Chapter 4 classical methods in techniques of analytical chemistry titrimetric methods of analysis
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Chapter 4 : Classical Methods in Techniques of Analytical Chemistry : Titrimetric Methods of Analysis. Pn Syazni Zainul Kamal PPK Bioproses. CO4: ABILITY TO DIFFERENTIATE AND CALCULATE CONCENTRATION OF ANALYTES OF VARIOUS TITRIMETRIC METHODS (ACID-BASE, COMPLEXATION, REDOX, PRECIPITATION).

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Chapter 4 classical methods in techniques of analytical chemistry titrimetric methods of analysis

Chapter 4 : Classical Methods in Techniques of Analytical Chemistry : Titrimetric Methods of Analysis

Pn Syazni Zainul Kamal

PPK Bioproses


Pn syazni zainul kamal ppk bioproses

  • CO4: ABILITY TO DIFFERENTIATE AND CALCULATE CONCENTRATION OF ANALYTES OF VARIOUS TITRIMETRIC METHODS (ACID-BASE, COMPLEXATION, REDOX, PRECIPITATION)


What is titrimetry

What is Titrimetry??

  • Any method in which volume is the signal.

    Also known as volumetric method.


Types of titrimetric methods

Types of Titrimetric Methods

  • Classified into four groups based on type of reaction involve;

  • Acid-base titrations

  • Complexometric titrations

  • Redox titrations

  • Precipitation titrations


Acid base titration

Acid base titration

  • Neutralization - acid is reacted with equivalent amount of base (reaction in acid base titration)

  • Titration curve – the independent variable (X) is the volume of the titrant, while the dependent variable (Y) is the pH of the solution (which changes depending on the composition of the two solutions).

  • Titrant: A reagent of a known concentration

  • Equivalence point: the point at which all of the starting solution, eg an acid, has been neutralized by the titrant, eg a base.


Pn syazni zainul kamal ppk bioproses

titrant


Pn syazni zainul kamal ppk bioproses

  • End point/ equivalence point – signals the completion of the reaction (the point at which all of the starting solution, usually an acid, has been neutralized by the titrant, usually a base).


Strong acid vs strong base easy titration

Strong acid vs strong base (easy titration)


Pn syazni zainul kamal ppk bioproses

  • In the case of strong acid and strong base titration, both the titrant & analyte are completely ionized

  • Example 1 : titration of 0.100 M HCl with 0.100M NaOH :

    H+ + Cl- + Na+ + OH- H2O + NaCl


Pn syazni zainul kamal ppk bioproses

  • Before titration started – only have HCl.

    [H+]=[HX]

  • Titration proceed – part of H+ is removed from solution as H2O. [H+] decrease gradually. pH increase

    [H+]=[remaining HX]

  • At equivalence point - we have solution of NaCl. Neutralization complete. HCl had been neutralize by NaOH.

    [H+]= √Kw

  • Excess NaOH added – [OH-] increase. pH is determine by concentration of OH-

    [OH-]=[excess titrant]


Pn syazni zainul kamal ppk bioproses

Equation governing a strong acid (HX) titration

Fraction F titrated Present Equation

F=0 HX [H+]=[HX]

0<F<1 HX/X- [H+]=[remainingHX]

F=1 X- [H+]= √Kw

F>1 OH-/X- [OH-]=[excess titrant]


Pn syazni zainul kamal ppk bioproses

  • A 50.0ml aliquot of 0.100 M HCl is titrated with 0.200 M NaOH. Calculate the pH of the solution after addition of 0.0, 5.0, 10.0, 25.0, 35.0, 45.0, 55.0 of NaOH

Molarity = total moles of solute (mol)/vol. of solution (ml)


Pn syazni zainul kamal ppk bioproses

Solution :

a) Addition 0f 0.0ml NaOH

[H+]=[HX]

pH = - log [H+]

= - log 0.100

= 1.00


Pn syazni zainul kamal ppk bioproses

b) Addition of 5.0ml NaOH

[H+]=[remaining HX]

