The total volume = 250 mL
Mass of Na2CO3 = 2.4134 grams
So moles added = Mass / Molecular weight = 2.4134 / 106 =
0.0228
pK1 = 6.351
K1 = 4.46 X 10^-7
pKa2 = 10.329
Ka2 = 4.68 X 10^-11
        Na2CO3 + H2O
--> NaHCO3 + OH-
initial  Â
0.0228Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â
0
final    Â
0.0288-x                     Â
x            Â
x
Kb1 = Kw/Ka2 = (1.00 x 10^-14)/(4.68 x 10^-11) = 2.14 x
10^-4
Kb = 2.14 x 10^-4 = x^2/(0.0228 - x)
x^2 = 2.14 x 10^-4 (0.0228 - x)
x^2 - 4.88 X 10^-6 + 2.14 x 10^-4x = 0
on solving
x = 0.00210 = [OH-]
pOH = -log [OH-] = 2.68
pH = 14 - 2.68 = 11.32
