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 1.135 was achieved (assuming no volume change). To this buffer 1.180 moles of NaOH were added (assume no volume change). What is the final pH of this solution

Answer 1

The final pH of this solution is 7.69

Step-by-step explanation:

having the chemical reaction:

H2SO3 = HSO3- + H

Ka = 1.5x10^-5

pKa = 1.82

H2SO3 + OH- = H2O + HSO3-

1 ? 0

1-x 0 +x

the pH is equal to:

pH = pKa + log([HSO3-]/[H2SO3])

1.135 = 1.82 + log(x/(1-x))

Clearing x:

x = 0.17

Having the reaction:

H2SO3 + OH- = H2O + HSO3-

0.83 1.180 0.17

0 0.83 1

HSO3- = SO3-2 + H+

Ka2 = 1x10^-7

pKa2 = 7

HSO3- + OH- = SO3-2 + H2O

1 0.83 0

0.17 0 0.83

pH = pKa2 + log([SO32-]/[HSO3-]) = 7 + log(0.83/0.17) = 7.69