Question

a buffer solution based upon sulfurous acid (H₂SO₃) was created by treating 1.00 L of a 1.00 M sulfurous acid solution with NaOH until a pH of 1.323 was achieved (assuming no volume change). to this buffer 1.410 moles of NaOH were added (assume no volume change). what is the final pH of this solution? For this problem we can assume the 5% assumption is valid. (K_a1=1.5E-2 K_a2=1.0E-7) please explain how you solved this if possible.

Answer 1

Given pH = 1.323 = -log[H+]

Using Hendersen-Hasselbalck equation,

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

1.323 = pKa + log[base]/[acid]

pKa1 is closest to pH, so

1.323 - 1.824 = log[base]/[acid]

[acid] = 0.316[base]

[acid] = H2SO3

[base] = HSO3^-

the total concentration remains the same,

[acid] + [base] = 1.0 M

Substituting from above,

0.361[base] + [base] = 1.0

[base] = 0.735 M

[acid] = 0.361 x 0.735 = 0.265 M

Since volume = 1.0 L

moles of acid = 0.265 moles

moles of base = 0.735 moles

Added 1.410 moles of NaOH

new moles of base = 0.735 + 1.410 = 2.145 moles

molarity of [base] = 2.145 M

acid gets neutralized and remaining base = 1.410 - 0.265 = 1.145 moles of base = 1.145 M

total base = 2.145 + 1.145 = 3.29 M

Kb = Kw/Ka1 = 1 x 10^-14/1.5 x 10^-2 = 6.67 x 10^-13

let x amount of base is hydrolyzed then,

Kb = [baseH][OH-]/[base-]

6.67 x 10^-13 = x^2/3.29

x = [OH-] = 1.48 x 10^-6 M

[H+] = Kw/[OH-] = 1 x 10^-14/1.48 x 10^-6 = 6.751 x 10^-9

pH = -log[H+] = -log(6.751 x 10^-9) = 8.17

So the pH of solution is 8.17