Chem 222

#19 ReviewMar 15, 2005

Announcement

• Midterm Exam on the next Thursday this room in the same time (50 mins). Come on time.

• Answers for Quiz 4 is uploaded at the web site. Please check it.

Exp 13 Iodimetric Titration of Vitamin C

Step 2 (Standard Solution KIO3)CKIO3 = mass/(MW VKlO3)

Step 3 (Preparation of Standard I3-): IO3

- + 8I- + 6H+ ↔ 3I3- + 3H2O (16-18)↑ ↑

(a little excess) from H2SO4

IO3- : I3- = 1: (Q1) CI3

- = (Q1) CKIO3

Ascorbic acid (Vitamin C)

+ I3-+H2O→

16-7 Method Involving Iodine

Amylose(Starch)-I2 complex Purple (Detector of I3-)

I2 is only slightly soluble in H2O (1.3 mM), but it’s solubility is enhanced with I3-.

I2(ac) + I- ↔ I3- K = 7 × 102

Step 3 Standardization of thiosulfate

(S2O32-) solution by titration

I3- + 2S2O32- ↔ 3I- +

Thiosulfate

CI3-VI3-/CS2O32-VS2O3

2- = [Q1]

CS2O3 = ?

Prepared from KI & KIO3- using (16-18)↓

Analysis of Vitamin C(Backtitration with S2O3

2-)

+ I3-+H2O→

Ascorbic acid (Vitamin C)

Add Excess I3-

↓

I3- + 2S2O32- ↔ 3I- +

Thiosulfate↑Excess I3- is titrated with S2O3

Moles of I3- = Moles I3- reacted with S2O32-

+ Moles of I3- reacted with Vitamin C

(p365)

(16-19)

Exp 13 Iodimetric titration of Vitamin C

• Starch is easily degraded into glucose, which is a reducing agent. To prevent this, we use HgI2 as preservative..(p356)

Iodimetry: I3- +2e- 3I- (I2 +2e- 2I- )

Ascorbic acid:Ascorbate + H2O dehydroascorbate

+ 2H+ + 2e-

Iodometry: 3I- I3- +2e- (2I- I2 + 2e-)

Standardization: IO3

- + 8I- + 6H+ 3I3- + 3H2O iodate ↑ A slight excess of KI

I3- + 2S2O32- 3I- +

Thiosulfate

From Last Year’s Midterm Sample Quiz

A certain procedure required the preparation of ~0.05 M HCl and ~1.5 M NaOH. The molarities of these solutions were determined by standardization with Na2CO3 and KHP, respectively. Five titrations of Na2CO3 with ~1.5 M HCl and KHP with ~0.05 M NaOHwere performed, and the calculated molarities for the NaOH and HCl solutions were as follows: 0.05012, 0.05009, 0.05015, 0.05011, and 0.05001 mol/L for HCl solution; 1.5102, 1.5097, 1.5099, 1.5101, and 1.5096 mol/L for the NaOH solution.

a Perform the Q test to determine if it is necessary to discard any of the calculated molarities for HCl or NaOH.

Q (90% confidence) Number of Observations0.76 40.64 50.56 6

continued

b. Using the mean molarities of the HCl and NaOH solutions

mean, standard deviation

c. calculate the H+

concentration and the pH in each solution. Kw=1.01*10-14

For 0.05 M NaOH[H+] = Kw/[OH-] pH = pKw – [Q1]

Significant figures for LogX 10X

(Ch3 P48)

• Log0.10 = [Q1]a) -1 b) -1.0 c) -1.00 d) -1.000

• 106.142 = a) 1 × 106

b) 1.4 × 106

c) 1.39 × 106

d) 1.387 × 106

e) 1.3868 × 106

Ch6. Chemical Equilibrium

• 6-1 The equilibrium constantv1

aA + bB cC + dD [6-1]v2

v1 = k1[A]a[B]b & v2 = k2[C]c[D]D,where [A] stands for the “concentration” of A.

If the reaction reaches equilibrium, v1 = v2 k1[A]a[B]b = k2[C]c[D]d

Namely, K ≡ k1/k2 = [6-2]←The left side in [6-1] in the denominator

In evaluating an equilibrium constant, 1. Concentrations of solutes should be

expressed as M (moles/L).2. Concentration of gases should be

expressed in bars3. Concentration of pure solids, pure

liquids, solvents are omitted because they are unity.

←→

ba

dc

BADC

][][][][

6-3 Solubility Product (p104)• The solubility product is the equilibrium

constant for the reaction in which solid salt dissolves to give its constituent ions.

Hg2Cl2(s) Hg22+ + 2Cl-

Ksp = [Hg22+][Q1] [Cl-][Q2] = 1.2 × 10-18

←→

x (2x)2 = 1.2 × 10-18

4x3 = 1.2 × 10-18 x = 6.7 × 10-7 M

What happens if NaCl is already dissolved at 0.030 M? 2x << 0.030

Home work• You have 1L of 0.012 M Hg2

2+

solution. How many moles of NaCl should be added if you want to precipitate 99.99 % Hg2

2+. Assume volume does not change by adding NaCl.

