Question

In: Chemistry

A buffer solution based upon sulfurous acid was created by treating 1L of 1M sulfurous acid...

A buffer solution based upon sulfurous acid was created by treating 1L of 1M sulfurous acid with NaOH until a pH of 1.347 was achieved (assuming no volume change). To this buffer 1.170 moles of NaOH were added (assume no volume change). What is the final pH of this solution? Assume 5% assumption is valid.

Expert Solution

moles of H2SO3 = 1 x 1 = 1

H2SO3 + NaOH --------------------> NaHSO3 + H2O

1 x 0

1-x 0 x

pH = pKa1 + log [NaHSO3 ]/[H2SO3 ]

1.347 = 1.85 + log ( x / 1-x)

x / 1- x = 0.314

x = 0.314 - 0.314 x

x = moles of NaHSO3 = moles of NaOH

moles of NaOH = 0.240

moles of H2SO3 = 1

on addition of 0.170 moles NaOH :

pH = 1.85 + log [0.240 + 0.170 / 1.0 -0.170]

pH = 1.54

final pH = 1.54

note : NaOH moles i think 0.170 moles not 1.170. if 1.170 moles NaOH is taken buffer will fail. verify once

queen_honey_blossom answered 2 years ago

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