Question

Demonstrate through calculations and explanation that a HNO3/NaNO3 buffer would be a poor choice for a buffer at pH=3, particularly towards the addition of a base. Ka (HNO3)=24.2

Answer 1

Solution:

In order to function as a buffer, it must have two components:
a) a weak acid and its conjugate base or
b) a weak base and its conjugate acid
HNO₃ is a strong acid, therefore HNO₃ and NaNO₃ cannot function as a buffer. Hence it is a poor choice for a buffer

pH of a acidic buffer is calculated as,  pH = pKa + log [salt] / [acid] , since Ka = 24.2, hence pKa = -log (24.2) =0.0138

pH = 0.0138 + log [NaNO3]/ [HNO3]

and when we maintain the pH=3, then

3=0.0138 + log [NaNO3] / [HNO3] or log [NaNO3] / [HNO3] = 2.98,

It means the concentration of NaNO3 should be more than hundred times higher than HNO3, which is not possible.