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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
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