Question

Calculate the ph of 1.00 l of a buffer that is 1.00 m in acetic acid and 1.00 m in sodium acetate after the addition of 0.550 mole of naoh.

Answer 1

initial moles of sodium acetate = acetic acid = 1 m * 1 L = 1 moles 

[ acetate ] after adding NaOH = 1 mol+ 0.55 mol = 1.55 M

[acetic acid] after adding NaOH = 1 mol - 0.55 mol = 0.45 M

According to Handerson-Hasselblach equation:

PH = Pka + ㏒ [ acetate]/[acetic acid]

when Pka = 4.74

so, by substitution:

∴PH = 4.74 + ㏒[1.55/0.45]

∴ PH = 5.28