Question

The Ka’s for H₂SeO₂ are K1 = 2.7 x 10^-3 K2 = 2.5 x 10^-7 Calculate the pH of a solution formed when 30.0 mL of 0.25 M KSeO3 is mixed with a) 1.5 mL of 0.20 M KOH b) 20.0 mL of 0.25 M HCl c) 25.0 mL of 0.20 M K₂SeO₃

I know how to calculate pH's however, I do not know when to use K1 vs K2, can you explain please?

Answer 1

I think the species in question would be H2SeO3 and not H2SeO2

and, the base form for it would be KHSeO3 and not KSeO3

So taking these we will do the calculations,

(a) 1.5 ml of 0.2 M KOH

moles of KHSeO3 = 0.25 M x 0.030 L = 7.5 x 10^-3 mols

moles of KOH = 0.2 M x 0.0015 L = 3 x 10^-4 mols

remaining KHSeO3 = 7.2 x 10^-3 mols

[K2SeO3] = 3 x 10^-4/0.0315 = 9.524 x 10^-3 M

[KHSeO3] = 7.2 x 10^-3/0.0315 = 0.229 M

this is buffer

pH = pKa2 + log([base]/[acid])

      = 6.60 + log(9.524 x 10^-3/0.229)

      = 5.22

(b) 20 ml of 0.25 M HCl added

moles of KHSeO3 = 0.25 M x 0.030 L = 7.5 x 10^-3 mols

moles of HCl = 0.25 M x 0.020 L = 5 x 10^-3 mols

remaining KHSeO3 = 2.5 x 10^-3 mols

[H2SeO3] = 5 x 10^-3/0.050 = 0.1 M

[KHSeO3] = 2.5 x 10^-3/0.050 = 0.05 M

pH = pKa1 + log([base]/[acid])

      = 2.57 + log(0.05/0.1)

      = 2.27

(c) 25 ml of 0.2 M K2SeO3 added

[KHSeO3] = 0.25/0.055 = 4.545 M

[K2SeO3] = 0.2/0.055 = 3.636 M

pH = pKa2 + log([base]/[acid])

      = 6.60 + log(3.6364/4.545)

      = 6.50