Question

A 31.8 mL sample of 0.3474 M selenious acid, H₂SeO₃, is titrated with 0.3474 M potassium hydroxide.

What is the pH after 15.9 mL, 31.8 mL, 47.7 mL and 63.6 mL of KOH have been added?

step	mL KOH added	pH
1	15.9
2	31.8
3	47.7
4	63.6

The first and second ionization constants for H₂SeO₃ are 2.4×10^-3 and 4.8×10^-9

Answer 1

step	mL KOH added	pH
1	15.9	2.62
2	31.8	5.47
3	47.7	8.32
4	63.6	10.69

Explanation

(a.) 15.9 mL KOH added

Given : initial concentration of H2SeO3 = 0.3474 M

initial volume of H2SeO3 = 31.8 mL

initial moles of H2SeO3 = (initial concentration of H2SeO3) * (initial volume of H2SeO3)

initial moles of H2SeO3 = (0.3474 M) * (31.8 mL)

initial moles of H2SeO3 = 11.05 mmol

Similarly, moles of KOH added = 5.525 mmol

Balanced reaction : H2SeO3 + KOH = KHSeO3 + H2O

moles of KHSeO3 formed = moles of KOH added

moles of KHSeO3 formed = 5.525 mmol

moles of H2SeO3 remaining = (initial moles of H2SeO3) - (moles of KHSeO3 formed)

moles of H2SeO3 remaining = (11.05 mmol) - (5.525 mmol)

moles of H2SeO3 remaining = 5.525 mmol

Ka1 = 2.4 x 10^-3

pKa1 = -log(Ka1)

pKa1 = -log(2.4 x 10^-3)

pKa1 = 2.62

According to Henderson - Hasselbalch equation,

pH = pKa + log([conjugate base] / [weak acid])

pH = pKa1 + log(moles of KHSeO3 formed / moles of H2SeO3 remaining)

pH = 2.62 + log(5.525 mmol / 5.525 mmol)

pH = 2.62 + log(1)

pH = 2.62 + 0

pH = 2.62