# DETERMINATION Acids have some characteristics which are as follows-

DETERMINATION OF pH OF A POLYPROTIC ACID

To determine the pH of a monoprotic acid
we just take the concentration of H+ ions (at equilibrium in case of
a weak acid) formed in the solution and put the H+ value in the
below formula and easily get the pH value.

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pH=-log10H+

but in case of di, tri or more generally
polyprotic acid the pH determination method is something lengthy. For this we
need the step wise dissociation constant (Ka) values. Polyprotic
Acids have some characteristics which are as follows-

a. can give more than one proton.

b. dissociation of 2nd and
consecutive protons are more difficult as after dissociation the acid turns to
a negatively charged ion and it is increasingly difficult to lose proton from a
negatively charged species compared to a neutral molecule.

c. having n (n>1) no. of ionizable
proton, the 1st dissociation step should be faster than all other
steps i.e. Ka1 should be greater than that of Ka2, Ka3
and so on.

We can discuss an example-

Q. Calculate the pH of 0.05 (M) H2SO4 solution.
(Given Ka1= Very Large, Ka2=1.2 X 10-2)

Approach: As here it is given that Ka1
is very large so we can easily assume that H2SO4 is fully ionized to H+ and
HSO4- ion. So, we can write it as-

H2SO4   ?      H+      +      HSO4-

0.05 (M)      0.05 (M)    0.05 (M)

That means the concentration of H+
ion, H+ after first dissociation is 0.05 (M).

Now coming to 2nd dissociation
step. But it is involving comparatively low Ka value than the 1st
one, (Ka2 = 1.2 X 10-2). So, it can be treated as
equilibrium as follows-

HSO4-   ?      H+      +      SO42-

Suppose x (M) HSO4-
has dissociated. So, at equilibrium the picture will be like below-

HSO4-   ?      H+      +      SO42-

At equilibrium
0.05 – x      0.05 + x           x              (As already 0.05 M H+

ion has been produced

in the 1st step)

Thus,

So, the total H+ ion
concentration, H+ at equilibrium is (0.05+0.0085) (M) = 0.0585 (M)

Therefore, the pH value will be-

pH = -log10 H+ = -log10 (0.0585) =
1.232

Below, there is a similar problem which is
open to all of you. Try to solve it own.

Q. Calculate the pH of 0.02 (M) H3PO4 solution.
(Given Ka1 = 7.5 X 10-3, Ka2 = 6.2 X 10-8
and Ka3 = 4.8 X 10-13).