Initial mmol H+ = 0.100 M x 50.0 ml = 5.00 mmol

mmol OH- added = 0.200 M x 5.0 ml = 1.00 mmol

mmol H+ left = 4.00 mmol in 55.0ml

[H+] = 4.00 mmol = 0.0727 M

(5.0 + 50.0)ml

pH = - log [H+] = - log 0.0727 = 1.14


Pn syazni zainul kamal ppk bioproses

c) Addition of 25.0 ml NaOH

Initial mmol H+ = 0.100 M x 50.0 ml = 5.00 mmol

mmol OH- added = 0.200 M x 25.0 ml = 5.00 mmol

The equivalence point is reached. All H+ had reacted with OH-. The H+ now is from H2O.NaCl does not contribute to any pH value.

[H+]= √Kw

= √1.00x10-14 = 1.00x10-7

pH = - log 1.00x10-7 = 7.00


Pn syazni zainul kamal ppk bioproses

c) Addition of 35 ml NaOH

Initial mmol H+ = 0.100 M x 50.0 ml = 5.00 mmol

mmol OH- added = 0.200 M x 35.0 ml = 7.00 mmol

OH- excess 2.0 mmol

[OH-]=[excess titrant]

[OH+] = 2.0 mmol = 0.0235 M

(35.0 + 50.0)ml

pOH = - log [OH+] = - log 0.0235 = 1.63

pH= 14.00 – 1.63 = 12.37


Pn syazni zainul kamal ppk bioproses

STRONG BASE

VERSUS

STRONG ACID


Pn syazni zainul kamal ppk bioproses

Equation governing a strong base (BOH) titration

Fraction F titrated Present Equation

F=0 BOH [OH-]=[BOH]

0<F<1 BOH/B- [OH-]=[remainingBOH]

F=1 B+ [H+]= √Kw

F>1 H+/B+ [H+]=[excess titrant]


Assignment 1

Assignment 1

Question 1

  • A 50.0ml aliquot of 0.100 M NaOH is titrated with 0.200 M HCl. Calculate the pH of the solution after addition of 0.0, 5.0, 10.0, 25.0, 35.0, 45.0, 55.0 of HCl. Plot the titration curve in a graph paper.


Pn syazni zainul kamal ppk bioproses

WEAK ACID

VERSUS

STRONG BASE


Weak acid vs strong base

Weak acid Vs Strong base

  • Titration of 100 ml 0.1 M acetic acid with 0.1 M sodium hydroxide

  • Neutralization reaction :

    HOAc + Na+ + OH- H2O + NaOAc

  • Acetic acid is neutralized to water and equivalent amount of sodium acetate.


Pn syazni zainul kamal ppk bioproses

  • Before titration started – only have HOAc.

    [H+]= √Ka . CHA

  • Titration started - some of HOAc convert to NaOAc & buffer system set up.

  • Titration proceed – pH increase slowly as the ratio [OAc-]/[HOAc] change.

    pH=pKa + log CA-

    CHA

  • Midpoint of titration – [OAc-] = [HOAc] (pH=pKa)


Pn syazni zainul kamal ppk bioproses

  • At equivalence point – we have solution of NaOAc. NaOAc is a bronsted base (it hydrolyze in water to form OH- and undissociate HOAc)

    [OH-]=√Kw . CA-

    Ka

  • Excess NaOH added – pH is determine by concentration of OH-


Pn syazni zainul kamal ppk bioproses

Titration of HOAc with NaOH


Pn syazni zainul kamal ppk bioproses

Equation governing a weak-acid (HA)

Fraction F

titrated Present Equation

F=0 HA [H+]= √Ka . CHA

0<F<1 HA/A- pH=pKa + log CA-

CHA

F=1 A- [OH-]=√Kw . CA-

Ka

F>1 OH-/A- [OH-]= [excess titrant]


