Topic: Henderson Hasselbalch Equation only for Weak acids? (Read 3728 times)

SinkingTako · « **on:** November 09, 2012, 07:08:05 AM »

Is Henderson Hasselbalch Equation only suitable for weak acids?

I have a question as follows:

0.5 mol of HA (Ka = 1.5×10-3) was added to 3.0 L of water. After equilibrium was reached, 1.0 g of NaOH was added to the solution. What is the new [H+]?

Okay, so if I find that there is 0.025mol of NaOH, then [NaA] formed is 8.3333 x 10^-3M

As for HA, the new concentration after NaOH reaction is 0.15833M.

Assuming that NaA will dissociate fully, [A-] will be 8.3333 x 10^-3M. If I now just use the
Henderson Hasselbalch Equation ([salt] is 8.333 x 10^-3, [acid]=0.15833), the pH of the final solution will be 1.54. However, looking at the original solution, the pH is 1.82

If I use the ICE table, do brute force using Ka, then the pH I obtain is 1.95, which is still reasonable...

Thus, what is wrong with this calculation? Or is it that in this case, in order to use the equation, the [salt] is the sum of [A-] from both the NaA formed and the dissociation of HA?

Thanks in advance!