Question

What will the pH of a solution be when 0.6063 L of NaOH (pH = 14.411) is added to 0.295 L of 2.51 M sulfurous acid (H₂SO₃)?

Answer 1

moles of H2SO3 = 0.295 x 2.51 = 0.74045

pH = 14.411

pOH = - 0.411

[OH-] = 2.576 M

moles of NaOH = 2.576 x 0.6063 = 1.562

H2SO3   +    NaOH    ----------------> HSO3-   +   H2O

0.74045      1.562                               0               0

    0                0.8216                       0.74045         

HSO3-   +   NaOH     ---------------> SO32- +   H2O

0.74045      0.8216                              0              0

     0            0.08112                        0.740

here strong base remains.

[NaOH] = 0.08112 / (0.6063 + 0.295) = 0.090 M

pOH = -log (0.090) = 1.045

pH = 12.95