منتدى علوم المنصورة
المحاضره الخامسه فى الكيمياء التحليليه Ezlb9t10


منتدى علوم المنصورة
المحاضره الخامسه فى الكيمياء التحليليه Ezlb9t10

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descriptionالمحاضره الخامسه فى الكيمياء التحليليه Emptyالمحاضره الخامسه فى الكيمياء التحليليه

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Lecture Five covers the following points:-


 Types of neutralization reaction
 Types of indicator used
 Which one will be used in Quantitative titration



Types of neutralization reaction


 Strong Acid + Strong Base = Salt + Water
 HCl + NaOH = NaCl + H2O



 Strong Acid + Weak Base = Salt + Water
 HCl + NH4OH = NH4Cl + H2O


 Weak Acid + Strong Base = Salt + Water
 CH3COOH + NaOH = CH3COONa + H2O



 Weak Acid + Weak Base = Salt + Water
 CH3COOH + NH4OH = CH3COONH4 + H2O


Neutralization Curves


 Preequivalence
 Equivalence
 Postequivalence

Titrating a strong acid with a strong base:



We will be calculate hypothetical titration curves of pH versus volume of titrant
 50 ml of 0.05 M HCl with 0.1 M NaOH

 Initial point:
 pH= - log [H3O+]
 = - log 0.05 = 1.3

 Preequivalance:
 After adding 10 ml of NaOH
 CHCl = no. mmol HCl –no. mmol NaOH/ total V
 = (M x V)HCl _ ( M x V )NaOH / Vtotal
 = (0.05 x 50 ) _ ( 0.1 x 10 ) / 60
 = 2.5 x 10-2
pH = - log (2.5 x10-2) = 1.6
 After addition of 24.9 ml NaOH
 CHCl = (M x V )HCl - ( M x V )NaOH / Vtotal
 = ( 0.05 x 50 ) – ( 0.1 x 24.9) / 74.9
 = 2.5 - 2.49 / 74.9 = 0.01/ 74.9
 = 1.33 x10-4
 pH = - log ( 1.33 x 10-4 ) = 3.87

 Equivalence point:
 [ H3O+] = ( Kw )1/2 = ( 1 x 10-14)1/2 = 1 x 10-7
 pH = - log ( 1 x 10-7 ) = 7
 Postequivalence :
 After addition of 25.1 ml NaOH
 CNaOH = (M x V )NaOH - ( M x V )HCl / Vtotal
 = ( 0.1 x 25.1 ) – ( 0.05 x 50) / 75.1
 = 0.01/ 75.1 = 1.33 x 10-4
 pOH = - log ( 1.33 x 10-4 ) = 3.87
 pH = 14 – 3.87 = 10.13
 After addition of 40 ml NaOH
 CNaOH = (M x V )NaOH - ( M x V )HCl / Vtotal
 = ( 0.1 x 40 ) – ( 0.05 x 50) / 90
 = 4 – 2.5 / 90 = 1.5 / 90 = 0.0166
 pOH = - log ( 0.0166 ) = 1.77
 pH = 14 – 1.77 = 12.23


Acid –Base Indicator


Many substances both naturaly occurring and synthetic, display colors that depend upon the pH of the solutions in which they are dissolved.
An acid – base indicator is a weak organic acid or weak organic base whose undissociated form differs in color from its conjugate base or its conjugate acid form. For example, the behavior of an acid type indicator, HIn , is described by the equilibrium:

HIn + H2O = In- + H3O+

Here , internal structural changes accompany dissociation and cause the color change. The equilibrium for a base type indicator In, is:

In + H2O = InH+ + OH-
In the following paragraphs we focus on the behavior of acid type indicators. The discussion, however, can be readily extended to base type indicator as well.
THE EQUILIBRIUM CONSTANT EXPRESSION FOR DISSOCIATION OF AN ACID INDICATOR TAKES THE FORM:

Ka = [H3O+] [ In- ] / [ HIn]

Rearranging leads to

[H3O+] = Ka [HIn ] / [ In- ]
- log [H3O+] = - log Ka - log [HIn ] / [ In- ]
pH = pKa - log [HIn ] / [ In- ]
The human eye is not very sensitive to color differences in a solution containing a mixture of In- and HIn , particularly. When the ration [In-] / [HIn] is greater than about 10 or smaller than 0.1. Consequently, the color imparted to a solution by a typical indicator appears to the average observer to change rapidly only within the limited concentration ratio of approximately 10 to 0.1.
To obtain the indicator pH range:
pH ( acid color) = pKa - 1
pH ( basic color) = pKa + 1

descriptionالمحاضره الخامسه فى الكيمياء التحليليه Emptyرد: المحاضره الخامسه فى الكيمياء التحليليه

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 After addition of 40 ml NaOH
 CNaOH = (M x V )NaOH - ( M x V )HCl / Vtotal
 = ( 0.1 x 40 ) – ( 0.05 x 50) / 90
 = 4 – 2.5 / 90 = 1.5 / 90 = 0.0166
 pOH = - log ( 0.0166 ) = 1.77
 pH = 14 – 1.77 = 12.23
ممكن تفهمينى اية الفرق بين  Equivalence pointو Preequivalance؟
وعرفت ان point فى اى حالة ازاى؟
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