• Ksp (1.2 × 10-18) = [Hg22+][Cl-]2.

6-8 pH & pOH• pOH = -log[OH-]• pH + pOH = -log[H+][OH-]

= -logKW = 14.00 at 25 °C

So when pH = 5, pOH = 14 -5 = 9.

[OH-] = 10-9

Solution is acidic if [H+] > [OH-]Solution is basic if [H+] < [OH-]

Strengths of Acids and BasesStrong Acids and BasesA strong acid or base is completely

dissociated in aqueous solution.HCl(aq) H+ + Cl- (~ 100%)KOH(ac) K+ + OH-(~ 100%)

Remember all the acids and bases in Table 6-2. Weak acids and bases

HA H+ + A- a weak acid

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

B + H2O BH+ + OH- a weak baseKb = [BH+][OH-]/[B]

←→

←→

←→

←→

7-4 The precipitation Titration Curve

• The titration curve is a graph showing how the concentration of one of the reactions varies as a titrant is added

• Because concentration of the reagent varies over many orders, it is most useful to plotpfunction: pX = -log10[X]

Q. How much is pAg+ for 0.001 M Ag+ solution?pAg+ = -Log(0.001) = (Q1)

Consider the titration of 25.00 mL of 0.1000 M I- with 0.05000 M Ag+.Ksp=[I-][Ag+]=8.3×10-17

I- + Ag+ AgI(s) (7-9)

Q1 . What volume of Ag+ titrant is needed to Reach the equivalence point?

CIVI = CAgVAge

VAge = (0.1000 M)[Q1] mL/0.05000 M

= 50.00 mL

titrant

Q2. How much is [Ag+] when 49.0 mL and 52 mL of Ag+ is titrated ?

Stream Line Calculation (p136)

• [I-] =

• [Ag+] =

)00.7400.25)(1000.0(

00.5000.1

mLmLM

mLmL

Fraction Remaining

Original [I-] Dilution Factor

Try Ex. in p 136

Total volume(25 + 49) mL

↓(50 - 49) mLBefore Equivalence Point at VAg = 49 mL

After Equivalence Point at V =52 mL

)00.7700.2)(0500.0(mLmLM

Total volume(25 + 52) mL

Original [Ag+]

What is this ?

VAge

Titration curve

↑

For 1:1 stoichiometry of reactants, the equivalence pointcan be indicated by

Excess I- Excess Ag+

pAg+

VAg+ (mL)

Home work

Consider the titration of 25.00 mL of 0.1000 M I- with 0.05000 M Hg2

2+

solution. Ksp of Hg2I2 is 1.1 × 10-28.Q1) How much volume of Hg2

2+

solution is needed to reach the equivalence point?

Q2) What is the concentration of [Hg2

2+] when you titrated 20 mLand 30 mL of Hg2

2+?

Standardization of KMnO4Q 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

7-2 Titration Calculation (p130)

Titration Mixture: Ex. (7-2 p132)Home work

y g

(1.372 – y) g

Home work1) 15.00mL of a solution containing unknown

amount of HCl is titrated by adding 60.00mL of 7.654 M NaOH, and then the excess NaOH is titrated with 13.02mL of a KHP solution (MW=204.221). The KHP solution was prepared by dissolving 0.6500g of KHP in 40.00mL of H2O.

Back titration Moles of H+ = Moles of OH-

Left = Moles of H+ from HCl+ Moles of H+ from KHP

a) Calculate the Molarity of the KHP solution

b) Calculate the number of moles of NaOHadded.

c) Calculate the moles of excess NaOH.d) Calculate the number of moles of HCl in the

initial solution.

Back titration HCl moles = CHCl VHCl; (CHCl is unknown) NaOH moles = CNaOHVNaOH

The moles of excess NaOH isCNaOHVNaOH - CNaOHVe NaOH= CNaOHVNaOH - CHCl VHCl ( Use eq. 1)

(2)

This excess moles of NaOH is titrated by KHP (see p210 about KHP). 1 moles of KHP is

required for 1 moles of excess NaOH.So Excess NaOH moles = CKHPVKHP (Answer for c)

(3)

This gives an equation about excess NaOHmoles:

CKHPVKHP = (CNaOHVNaOH - CHCl VHCl) (3) = (2)

CHCl VHCl = (CNaOHVNaOH - CKHPVKHP) (Answer for d)

Ch 13 EDTA TitrationEDTA: EthylenDiamineTetraAcetic acid

EDTA4- + M2+ EDTA-M2+

• EDTA forms a strong [Q1]: 1 complex with most metals

♦Cyanide (CN-) is called a monodendate.EDTA is called [Q2].

Using the conditional formation constant; Ex. (p265)

Q1. The formation constant for CaY2- is 4.9 ×1010. Calculate the concentrations of free Ca2+ in a solution of 0.10 M CaY2- at pH 10.00 and pH6.00. You can use the table below.

