Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

pH logH 3 O log( 1. 310

3 )     2. 87

Case 2: 10.0 mL 0.10 M NaOH solution has been added – buffer system

Now strong base is added to a weak acid. The following reaction runs completely:

CH 3 COOH (aq) + OH- (aq)  CH 3 COO-(aq) + H 2 O (l)

The initial and end-conditions are determined for each of the important components:

n(CH 3 COOH) 0 = 0.10 M·50 mL = 5 mmol

n(CH 3 COO-) 0 = 0 mmol

[OH-] 0 = 0.10M·10mL = 1 mmol (The autoprotolysis of water is neglected)

The reaction runs completely which is why the end-conditions are:

n(CH 3 COOH) 0 = 0,10 M·50 mL = (5 – 1) mmol = 4 mmol

n(CH 3 COO-) 0 = 1 mmol

[OH-] 0  0 mmol (The autoprotolysis of water is neglected)

We now have a corresponding weak acid base pair in the same order. The pH value may be determined

using the buffer equation:

 



4. 14

1

4. 74 log^4

( )

log log ( )

3

3

3

3

(^) nCHCOO
pK nCHCOOH
CHCOO
pH pK CHCOOH
a a
Case 3: 25.0 mL 0.10 M NaOH solution has been added – buffer system
The procedure is completely analogous case number 2. The following reaction running completely to the
right is once again written as follows:
CH 3 COOH (aq) + OH- (aq)  CH 3 COO-(aq) + H 2 O (l)
The initial and end-conditions are written for the different components:
n(CH 3 COOH) 0 = 0.10 M·50 mL = 5 mmol
n(CH 3 COO-) 0 = 0 mmol
[OH-] 0 = 0.10M·25mL = 2.5 mmol (the autoprotolysis of water is neglected)
The reaction runs completely which is why the end conditions becomes:
n(CH 3 COOH) 0 = 0.10 M·50 mL = (5 – 2.5) mmol = 2.5 mmol
Acids and bases