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

pH of buffer is 4.81

Step-by-step 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
`([salt])/([acid])`

`pH = 4.74 +log(0.108)/(0.092) =4.81`