Question

To prepare 1.0 L of a pH 5.00 buffer find the volumes of each solution below. Find the solution and the volume of that solution.

a. 0.10 M HN3

b. 0.10 M NaN3

Answer 1

let volume of HN3 be v in L
then volume of NaN3 will be 1-v

number of mole = (molarity)*(volume in L)
number of mole HN3 = 0.1v
number of mole of NaN3 = 0.1(1-v)

since, final volume is 1 L
so, final molarity = number of mole

use,
pH = pKa + log([NaN3]/[HN3])
pKa for HN3 = 4.72
so,
5.00 = 4.72 + log{(0.1(1-v))/0.1v}
log{(1-v)/v} = 0.28
(1-v)/v = 1.91
1-v = 1.91v
2.91v = 1
v = 0.344 L

volume of HN3 = v
= 0.344 L
volume of NaN3 = 1.0 - v
= (1.0 - 0.344) L
= 0.656 L