[SodiumBicarbonate]

Hi,

I'm asked to calculate the pH value of a 0.320 M sodium bicarbonate solution, given the following K_a values:

1. H₂CO₃ H⁺ + HCO₃^-, K_a1 = 4.45×10^-7

2. HCO₃^- H⁺ + CO₃^-, K_a2 = 4.7×10^-11

Here's my idea:

I assume sodium bicarbonate dissociates completely in water, so we start with 0.320 M of HCO₃^-, and then the system reaches an equilibrium. In order to use the given data, I look at this equilibrium as if it was the result of a two-step ionization process of the biprotic acid H₂CO₃. For calculating the pH value purposes, the second step is negligible, so we focus on the first step.

Now I consider the following reaction: HCO₃^- + water H₂CO₃ + OH^-, with K_b = K_w / K_a1 = 2.247×10^-8. Say x M of sodium bicarbonate have reacted with water, so we have 0.320 - x ≈ 0.320 M sodium bicarbonate, x M H₂CO₃ and x M OH^-.

We get 2.247×10^-8 = x² / 0.320, so x = 0.0000848, pOH = -log(x) = 4.072 and pH = 14 - pH = 9.928.

Am I right? Is this a valid argument? Is there any simpler way to do it (for instance, without calculating the pOH first)? Does it make sense I didn't use K_a2 at all?

Thanks in advance!

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

[SodiumBicarbonate]

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

Oh, great, thanks!
Now I wonder where exactly my argument fails...

[Borek]

You can't ignore second dissociation step.

[SodiumBicarbonate]

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

Actually, there's something I don't quite understand. Equation 12.4 says [H⁺] = [A²] - [H₂A], but there's a second source of H⁺, the autoionization of water. How do we know its effect is negligible? I mean, if K_a1 = 4.45×10^-7, as in my case, aren't we dealing with H⁺ concentration so small that the autoionization of water can no longer be ignored?

Again, thanks in advance. I really appreciate your help.

[Borek]

Please reread the page, it addresses your concerns.

At the very beginning it is stated that we ASSUME water autoionization can be neglected. As every assumption, it doesn't have to be true always.

Then, at the bottom, there are tables that show how the equation behaves depending on the concentration of the amphiprotic salt. The latter shows how the assumption becomes more and more problematic for diluted solutions.

Contrary to what it may look it is not concentration of H⁺ that tells us if we can ignore water autoionization, but concentrations of substances involved in the H⁺ production and consumption. If they are large enough, water autoionization is not capable of overcoming whatever they do. Your problem asks about pH of the solution that 0.320 M in HCO₃^-, so we are on the safe side. It it were 3.2×10^-5 situation would be completely different.

[SodiumBicarbonate]

Please reread the page, it addresses your concerns.

At the very beginning it is stated that we ASSUME water autoionization can be neglected. As every assumption, it doesn't have to be true always.

Then, at the bottom, there are tables that show how the equation behaves depending on the concentration of the amphiprotic salt. The latter shows how the assumption becomes more and more problematic for diluted solutions.

Contrary to what it may look it is not concentration of H⁺ that tells us if we can ignore water autoionization, but concentrations of substances involved in the H⁺ production and consumption. If they are large enough, water autoionization is not capable of overcoming whatever they do. Your problem asks about pH of the solution that 0.320 M in HCO₃^-, so we are on the safe side. It it were 3.2×10^-5 situation would be completely different.

You're right, we have 0.320 M of HCO₃^-, but on the other hand, K_a1 = 4.45×10^-7, in my case, which is way smaller than NaH₂PO₄'s K_a value of 0.007 from ChemBuddy's example. So, if I understand what's going on, in my case, reaction 1 will tend much more toward H₂CO₃ in comparison to ChemBuddy's example, which means much lower concentration of H⁺. Am I right? Assuming I am, it's still not completely clear to me why it's legitimate to neglect the water autoionization effect on equation 12.4.