quiz 5 on Nov 16 (Tue)Ch 11, 12, 13, (14)• Final Schedule Tuesday Dec 7 (6-8 pm) OK?• The area to be covered in the final

includes all the area we studied• Midterm Exam Answers are Upat the web site• Your lab notebook will be graded at

the end. Organize neatly, use a pen, but do not modify it (modified notebook will not accepted).

Grade for Lab Reports up to KH3-14

0

5

10

15

20

25

30

50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 90-95 95-100

Point Range (%) up to KH3-14

# of

stu

dent

s

Series1

Read the summary p211

~ (K1K2)1/2

11-2 Diprotic Buffers

You can use either or both of the equations

pH = pK1 + Log[HA-]/[H2A]pH = pK2 + Log[A2-]/[HA-]

Ex. Find the pH of a solution prepared by dissolving 1.0 mmol of KHP and 2.0 mmolof Na2P in a 1L of H2O.

pH = pK2 + Log{[P2-]/[HP-]}

pH = 5.408 + Log{1/2}

A2-HA-H2A

• Ex2 . How much volume of 0.800 M of NaOH should be added to 0.80 L of 0.50 M KHP solution to give a pH of 6.4. pK1= 3.0 & pK2 = 5.4.

Initial moles of KHP is 0.80 L × 0.50 M= 0.40moles. Assume x moles of OH- is added into solution.

• HP- + OH- P2- +H2O 0.40 x

0.40-x x

pH = pK2 + Log[P2-]/[HP]

[P2-]/[HP] = 10{pH-pK2} =10{6.4-5.4}

x/(0.4 – x) = 101.0 = 10

X = 4 -10x x =4/9 = 0.44

V (L) × 0.800 (M)= 0.44 moles V = 0.44/0.800 = 0.55 L

11-4 Which is the principal species? (p214)

(1) What is the principal form of benzoic acid at pH 8

pH = pKa + Log[A-]/[AH]

(2) How much is the fraction of ?

Example (p214)♦What is the predominant form of

ammonia in a solution at pH 7.0 (Ka= 9.24) ?

♦ Approximately what fraction is in this form?

7.0 = [Q1] + Log[NH3]/[NH4+]

[NH4+]/[NH3] = 10[Q1]

11-3 Polyprotic Acids and Bases

You can obtain pH from F Try Example in P213

pH ~(Q1)

pH ~(Q2)

Example (p215)

♦What is the principal form of Arg at pH = 10? What is the second major species?

H2Arg+H3Arg2+

HArgArginine

Arg-

(8.99 + 12.48)/2 = 10.74

10 = 8.99 + Log[HArg]/[H2Arg+]10 = 12.48 + Log[Arg-]/[HArg]

11-5 Fraction Composition Equation

HA H+ + A-; Ka = [H+][A-]/[HA]

αA- =

Q. How much is the fraction of A- for a given pH?

αA- =

][][][

−

−

+ AHAA

a

a

a KHK

AKAHA

AHAA

+=

+=

+ +−−+

−

−−

−

][][/]][[][

][][][

From [HA]= [A-][H+]/Ka

αHA = ][][

][−+ AHA

HA

(11-18)

αHA = (Q1) (11-17)

Monoprotic Systems

Diprotic Systems• For the monoprotic systems.

• K1 = [H+][HA-]/[H2A] [H2A] = [H+][HA-]/K1

• K2 = [H+][A2-]/[HA-] [A2-] = K2[HA-]/[H+]

αH2A =

DH

KKKHHH

HHAKHAKHAHKHAH

AHAAHAH

2

2112

221

1

2

2

][][][][

]/[][][/]][[/]][[

][][][][

+

++

+

+−−−+

−+

−−

=++

=

++=

++

αHA = αH2A K1/[H+] =[H+]K1/D

αA = αHA K2/[H+]= K1K2/D

(11-19)

(11-21)(11-20)

11-6 Isoelectric and Isoionic pH

• Isoionic point is the pH when neurtral polyprotic acid HA is dissolved in water.Majority is HA.

Only ions are H2A+, A-, H+, OH-

In general, [H2A+] ≠ [A-]

♦Iosoelectric point is the pH at which concentration of H2A and A- are equal. Average charge Zero

pH = (pK1 + pK2)/2

Isoionic point:

Isoelectric point:

Isoelectric Focusing

Ch 12 Acid-Base Titration (p225)

• 12-1 Titration of Strong Acid with Strong Base

• Titration of 50.0 mL of 0.0200 M KOH with 0.100 M KBr

At Equivalence PointVe × 0.100 = 50.00×0.02

Ve = 10.0 mL

pH = ?

• Titration of 50.0 mL of 0.0200 M KOH with 0.100 M KBr

• Region 1: Before the equivalence pointWhen V = 3 mL, what is the pH?

[OH-]

• Region 2: After the equivalenceBeyond the equivalence point, we add excess HBr. When V = 10.50 mL[H+]

MM

VVVC

VVV

OH

OHIniOH

e

e

0132.00.30.50

0.50)0200.0(0.100.30.10

)(

=+

−

=

+−

= −

↑Fraction of Remaining OH-

MmL

mLM

VVVVC

OHeIniH

4102.8)5.100.50(

15.0)1000.0(

1)(

−

−

×=+

=

+−=

↑Moles ofExcess H+

12-2 Titration of weak acid with strong base

Titration of 50.0 mL of 0.02000 M MES with 0.100 M NaOH

Ve 0.100 M = 50.0 mL× 0.0200M Ve = [Q1 ] mL

Home Work

• 11-A, 11-B, 11-C, 11-D, 11-2, 11-4 thru 11-6, 11-14 thru 11-18, 11-22, 11-23, 11-29

• 12-A thru 12-C, 12-2 thru 12-6,• Read the rest of Ch 12