Chem 222

#29 ReviewApr 28, 2005

Announcement• Please meet me after the class if

you have any conflicts with the final exam schedule.

• You can expect similar questions with Quiz 6 in the final exam.

• If you have not received your notebook, reports, or quiz 5 back, please contact your section TA

• Your section TA is supposed to notify your current grade by this Saturday. If not please contact TA.

• I have a office hour this Friday and next Monday. I will be also available at T/R in class time.

What is next ?

• Chem 343 Physical Chemistry Lab(Instrumental analysis by UV/Vis,FT-IR, GC-Mass, NMR, Gas

effusion, Flash photolysis)

• Research in Lab- Choose the area of your interest- Longer term is better (1 year)

Final Exam

• 250 points ~ 40 points extra

Curve ?• If top 14 students gain 90 % Linear scaling factor ~1.10• If top 14 students gain 97 % Scaling factor ~1.0

3. KHP is a salt of the intermediate form of phthalic acid. What is approximate PH of 0.15M of KHP? (pK1 = 2.950, pK2 = 5.408)

a PH = 2.2b PH = 4.2c PH = 6.2d PH = 8.2e PH =11.2

4. The amino acid, Arginine, has the following forms. H3Arg2+ ------- H2Arg+ -------- HArg ---------- Arg-

Pk1=1.82 Pk2 = 8.99 Pk3 = 12.48

What are the principal form and the second most abundant form of arginine at PH=10.0?

Principal the second most abundant form

a) H2Arg+ H3Arg2+

b) H2Arg+ HArgc) H2Arg+ Arg-

d) HArg Arg-

e) HArg H2Arg+

5. There are two different acids in each beaker. One beaker contains weak acid (0.0200 M, 100.0 mL, pKa = 6.00). The other contains strong acid (0.0200 M, 100.0 mL). These acids are titrated with strong base, 0.0200 M NaOH. Choose the answer that gives closest PHs for weak acid and strong acid at the equivalence point?

PH of weak acid PH of strong acida) 6.12 7.00b) 7.00 6.12 c) 7.00 7.00d) 7.00 9.00e) 9.00 7.00

AH + OH- A- + H2O

For week acids A- + H2O AH + OH-F’ x x

Ve=100 mLx = (F’ Kb)1/2

= {0.0200 M (100 mL /200 mL) 10-8.00}1/2

= 1.00 × 10-5 M pOH = 5

• When V =0,[Ag+] = Ksp/[X-] (1)pAg+ = -Log[Ag+] (2)[X-] = 0.1 M

Higher pAg+

Q1. (a. Higher b Lower) [Ag+]Q2. (a. Higher b Lower) Ksp

• When V = Ve,[Ag+] = [X-] [Ag+] = (Q3)

Higher pAg+

Q4. (a. Higher b Lower) [Ag+]Q5. (a. Higher b Lower) Ksp

6. When you are titrating 0.1 M KI, KCl and KBrsolutions with 0.05 M AgNO3, you obtained the following titration curve. Then, Ksp of AgI, AgCl and AgBr will be (p138)

a. Ksp(AgI)> Ksp (AgCl)> Ksp(AgBr)b. Ksp(AgCl) > Ksp(AgBr) > Ksp(AgI)c. Ksp(AgBr) > Ksp(AgI) > Ksp(AgCl)d. AgI, AgCl and AgBr all have same Ksp

e. None of the above

[Ag+] = Ksp/[X-]

How much is [X-] in the equivalence point?

8. From the following reduction potentials:I2 (s) + 2e- → 2 I - Eº = 0.535 VI2 (aq) + 2 e- → 2 I - Eº = 0.620 VI3

- + 2 e- → 3 I - Eº = 0.535 V

Choose all the correct statement(s). No points unless you answer all correctly. (Hint: E0 = (0.05916/n)LogK ).

a) The equilibrium constant of reaction:I2 (aq) + I - → I3

- is 300. b) According to the equilibrium constant of I2 (aq) + I - → I3

- , 0.1 g of molecular iodine in solid forms can be dissolved in 100 ml water completely.

c) The reaction: I2 (aq) + I - → I3 - is

spontaneous at standard condition.d) The reaction: I2 (s) + I- → I3- is

spontaneous at standard condition.e) The equilibrium constant of reaction I2 (s) + I- → I3- is 1.0.

