Question

Suppose you are given 0.50 liter of 0.500 m acetic acid, and 0.50 liter of 0.250 liter sodium acetate. what is the maximum volume of buffer solution that you can make if the buffer must have a ph of 4.58? ka for acetic acid is 1.8 x 10-5.

Answer 1

First, we are going to use H-H equation :

when PH = Pka + ㏒ [acetate / acetic acid]

when we have Ka value = 1.8 x 10^-5 so, we can get the Pka value:

Pka = - ㏒Ka

= - ㏒ 1.8 x 10^-5

= 4.74

by substitution:

∴ 4.58 = 4.74 + ㏒[acetate / acetic acid]

∴acetate/acetic acid = 0.692

so we need moles of acetate = volume *molarity

= 0.5 L * 0.25 M

= 0.125 moles

when the acetate/ acetic acid ratio = 0.692 and moles acetate = 0.125 moles

∴ moles acetic acid = 0.125 moles / 0.692 moles = 0.182 moles

∴ volume use for 0.182 moles = 0.182 / 0.5 = 0.364 L

∴volume of buffer solution = 0.5 L + 0.364L

= 0.864 L