• Q2. How does [Ca2+] depend on pH?

Also try 13-5 in p279

14-1 Basic Concept of Oxidization and Reduction

• A species is said to be oxidizedwhen it loses electrons.

• It is reduced when it gains electrons.

Fe3+ + V2+ Fe2+ + V3+

Reduced Oxidized

• A reducing agent, also called a reductant, gives electrons to another substances.

Fe3+ + V2+ Fe2+ + V3+

Oxidizing Reducingagent agent

Nernst Equation (p293 Ch 14-4)• aA(oxidized) + ne- bB(reduced)

E = E0 – (RT/nF)Log ([B]b/[A]a)= E0 – (0.05916/Q1)Log([B]Q2/[A]a)

(14-13)

Please memorize the equation(constants should be provided)

Q. Write a Nernst’s equation forbB(reduced) aA(oxidized) + ne-

E = E0 – (0.05916/n)Log(Q3)

E = -∆G/nF

∆G: Change in Free Energy from Oxidized State to Reduced StateE > 0 Ox +ne- Red

16-1 The shape of redox titration

K = 1016 in 1M HClO4 The reaction be almost 100 %

Fe2+ in 1M HClO4

Ce4+

Pt Wire

Stirring bar

Calomel Reference Electrode

At the Pt Electrode

Potentiometer(measure voltage)

At the Calomel Reference Electrode

(Note: only small amount reacts)

Ce4+ + Fe2+ Ce3+ + Fe3+

Ceric Ferrous Cerous Ferric

Titrant Analyte

Pot

entia

l (V

ver

sus

S. C

. E)

Before equivalence Excess Fe2+ & Fe3+

Little Ce4+

How we calculate the titration?Ce4+ + Fe2+ Ce3+ + Fe3+

Titrant Analyte

After equivalence Excess Ce4+ & Ce3+

Little Fe2+

What must be changed ?

In the example question, Fe2+ is titrated to Ce4+ solution

Before the end point (V < Ve)(Most Fe2+ is consumed)• [Ce4+]/[Ce3+] = (Ve-V)/V

After the end point (V > Ve)(Most Ce4+ is consumed)• [Fe2+]/[Fe3+] = (V-Ve)/Ve

Try Ex. in p351 & Demonstration in p352!

P352 Demonstration

MnO4- + 5Fe2+ + 8H+

Titrant Analyte Mn2+ + 5Fe3+ + 4H2O

Dissolve 0.392 g of Fe(NH4)(SO4)2 6H2O (FM 392.13; 1.00 mmol) in 500 mL of 1 M H2SO4. Titrate the well-stirred solution with 0.020 M KMnO4 .

Q1. How much KMnO4 solution is required to reach the equivalence point?

1.00 mmol/5 = 0.020 M × Ve Ve = 10 mL

Q2. Calculate the cell voltage when volume of the titrant, V, is 5 mL, 10 mL, 20 mL

At V = 5 mL, ♦ the fraction of Fe2+ oxidized into Fe3+ is V/Ve♦ The fraction of remaining Fe2+ is (Ve -V)/Ve

At V =10 mL, what is the ratio between [MnO4-] and

[Fe2+]?

At V =20 mL,♦ the fraction of Mn2+ is Ve/V.♦ the fraction of MnO4

- is (V-Ve)/V.

Q1

Q2

Q3 Q4

E = h[Q5] = h[Q6]/λ

What are the Diagrams?

A compound has a molar mass of 292.16 was dissolved in a 5-ml volumetric flask. A 1.00 ml aliquot was withdrawn, placed in a 10-ml volumetric flask, and diluted to the mark. The absorbance at 340 nm was 0.427 in a 1.000-cm cuvet. The molar absorptivity for this compound at 340 nm is 6130 M-1cm-1.

a. Calculate the concentration of the compound in the cuvet.

A =εCb C = A/εb

b. Calculate the concentration of a compound in the 5-ml flask.Dilution: C0V0 = CV C0 = CV/V0

c. How many milligram of compound were used to make the 5-ml solution?

C0: Molarity ?

Atomic Emission

Fe + hot plasma Fe* Fe + hν

Atomic Emission Spectrum

Multi-element Analysis

Excited State

Extremely Narrow Line0.001 nm Resolution

Ch 21 Atomic Spectroscopy

• 21-1 Overview2000-3000 K

Why do we need a laser ?

Memorize the diagrams

λ

Standard Addition (Exp 20)

In environmental or biological samples, there is often interference by other species.A = kC A = k’C

So it is difficult to estimate k’ with standard Samples in H2O. What do you do in standard addition?

Standard Addition in Atomic Absorption (p510)

Slope 0.0188

Slope 0.0308

How to get Henderson Equation?

• Henderson-Hasselbalch EquationExpecting pH is also simple

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

-Log Ka = -Log[H+] - Log{Q1}

pKa = pH - Log {Q1}

pH = pKa + Log][][

HAA−

Remember the derivation

Kb = {Q2}/[B] Ka ?

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}