• K = 10(E0*n/0.05916)

where E0 = E+0 – E-

0

I2 (aq) + 2 e- → 2 I - E+º = 0.620 V- I3

- + 2 e- → 3 I - E-º = 0.535 V------------------------------------------------------------------------

I2 (aq) + I - → I3 – E0 = 0.085 V

K = 747 ≠300 E0> 0: Spontaneous

I2 (s) + 2 e- → 2 I - E+º = 0.535 V- I3

- + 2 e- → 3 I - E-º = 0.535 V------------------------------------------------------------------------

I2 (s) + I - → I3 – E0 = 0.00 V

K = 1 Not spontaneous

10. The following is the unbalanced reaction:

Cr2O72- + Fe2+ → Cr3+ + Fe3+

Which statement(s) is correct for the reaction? No points unless you answer all correctly.

a) The balanced reaction is: 1Cr2O7

2- + 3Fe2+ + 14H+ → 2Cr3+ + 3Fe3+

+ 7H2Ob) The balanced reaction is: 1Cr2O7

2- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+

+ 7H2O

c) Cr2O72- acts as a strong oxidant in this

reaction. Fe2+ Fe3+ + e-

d) Overall reaction, Cr2O72- was oxidized by

Fe2+

Cr(VI)2O72- 2Cr3+

e) Fe2+ was reduced by Cr2O72-.

Q 11

Cd electrode Ag electrode

Cd(NO3)2 (aq) Ag(NO3) (aq)

Salt bridge

Switch P

Switch R

Light bulb

+-

Consider the following statements about the above setup with 1M of electrolyte (AgNO3 and Cd(NO3)2)A. When switch P is closed or turned on (switch R is open or turned off) the setup works as a galvanic cell and the light turns on.B. When switch R is closed or turned on (switch P is open or turned off) the silver ion is reduced.C. When the switch P is closed or turned on (switch R is open) Cd electrode functions as an anode.

From the above statements the incorrect statements area) A, B and C b) A and Cc) A and B d) B and C e) Other than a)-d)

:Battery 1.5V Cd2+ + 2e- Cd (s) Eo = -0.402 V

Ag+ + e- Ag (s) Eo = 0.799 V

12 Assume that the light bulb in the above circuit is replaced by an analog voltmeter and switch P is closed or turned on (switch R is off). Both the electrolyte solutions are 1.00000 M. The reading in the voltmeter could be

a) -1.001 V b) 0.397 Vc) 1.201 V d) 2.000 V

13 The potential of a galvanic cell could be affected by

A. Temperature of the electrolytes in half cellsB. Concentration of the electrolytes (ions) in half cells

C. What kind of electrodes are useda) A, B and C b) A and B

c) only B d) B and C

Final Exam

5 Sections• Each section contains 6 multiple

choices and 1 long question• Multiple choice 3 each×6 = 18• Long question 40 each• 58 × 5 = 290 • You can gain up to 250

5 sections• Ch 0-6• Ch 7, 13• Ch 10-12• Ch 16, 18, 23, 25, 26 • Ch 14, 15

Ch 0-6

• Significant figures • Sig fig in arithmetic (Log, etc)

Log(3.39×10-5) (see Ch 3)• Q-test• Quiz 6

• Definition and calculation of pH• Ksp

Ksp for CuCl2 isKsp = [Q1 ]

Long q from Ch 0-6

Ch 6-3, 6-4, 6-5• Solubility, Common Ion, Separation

by precipitation

When pKsp = 3.00, Ksp = [Q1]

How many sig fig do you need?

How much Hg2I2 can you dissolve in 1L of solution when Ksp = 1.1 ×10-28 ?Ksp = [Hg2

2+][Q1]= x (Q2) x = ??

Ch 7(titration), Ch 13 (EDTA)

• Word definitions in 7-1 (end point, equivalence point, etc)

• Titration curves in Ch 7• Spectroscopic titration 7-3

Dilution effect• Precipitation titration

Ch 7, 13 Long q

• EDTA titration (13-3)- Conditional formation constant- pMetal in a titration curve

• One topic from the previous list

Ch 10, 11, 12Acid base & titration

• How to calculate pH for - strong acid/base [H+] [OH-]- weak acid/base Ka/Kb

- buffer pH = pKa + Log(Q1)a. [A-]/[HA] b. [HA]/[A-]

• Polyprotic acid

Mixing a weak acid and its conjugated base

• HA H+ +A- pKa = 4.00 Ka = 10[Q1]

F- x x x F = 0.01 M

x2/(F-x) = Ka x =3.1×10-3

Fraction of dissociationα =[A-]/{[A-] +[HA]}

= x/F = 0.031 =3.1 %

What happens if you add A- to H2OHA H+ +A-

How much fraction of A- reacts with water in a solution containing 0.10 M of A-?

