Question

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH of this buffer?

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

What is the pH of the buffer after the addition of 0.02 mol of HI?

Answer 1

pKa of propionic acid = 4.87

initial conc of
propionic acid = 0.18/0.25 = 0.72 M
sodium propionate = 0.26/0.25 = 1.04 M

when only propionic acid and sodium propionate are present

pH = pKa + log([salt]/[acid]) = 5.03

when 0.02 mol NaOH is added
conc of
propionic acid becomes 0.64 M
sodium propionate becomes 1.12 M

new pH = 5.11

when 0.02 mol HI is added
conc of
propionic acid becomes 0.8 M
sodium propionate becomes 0.96 M

new pH = 4.95

Answer 2

Ka = 1.3 x 10^-5
pKa = 4.9

[acid]= 0.18 / 1.2 =0.15 M
[salt]= 0.26 / 1.2 = 0.217 M
pH = 4.9 + log 0.217/0.15=4.64

moles acid = 0.18- 0.02 = 0.16
[acid]= 0.18/ 1.2=0.15 M
moles salt = 0.26 + 0.02 = 0.28
[salt]= 0.28/ 1.2=0.234
pH = 4.9 + log 0.234/ 0.15 = 5.09

moles acid = 0.18 + 0.02 = 0.20
[acid]= 0.20/ 1.2 = 0.167 M
moles [salt]= 0.26- 0.02= 0.24
[salt]= 0.24/ 1.2 = 0.2
pH = 4.9 + log 0.2/ 0.167= 4.978