Question

Sodium carbonate (1.3579 g) was dissolved in deionized water to give a solution with a total volume of 100.0 mL. What was the pH of the resulting solution?

Answer 1

mass of Na2CO3 = 1.3579 g

molar mass of Na2CO3 = 106 g/mol

moles of Na2CO3 = mass / molar mass

                            = 1.3579 / 106

                            = 0.0128

volume = 100 mL = 0.1 L

molarity = moles / volume

               = 0.0128 / 0.1

               = 0.128 M

Ka2 = 4.7 x 10^-11

Kb1 = Kw / Ka2

Kb1 = 1.0 x 10^-14 / 4.7 x 10^-11

        = 2.13 x 10^-4

CO3-2 + H2O -------------------> HCO3- + OH-

0.128                                           0               0 --------------> initial

0.128-x                                       x                  x -----------------> equilibrium

Kb1 = [HCO3-][OH-]/[CO3-2]

2.13 x 10^-4 = x^2 / 0.128-x

x^2 + 2.13 x 10^-4 x - 2.72 x 10^-5 = 0

x = 5.11 x 10^-3

[OH-] = x = 5.11 x 10^-3 M

pOH = -log [OH-]

pOH = -log ( 5.11 x 10^-3 )

pOH = 2,29

pH + pOH = 14

pH = 11.71