Question

What mass of sodium sulfite must be dissolved in 250mL of solution to result in a solution with a pH of 9.691? (H2so3aq k1=1.7x10-2 k2=6.2x10-6)

Answer 1

[H+] = 10^-pH = 10^-9.691 = 2.0370*10^-10

so

NaSO3 = Na+ + SO3-2

SO3-2 + H2O = HSO3- + OH-

assume this is mainly due to the first ionization of SO3-2 wit water

Kb = [HSO3-][OH-]/[SO3-2]

Kb = Kw/Ka = (10^-14)/(6.2*10^-6) = 1.612*10^-9

and

pOH = 14-pH = 14-9.691 = 4.309

[OH-] = 10^-pOH = 10^-4.309 = 0.00004909078

[HSO3-] = [OH-] = 0.00004909078

s

[SO3-2] = M - 0.00004909078

solve for M

Kb = [HSO3-][OH-]/[SO3-2]

1.612*10^-9 = (0.00004909078)(0.00004909078)/M-0.00004909078)

M = (0.00004909078^2)/(1.612*10^-9) + 0.00004909078

M = 1.49502 M

so

mol = MV =1.49502*0.25 = 0.373755 mol of Na2SO3 required

mass = mol*MW = 0.373755*126.043 = 47.109 g of Na2SO3 required for 250 mL for pH = 9.691 approx