A- +H2O HA + OH- pKb =[Q2] F’ –y y y

y2/(F’-y) = Kb y =3.1× 10-6 & α = 3.1 × 10-5

10-4 Weak base equilibria

• B + H2O BH+ + OH-

x x[Q1]

bKxFx

BOHBH

=−

=−+ 2

][]][[

Ex. (p188) Find the pH of 0.10 M Ammonia with pKb = 4.756.

NH3 + H2O NH4+ + OH-

(F-x) x x

X2/(F-x) = Kb =1.75 ×10-5

x = ([ Q2 ] × 1.75 ×10-5)1/2

= 1.3 × 10-3 M[OH-] = (Q1 ) [H+] = (Q2)/[OH-]

Ex. Effect of adding acid to a buffer

• We prepare 1L of solution containing 0.102 M of tris and 0.0296 M of trishydrochloridewith pKa of 8.075.

pH = pKa + Log[B]/[BH+]

pH = (Q1) +Log(Q2 /0.0296) = 8.612

Let’s add 12 mL of 1M HCl to the solution

Ex. How to prepare a buffer solution (p194)

• How many milliliters of 0.500 M NaOHshould be added to 10.0 g (0.0635 moles) of tris hydrochloride to give a pH of 7.60 in a final volume of 250 mL? pKbof tris is 5.925.

• BH+ + OH- BInitial 0.0635 x Final 0.0635 –x x

pKa = (Q2) – pKb

[Q1] = pKa + Log[B]/[BH+][B]/[BH+] = 10{(Q1) – pKa}

x / (0.0635 – x) = 10{(Q1) – pKa}

Diprotic systems

R pKa1 R pKa2 R| 2.33 | 9.74 |

H3N+CHCO2H ↔ H3N+CHCO2- ↔H2NCHCO2

-

H2L+ HL L-

Equilibrium in Diprotic SystemsH2L+ ↔ HL + H+ Ka1≡K1

HL ↔ L- + H+ Ka2 ≡K2

L- +H2O ↔ HL + OH- Kb1=KW/[Q1]HL +H2O ↔ H2L- + OH- Kb2 =[Q2]

How much is the pH of 0.05 M of L2H solution for L of pK1 = 2.33 ?

Simplified Calculation for the Intermediate Form

• [H+] = 2/1

1

112

++FKKKKFK

a

waaa

Two additional assumptions

Ka2F >> Kw

[H+] =

Ka1 << F

[H+] =

2/1

1

12

+FKKFK

a

aa

( ) 2/121

2/112

aaaa KK

FKFK

=

pH = -Log[H+] = -Log(Ka1Ka2)1/2

= {-LogKa1 –LogKa2}/2 = [Q1] (11-12)

(Remember This )

Ch 10-12 Long q

• Acid/Base Titrations • Polyprotic acid/base• Study also previous quiz

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 HBr

At Equivalence PointVe × [Q1] = 50.00×0.02

Ve = 10.0 mL

pH = ?

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

Ch 16, 18, 21, 23, 25, 26

• Ch 16• Ch 18 Remember Equations Relationship of A, P, T Beer’s law• Ch 23, 25, 26

- Lecture notes- HPLC, plate #, resolution,

particle size-

• Quiz 6• Previous quiz

Ch 16, 18, 21, 23, 25, 26 Long q

• Chapter 16 (Redox titration)• Other

Balance the following half reactionsMnO4

- → Mn2+

Mo3+ → MoO22+

Combine the net reactionsExpect the potential in a titration

Standardization of KMnO4• Suppose that 0.3562g of Na2C2O4 is dissolved in a

250.0mL volumetric flask. If 10.00 mL of this solution require 48.36 mL of KMnO4 solution for titration, what is the molarity of KMnO4 solution ?

Oxalate concentration COX is

COX = 1.0633 mM

Lmolgg

VolumeMWWeight

2500.0)/00.134/(3562.0

solution of /OCNa of 422 =

moles 2moles [Q1]

KMnO4 of Oxalate of

44

==MnOMnO

OXOX

VCVC

MolesMoles

mmol8794.0000.5000.2

36.4810001063.0

52

7

3

44

=

××

==mL

mLMVVCC

KMnO

OXOXKMnO

Ch 14, 15

• Structure of pH electrode- What are involved as components

• pH combination electrode • Quiz 6• Reference electrode• Reaction spontaneous or not

Long QuestionNernst equationLine notationCell reactions