Question

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? For carbonic acid, pKa1 = 6.351 and pKa2 = 10.329.

Also:

What is the equilibrium concentration of H2CO3 in the solution

calculate the value of aHCO3−.

Write a mass balance and charge balance equation

please show all work!

Answer 1

concentration of Na2CO3 solution = 2.4134 g/106 g/mol x 0.250 L = 0.091 M

Na2CO3 + H2O <==> HCO3- + OH-

Ka1 = 4.45 x 10^-7 = x^2/0.091

x = [OH-] = 2.01 x 10^-4 M

pH = -log[OH-] = 3.70

pH = 14 - pOH = 10.3

[H+] = 5.01 x 10^-11 M

aHCO3- = [H+]Ka1/([H+]^2 + [H+]Ka1 + Ka1.Ka2)

              = (2.23 x 10^-17)/(2.51 x 10^-21 + 2.23 x 10^-17 + 2.08 x 10^-17)

              = 0.517 M

charge balance equation,

[CO3^2-] + [OH-] = 2[Na+] + [H+]

mass balance equation,

[CO3^2-] = 2[Na+]