Question

According to the label, Alka-Seltzer contains 1.916 g (0.0288 mol) of sodium hydrogen carbonate. When an alka-seltzer tablet dissolves in 150 mL of water, what will be the pH of the solution? Ka1=4.5E-7 and Ka2=4.7E-11

Answer 1

First, let's write and balance the overall reaction:

HCO₃^- + H₂O -------------> CO₃^2- + H₃O⁺

Now, let's calculate concentration of bicarbonate in this volume:
M = 0.0288 mol / 0.150 L = 0.192 M

Now, let's do the ICE table basing on the balance reaction:

r: HCO₃^- + H₂O -------------> CO₃^2- + H₃O⁺
i: 0.192 0 0
e: 0.192-x x x

Now, this compound came from the carbonic acid, (Ka1) so concentration of H+ is being given by:

[H⁺] = [(K1K2M + K1Kw) / K1+M]^1/2
[H⁺] = [(4.5x10^-7*4.7x10^-11*0.192 + 4.5x10^-7*1x10^-14) / 4.5x10^-7+0.192]^1/2 = 4.601x10^-9 M

pH = -log(4.601x10^-9)
pH = 8.34

The other way to do this, was to calculate pKa of both and then dividing by 2:

pH = pKa1 + pKa2 / 2
pH = -log(4.5x10^-7) - log(4.7x10^-11) / 2
pH = 8.33

Hope this helps