Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

n(CH 3 COO-) 0 = 2.5 mmol

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

We still have a corresponding weak acid base pair in the same amount and the pH value may be calculated

in terms of the buffer equation:

 



4. 74 log 1 4. 74

( )

( )

log log

3

3

3

3

(^) nCHCOO
nCHCOOH
pK
CHCOO
CHCOOH
pH pKa a
The point at the titration curve after the addition of 25.0 mL NaOH solution is special as it is half way to
the equivalence point. The original solution contained 50.0 mL of a 0.10 M CH 3 COOH solution
corresponding to 5.0 · 10-5 mol CH 3 COOH. It thereby takes 5.0 · 10-5 moles of OH- ions before the
equivalence point is reached is this requires in our case 50.0 mL of a 0.10 M NaOH solution. After having
added 25.0 mL NaOH we are half way towards the equivalence point. Thereby the pH value half way
towards the equivalence point always equal to the pKa value for the acid of the buffer system.
Case 4: 50.0 mL 0.10 M NaOH solution has been added – solution of a weak base
After the addition of a 50.0 mL NaOH solution we are now at the point of equivalence which means that
there are equal amounts and acetic acid and OH- ions. The primary contribution to pH in the solution is
thereby the weak base CH 3 COO- because the following reaction runs completely:
CH 3 COOH (aq) + OH- (aq)  CH 3 COO-(aq) + H 2 O (l)
Al acetic acid is now brought to acetate form (CH 3 COO-). The problem is now to determine pH in a
solution of a weak base with a concentration of 0.05 M (half of the initial concentration). The base
equilibrium and the corresponding base equilibrium constant Kb (5.6 · 10-10 M) are to be written:
CH 3 COO- (aq) + H 2 O CH 3 COOH(aq) + OH- (aq)
  
^
CHCOO
K M OH CHCOOH
b
3
5. 6 1010 3
Once again we look at the initial and end conditions:
[CH 3 COO-] 0 = 5.0 · 10-3 mol / (0.0500 L + 0.0500 L) = 5.0 · 10-2 M
[OH-] 0 * 0 M (the autoprotolysis of water is negelcted)
[CH 3 COOH] 0 = 0 M
and the end conditions are thereby:
[CH 3 COOH] = (5.0 · 10-2 – x) M
Acids and bases