Page 69 - A Working Method Approach For Introductory Physical Chemistry Calculations

Equilibrium ?I 53
K, = Kh Ka (remember the order: ‘WHAle’!) 1
~~
I In particular:
(b) Cation hydrolysis: CH+ + H20 S C + H30+. The equilibrium
constant is expressed by Kh = {[C][H30+])[CH+], since the activity, a,
of pure water is unity. An example of such a hydrolysis is: NH4+ +
H20 == NH3 + H30+, with Kh = {[NH3][H30’])/[NH4+]. Only the
conjugate acids of weak bases (such as NH3) undergo cation hydrolysis.
pFxq
In particular:

BUFFER SOLUTIONS
A buffer solution is a solution with an approximately constantpH. A
buffer contains an acid and its conjugate base in similar concentra-
tions. Such a solution changes pH only slightly when H30+ or OH- is
added. Therefore, buffers are used when the pH has to be maintained
within certain restricted limits.
In the reaction CH3C02H + H20 + CH3C0, + H30+, if both
the acid and anion are present in equal concentration, the equilibrium
can shift in either direction, to the right- (RHS) or to the left-hand
(LHS) side:

(a) If H30+ is added, the equilibrium shifts to the LHS to consume
.
H~O
(b) If OH- is added, the equilibrium shifts to the RHS as H30+ is
+
removed by OH- ion, in accordance with Le Chatelier’s
Principle.

For either case, the concentration of the buffer remains the same.
Therefore, the pH does not change to any great extent, remaining
practically constant.

How to Solve a Buffer Problem on Equilibrium
Buffer problems are solved using the standard Working Method
described in Chapter 4.
Example: Given that Ka(1) for the first ionisation of H3P04 = 7.5
x H3P04 + H20 + H30+ + H2P0,, determine the pH of
the buffer solution that contains equal volumes of 0.35 M H3P04,
and 0.25 M NaH2P04 respectively.