Dissociation and pH

Question

A buffer solution was obtained by 8.3g of acetic acid and 16.0g of sodium acetate. The buffer solution was diluted to 1litre. Calculate the pH of the solution if 0.7mol of 1ml HCl is added to it. The dissociation constant is 1.8 x10-5, molecular weight of acetic acid and sodium acetate are 60 and 82 respectively. 

Answer 1

We can use the Henderson Hasselbalch equation: pH = pKa + log [salt] / [acid]

pKa = -log Ka = 4.74

[salt] = 16.0 g NaAc x 1 mol / 82 g = 0.195 mol / 1 L = 0.195 M

[acid] = 8.3 g HAc x 1 mol / 60 g = 0.138 mol / 1 L = 0.138 M

Adding 0.07 mol HCl (in a volume of 1 ml) will not appreciably change the volume (we'll ignore it), but the HCl will react with the NaAc to reduce its concentration and to increase the concentration of HAc. It is assumed that you made a typo and we are adding 0.07 moles HCl and not 0.7 moles.

HCl + NaAc ==> NaCl + HAc

Final concentrations:

[NaAc] = 0.195 - 0.07 = 0.125 M

[HAc] = 0.138 + 0.07 = 0.208 M

pH = pKa + log [salt] / [acid]

pH = 4.74 + log (0.125/0.208) = 4.74 -0.22

pH = 4.52