Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

Where K of equilibrium is the Ka2: 6.5x10⁻⁸

HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.