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Sodium carbonate (2.4134 g) is dissolved in enough deionized water to give a solution with a...

Sodium carbonate (2.4134 g) is dissolved in enough deionized water to give a solution with a total volume of 250.0 mL. What is the pH of the resulting solution? Hint: For carbonic acid, pKa1 = 6.351 and pKa2 = 10.329. What is the equilibrium concentration of H2CO3 in the solution? calculate the value of alpha HCO3-

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Expert Solution

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    0.0228                         0              0

final      0.0288-x                       x              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

queen_honey_blossom answered 2 years ago
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