Acid Dissociation Constant Ka

From: ::Rose::
Date: 11 Jun 2008, 19:14

Q2. A solution of Phenol in (C6H5OH) in water has a concentration of 38g dm-3. The acid dissociation constant, Ka, of Phenol is 1.3 x 10 (to the -10) mol dm-3.
Write an expression for the acid dissociation constant of Ka, of Phenol.

Dissociation of Phenol:

HA + H2O = A- + H3O+

Ka = [H3O+][A-]/[HA][H2O]

Key points:
1) For every A-, there is one H3O+. Therefore [A-] is always equal [H3O+]

2) [H2O] is constant i.e. [H2O] = 1, because water is hardly altered by the ionisation of HA

Therefore the expression for the Ka of phenol is;
Ka = [H3O]2/[HA]

whereby Ka = 1.3 x 10-10 M (given by question itself)
[HA] = 0.4 M (see below)
[H3O]= 7.2 x 10-6 M (see below)
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Extra calculations for own understanding:

Converting g/dm3 into mol/dm3:

[HA] = 36 g/dm3
Divide this by the molecular mass of phenol (94 g/ml)

[HA] = 36g/dm3
---------
94g/mol
[HA] = 0.4 mol/dm3 (notice that the grams cancel out)

Finding the concentration of H30+:

Use the expression we already found above, just feed in the numbers

Ka = [H3O+]2/[Ha]
1.3 x 10-10 M = [H3O+]2/[0.4 M]
[H30+] = {(1.3 x 10-10)x(0.4)}to the power of 1/2
[H30+] = 7.2 x 10-6 M

If you want a more fancy eqn, use PhOH in place of HA, so

PhOH + H2O = PhO- + H30+

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