ch.10 acid-base titrations - yonsei universitychem.yonsei.ac.kr/~mhmoon/pdf/analchem/ch10.pdf10.1...

10.1

Anal. Chem. by Prof. Myeong Hee Moon

Ch.10 ACID-BASE TITRATIONS

Interests : How much of an acidic or basic substance is present by TITRATION ?

10-1. Titration of Strong Acid with Strong Base

Consider ! titration of 50.00 mL of 0.02000 M KOHwith 0.1000 M HBr

net rxn: H+ + OH- H2O 1/Kw = 1.00x1014

To reach the equivalence point,

molesH+ needed = molesOH- present

10.2


0.1000 M x Ve = Ve = mL

Before 10.00 mL OH- excessAfter " H+ excess

At each addition of H+

we can calculate [H+] or [OH-]***** Consider Volume Change ******then obtain pH or pOH


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10.4


Region 1. Before the equivalence point, when 3.00 mL of HBr added

mmoles H+ added = mLmmoles OH- consumed =

mmoles OH- left = = mmoles

[OH-] = = 0.0132 MThus, [H+] = 1.0 x 10-14 / [OH-] = 7.58 x 10-13 M

pH = 12.12


10.5


Region 2. At the equivalence point, [H+] = [OH-] in water.

All HBr and KOH are consumed and turn to water & KBr[H+] = [OH-] = 1.0 x 10-7 M pH = 7.0 only for strong acid-base

Region 3. After the equivalence pointBeyond the Eq., added HBr becomes excess.

The excess H+ directly affect the pH of solution.If 10.50 mL of acid added, 0.50 mL is excess.

mL.mL.

mL.M.]H[

50100050

50010008.26x10-4 M pH = 3.08


10.6


10-2 Titration of weak acid and strong base

we need to consider weak acid hydrolysispH calculationBuffer Concepts

Consider ! 50.00 mL of 0.02000 M MESwith 0.1000 M NaOH

MES : 2-(N-morpholino)ethanesulfonic acid, pKa = 6.15

1. Start with writing equation

O N+HCH2CH2SO3- + OH -

O N HCH2CH2SO3- + H2O

1/Kb = 1/(Kw/Ka) = 7.1 x 107

10.7


For Equivalence point :mmoles base = mmoles weak acid0.100 M x Ve = 0.0200 x 50.00 mL

Ve = 10.00 mL


Reg. 1. Before Base is added,weak acid exist only, weak acid hydrolysis with KaHA H+ + A- Ka = 10-6.15

F-x x xx = 1.19x10-4 M pH=3.93

10.8


Reg. 2. Before Equivalence Pointonce OH- is added, HA & A- mixture Buffer Remind Henderson-Hasselbalch eq.

need to know the ratio [A-]/[HA]]HA[

]A[logpKpH a

1) when 3.00 mL of OH- is added,HA + OH- A- + H2O

initial: 10 mmol 3mmolfinal: 7 ~0 ~3[HA] =

[A-] =

[A-] / [HA] = 3 / 7 pH = pKa + log = 6.15 + log (3/7) = 5.78 ]HA[

]A[


10.9


2) half of titration is finished. 5.00 mL added.[HA] =

[A-] =pH = pKa + log 1 = pKa

when Ve /2 pH = pKa


10.10


Reg. 3. At the Equivalence PointAt this point, all NaOH are consumed.

resulting solution is full with A-

Consider a weak base hydrolysisA- + H2O HA + OH-

F-x x x

Need to know F

=0.0167 M

xF

xKb

2

00100050

005002000

..

mL..F

x = 1.54 x 10-5 M = [OH-] get pOH & pH

pH = 9.18


10.11


Reg. 4. After the Equivalence PointWhen we add extra OH-,

base NaOH is much stronger than base A-

pH is totally controlled by OH-

When extra 0.10 mL of NaOH (total 10.10mL) is added,

[OH-] =

=1.66x10-4 M

pH = 10.22

mL..

mL.M.

10100050

10010000


10.12


a) Titration of 0.020M 50.0mL HA with 0.100M NaOH


10.13


10-3 Titration of Weak Base with Strong Acid

Reverse of the titration of weak acid with strong base.Titration : B + H+ BH+

1. before acid is added, consider Kb2. during titration, [B], [BH+] mixture

pH = pKa + log

(Pka = pKw - pKb)]BH[

]B[

3. at equilibriumBH+ B + H+

F-x x x Ka = get x (=[H+]) ---- pH

4. after equilibriumexcess H+ affect pH

10.14


10-4. Titrations of Diprotic Systems

Extension of the previous experiencesConsider the titration of 10.0 mL of 0.100 M B (dibasic base)

with 0.100 M HClB + H+ BH+ pKb1=4.0 Ka2BH+ + H+ BH2

+2 pKb2=9.0 Ka1

At first equivalence point,moles H+ added = moles B0.100 M x Ve = 0.100 M x 10.00 mL

Ve = 10.00 mL

At second equivalence point, Ve2 = 2 Ve1

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10.16


i) point A : Initial cond.B + H2O BH+ + OH-

0.100-x x xKb1 = x = 3.11x10-3 M = [OH-]

[H+] = pH = 11.49

x.

x

1000

2

]OH[

Kw

ii) point BAt any point between point A ~ C (first eq. point)

First Buffer Region [B], [BH+]At point C [B] = [BH+] Ka2 = Kw / Kb1

pH = pKa2 + log = 10.00 + log 1 = 10.00

for any point Va = 1.5 mL

]BH[

]B[


10.17


mmoles H+ = mmoles base = = [BH+] produced[B] = " left =

pH = pKa2 + log = 10.00 + log = 11.75]BH[

]B[

iii) point C : first equivalence pointB BH+ both acidic & basic

if K1 is small21

1

121 KKFK

KKFKK]H[ w

221 aa pKpK

pH


10.18


iv) point DSecond Buffer Regionwhen Va = 15.0 mL [BH+] = [BH2

2+]

pH = pKa1 + log = 5.00 + log 1 = 5.00v) point E

second equivalence point[BH2

2+] at initial cond ?

]BH[

]BH[

22

BH22+ BH+ + H+

0.0333-x x x

Ka1 = x = MpH = 3.24

vi) Beyond EpH is dependent only on extra amount of strong acid


10.19


10-5. Finding the end point with a pH electrode

10.20


10-6. Finding the end point with indicators

Acid - base indicators: acid or base having different colors

Choosing an indicator of which color changesnear an equivalence point of any reaction.

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10.22



10.23


10.24


10-7. Kjeldahl Nitrogen Analysis

1883, one of the most accurate & widely used for Nin protein, milk, cereal, flour

1. Digestion in boiling H2SO4organic C, H, N, NH4

+ + CO2 + H2O2. Neutralization : NH4

+ + OH- NH3 (g) + H2O3. Distil NH3 into HCl : NH3 + H+ NH4

+

4. Titration of unreacted HCl with NaOH : H+ + OH- H2O

10.25


ex) A typical protein has 16.2 % wt. Nitrogen. 0.500mL of protein sol. digested, NH3 (g) distilled into 10.00 mL of 0.02140 M HCl. Titration of unreacted HCl requires 3.26mL of 0.0198 M NaOH. Calculate conc. of protein (mg protein/mL) in the original sample. (p133)

10.26


Homework

Ch10 : 7, 17, 23, 25, 30

ch.10 acid-base titrations - yonsei universitychem.yonsei.ac.kr/~mhmoon/pdf/analchem/ch10.pdf10.1...

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