Question

A buffer is prepared by mixing 204.0 mL of 0.452 mol/L hydrochloric acid (HCI) and 500.0 mL of 0.400 mol/L sodium acetate (NaCHCO2, NaAc). What is the pH of the above buffer solution?

Answer 1

Answer:

pH of buffer is 4.81

Explanation:

The reaction between HCl and sodium acetate will product acetic acid. The presence of weak acid (acetic acid) and its salt will make a buffer.

The reaction will be:

$CH_{3}COONa + HCl --> CH_{3}COOH + NaCl$

Thus one mole each of HCl and acetate will give one mole of acetic acid.

The moles of HCl present = molarity X volume = 0.452 X 0.204 =0.092

moles of acetic acid formed on reaction of HCl with acetate = 0.092

Initial moles of sodium acetate present = molarity X volume = 0.4X0.5 =0.2

moles of sodium acetate left =0.108 mol

the pKa of acetic acid =4.74

The pH of buffer is calculated using Henderson Hassalbalch's equation :

pH = pKa + log$\frac{[salt]}{[acid]}$

$pH = 4.74 +log\frac{0.108}{0.092} =4.81$