Pn syazni zainul kamal ppk bioproses

Exercise 1

A 50.0 ml aliquot of 0.100 M acetic acid is titrated with 0.100 M NaOH. Calculate the pH of the solution after the addition of 0.0, 5.0, 25.0, 45.0, 50.0, 55.0 and 75.0 ml of NaOH. The dissociation constant for acetic acid is 1.75 x 10-5


Pn syazni zainul kamal ppk bioproses

Solution

  • Addition of 0.0 ml NaOH

    At 0 ml, we have a solution of only 0.100 M HOAc:

    HOAc H+ + OAc-

    Initial (M)0.100 0 0

    Equilibrium (M) 0.100 – x x x

    [H+][OAc-] = (x)(x)= 1.75 x 10-5

    [HOAc] 0.100

    x = [H+] = 1.32 x10-3 M

    pH = - log 1.32 x10-3 M = 2.88


Pn syazni zainul kamal ppk bioproses

b) Addition of 5.0 ml NaOH

Initial mmol HOAc = 50.0ml x0.100 M =5.00mmol

mmol NaOH added = 5.0ml x 0.100 M =0.50mmol

=mmol OAc- form

mmol HOAc left =4.50mmol

pH=pKa + log CA-

CHA

= - log 1.75 x10-5 + log 0.50 = 3.81

4.50


Pn syazni zainul kamal ppk bioproses

c) Addition of 50.0ml NaOH

Initial mmol HOAc = 50.0ml x0.100 M =5.00mmol

mmol NaOH added = 50.0ml x 0.100 M =5.00mmol

Equivalence point is reached. The solution contain salt of weak acid (conjugate base), NaOAc, hence the hydrolysis of salt will take place. All HOAc has been converted to OAc- (5.00 mmol in 100ml or 0.0500 M)

NaOAc Na+ + OAc-

OAc- + H2O HOAc + OH-

[OH-]=√Kw . CA-

Ka

=√1.00 x10-14 x 0.0500 = 5.35x10-6

1.75x10-5

pOH = -log 5.35x10-6 = 5.27

pH = 14.00 – 5.27 = 8.72


Pn syazni zainul kamal ppk bioproses

d) Addition of 55.0 ml NaOH

Initial mmol HOAc = 50.0ml x0.100 M =5.00mmol

mmol NaOH added = 55.0ml x 0.100 M =5.50mmol

Excess OH- =0.50 mmol in 105ml

At 55.0 ml, we have solution of NaOAc and excess added NaOH. So pH is calculated by concentration of excess OH-.

[OH-] = 0.50 mmol = 4.76x10-3

(50+55)ml

pOH = - log [OH-] = - log 4.76x10-3 = 2.32

pH = 14.00 – 2.32 = 11.68


Pn syazni zainul kamal ppk bioproses

WEAK BASE

VERSUS

STRONG ACID


Weak base vs strong acid

Weak base vs strong acid

  • Titration curve reverse of those for a weak acid versus strong base

  • Titration of 100ml 0.1M ammonia with 0.1M HCl

    NH3 + H+ + Cl- NH4+ + Cl-


Pn syazni zainul kamal ppk bioproses

  • Before titration started – only have 0.1 M NH3.

    [OH+]= √Kb . Cb

  • Titration started - some of NH3 convert to NH4+ & buffer system set up.

  • Titration proceed – pH decrease slowly as the ratio [NH3]/[NH4+] change.

    pH=(pKw – pKb) + log CB

    CBH+

  • Midpoint of titration – [NH4+] = [NH3]

    (pH=14 - pKb)


Pn syazni zainul kamal ppk bioproses

  • At equivalence point – we have solution of NH4Cl. Neutralization complete. NH3 had been neutralize by HCl.

    [H+]=√Kw . CBH+

    Kb

  • Excess NaOH added – pH is determine by concentration of H+


Assignment 11

Assignment 1

Question 2

  • Sketch a titration curve based on the pH values calculated from the addition of 0.0, 10.0, 25.0, 40.0, 50.0, 55.0 and 60.0 ml of 0.100M HCl into 50.0ml 0.100M NH3. Kb is 1.75x10-5.

  • Submit on : 16 march 2011 (Wednesday class)


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