Chem 222

#9 Ch 7Sep 21, 2004

Announcement

• The answer for quiz 2 with a grading guide line is at the web site

• The due date for report of Exp 5 is this W/R

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.

Q. Calculate the concentration of H+ and the pH in the following solutions at 25 °C: a. 0.056 M HCl, b. 0.25 M KOH, c. 5.60 M HCl. Kw = 1.01 × 10-14

a. [H+] = 0.056 M pH = -log[H+]= -log(0.056)= 1.252

b. [OH-] = 0.25 M [H+] = Kw/[H+] a. pH = -log[H+] = -log(5.60) = - 0.7482

←→

←→

Weak acids and basesHA H+ + A- a weak acid

Ka = [H+][A-]/[HA] = x2/(0.056 –x)

B + H2O BH+ + OH- a weak base

Kb = [BH+][OH-]/[B]

←→

0.056 M - x x x

←→

7-1 Titration

The color changes indicates “end point”, which is close to the “equivalent point”

The titrant functions as an “indicator”

At “equivalent point”

Indicator•BB (Blue Yellow)•MR (Yellow Red)

• pH meter (pH is detected byvoltage; see P 323)

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 5

KMnO4 of Oxalate of

44

==MnOMnO

OXOX

VCVC

MolesMoles

mmol8794.0000.5000.2

36.4810001063.0

52

7

3

44

=

××

==mL

mLMVVCC

KMnO

OXOXKMnO

Analysis of unknown

• CKMNO4 = 0.8794 mM

Calcium in a 5.00 mL urine sample was precipitated, was redissolved, and then required 16.17mL of standard MnO4- solution.

Find the concentration of Ca2+ in the urine.

Ca2+: C2O42-: MnO4

- = 5 : 5: 2

moles 2moles 5

KMnO4 of Ca of

44

==MnOMnO

CaCa

VCVC

MolesMoles

VCa = (5/2) × (CMnO4VMnO4)/VCa= (5/2) × (0.8794 mM × 16.17 mL)/5mL

Ex. (p132) (Related to Exp 8)

y g

(1.372 – y) g

Na2CO3 moles × 2 + NaHCO3 moles = HCl moles

{y/105.99 + (1.372 – y)/84.01} = CHCLVHCl = 29.11mL × 0.734 M= 0.02138 moles

(1/104.99 – 1/84.01)y + 1.372/84.01 = 0.02138

Do not forget to weigh your sample !Without 1.372 g you can never obtain the answer.

Spectrophotometric Titration (Lab. Exp 19)

• Absorbance: A = εbC

ConcentrationPathlength

(p411)

Fe3+ Binding Site of Transferrin

Apo-transferrin + 2Fe3+ (Fe3+)2transferrin

Show absorbance at 465 nm !

Titration of 2.00 mL of apotransferrinWith 1.79 × 10-3 M ferric nitrilotriacetate

Titration of 2.00 mL of apotransferrinWith 1.79 × 10-3 M ferric nitrilotriacetate

moles 1moles 2

protein of Fe of 3

==+

proteinprotein

FeFe

VCVC

MolesMoles

Apo-transferrin + 2Fe3+ (Fe3+)2transferrin

↑ 203 µL

mLLmM

VVCCprotein

FeFeprotein 00.22

20379.12 ×

×==

µ

To construct the above graph the dilution effect must be corrected.

initial

total

VVAbsObservedAbsCorrected =

Ex.

The absorbance measured after adding 125 uL of ferric nitrilotriacetate to 2.000 mL of apotransferrin was 0.260. Calculate the corrected absorbance that should be plotted in the Figure 7-5.

mLmL

VVAbsObservedAbsCorrectedini

total

000.2125.2260.0 ==

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. Consider the titration of 25.00 mL of 0.1000 M I- with 0.05000 M Ag+

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

Suppose we are monitoring Ag+ with an electrode. (7-9) is the reverse of the dissolution of

AgI(s) I- +Ag+ Ksp=[I-][Ag+]=8.3×10-17

←→

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

Almost 100 % forms AgI(s)

CIVI = CAgVAge

VAge = CIVI/CAg = (0.1000 M)25.00 mL/0.05000 M

= 50.00 mL

titrant

How does [Ag+] change before the end point?

At the equivalence point at VAge = 50.00 mL.

A. Before the equivalence point, we have an excess of I-. [Ag+] is obtained from

[Ag+] = Ksp/[I-] = 8.3×10-17/[I-]

[I-] =

AgI

AgAgII

VVVCVC

+

−=

+ volumeTotal

)Ag of (Moles - I-) Original of (Moles

If VAg = 49.00 mL

Moles of I- = (25.00 mL) (0.1000 M) - (49.00 mL)(0.05000 M)= 2.500 mmol - 2.450 mmol =0.050 mmol

Vtotal = (25.00 mL + 49.00 mL) = 74.00 mL[I-] = [Q1]/74.00 mL = 6.76 × 10-4 M See P135

[Ag+] = [Q2]/6.76 ×10-4 M = 1.23 × 10-13

pAg+ = -log[Ag+] = 12.91

Q. What is [Ag+] when VAg = 49.00 mL ?

At the equivalence point, what is [Ag+]?

• At equivalence point, VAg

eCAg = VICI

Little l- and Ag+ are left, and[Ag+] = [l-]. We define [Ag+] = x.

Then,Ksp = 8.3×10-17 = [Ag+][I-] = x2

x = 9.1×10-8 MpAg+ = -log(9.1 ×10-8 ) = 7.041

Q. What is [Ag+] when VAg = 51 mL

After the end point,Q. What is [Ag+] when

VAg = 51 mL

Now, all the I- ions are consumed, we have an excess of Ag+.

• Moles of Ag+ = (CAgVAg – CIVI) =

(51.00 mL)(0.05000 M) - (25.00 mL) (0.1000 M) =1.00 mL × 0.05000 M

= 0.05 mmol[Ag+] = 0.05 mmol/(total volume)

= 0.05 mmol/(51 mL + 25 mL)= 6.6 × 10 -4 M

pAg+ = 4 – log6.6 = 3.18

Titration curve

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

Excess I- Excess Ag+

Ion dependence(pAg+) e = -Log(Ksp

1/2)

12-1 Strong acid-strong base Titration

Excess OH- Excess H+

pH

Va (mL)

♦Excess OH- Calculate [OH-] [H+] =Kw/[OH-]

♦Equivalence Point Kw = x2 pH = 7.0

♦Excess H+

Calculate [H+] = (CHCLVHCl – CNaOHVNaOH)/(VHCl +VNaOH)

NaOH + HCl H2O + NaCl

P141 Calculation of titration from spread sheet

Known [M+] [X-] = Ksp/[M+] Then [M+] & [X-] VM

Home work

• Read Ch 13 1-3

• Ch7: 7-A, 7-B, 7-4, 7-7, 7-8, 7-9, 7-10, 7-14, 7-21, 7-23