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

The final pH of this solution is 7.69

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