Question

What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid to obtain a buffer solution of pH 4.74 ? What will be the pH of 0.01 mol of HCl is dissolved in the above buffer solution ? Compare the last pH value with the pH of 0.01 molar HCl solution. Dissociation constant of propanoic acid at 25°C is 1.34 × 10^–5. 

Answer 1

Using Henderson's expression

= 0.74 or [Salt] = 1.48 × 10^–2 M

Hence, amount of sodium propanoate to be added = 1.48 × 10^–2 × 96 g = 1.42 g 

The addition of 0.01 mol of HCl converts the equivalent amount of sodium propanoate into propanoic acid. Hence, we will have

pH = 4.87 + log (0.160) = 4.87 – 0.79 = 4.08 

(The pH of 0.01 molar HCl solution would be pH = – log (0.01) = 2)