Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 5- I:

Titration of weak acid with strong base

We have seen earlier how calculations of pH in solutions with strong acid and strong base are relatively

simple because strong acids and strong bases are completely dissociated. On the contrary, pH calculations

in cases where the titrated acid is weak is not as simple. In order to be able to calculate the concentration

of H 3 O+ ions after the addition of a given amount of strong base it is necessary to look at the weak acids

dissociation equilibrium. Calculations of pH curves for titration of a weak acid with a strong base involve

a series of buffer-related problems.

We look at the following problem: 50.0 mL 0.10 M CH 3 COOH is titrated with 0.100 M NaOH. We wish

to determine the pH value in the solution different places in the titration and on the basis of this to be able

to draw a titration curve. Acetic acid has a Ka value of 1.8 · 10-5 M and thereby a pKa value of 4.74.

Case 1: Not NaOH is added (solution of weak acid)

This problem is one we have met earlier. We are to determine pH in a solution of a weak acid. We are to

write the equation of equilibrium with corresponding expression of equilibrium.

CH 3 COOH (aq)  CH 3 COO-(aq) + H+ (aq)

  

CHCOOH

HO CHCOO

Ka M

3

1. 8105 3 3

^

We now look at the initial and equilibrium concentrations completely analogously earlier examples. The

initial concentrations are:

[CH 3 COOH] 0 = 0.10 M

[CH 3 COO-] 0 = 0 M

[H 3 O+] 0 * 0 M (the autoprotolysis of water is neglected)

And the final concentrations are thereby:

[CH 3 COOH] = (0,10 – x) M

[CH 3 COO-] = x M

[H 3 O+] = x M

which is why we get the following equation:

  



x M

x

x x

CHCOOH

HO CHCOO

Ka M^3

3

(^5331). 310
0. 10

1. 810
   ^
   (
   Thereby [H 3 O+] = 1.3 · 10-3 M which is why the pH value in the solution becomes:
   Acids and bases