Acids & Bases Chemistry Math Question 2

Question

Compute the pH of a solution formed by adding 11.6 g of baking soda (sodium bicarbonate) to enough water to make a 250.0 mL solution. The acid dissociation constants for carbonic acid are pK₁=6.35, pK₂=10.33

Answer 1

11.6 g NaHCO3 x 1 mol/84.007 g = 0.1381 moles

0.1381 moles/0.250 L = 0.552 M HCO3-

HCO3- + H2O ==> H2CO3 + OH-

Since HCO3- is acting as a base, we want the Kb which we can find from the pKa:

Ka = 1x10^-6.35 = 4.47x10^-7

Kb = 1x10^-14 / 4.47x10^-7 = 2.2x10^-8

Kb = [H2CO3][OH-] / [HCO3-]

2.2x10^-8 = (x)(x) / 0.138 - x and assuming x is small we can neglect it

2.2x10^-8 = x2 / 0.138

x² = 3.04x10^-9

x = 5.5x10^-5 (which is small compared to 0.138 so assumption was valid)

[OH-] = 5.5x10-5 M

pOH = -log 5.5x10-5 = 4.26

pH = 14 = 4.26 = 9.74

Since pKa2 is is so much greater than pKa1 (Ka2 much smaller than Ka1) the second dissociation of H2CO3 can be ignored as it will not contribute significantly to the pH.