Understanding dissociation constants

AI Thread Summary
In the discussion, there is clarification on the term "[HA]" in the equation Ka = [H+][A-]/[HA], which refers to the equilibrium concentration of the acid rather than its initial concentration. The initial concentration, sometimes called formal or analytical, is represented as [HA] + [A-]. A key point raised is the necessity of including both [H+] and [A-] in the numerator of the Ka expression. It is explained that using only one of these concentrations would be insufficient, as both are products of the dissociation reaction. Changes in the concentration of either [H+] or [A-] affect the equilibrium, thus necessitating their presence in the equation to accurately represent the acid dissociation constant.
sodium.dioxid
Messages
50
Reaction score
0
Ka = [H+][A-]/[HA]

Is the term "[HA]" referring to its initial concentration or its equilibrium concentration?
 
Chemistry news on Phys.org
All concentrations are equilibrium concentrations. Initial concentration of acid (sometimes called formal or analytical) is [HA]+[A-].
 
OK, another question: why is it necessary to have both [H+] and [A-] on the numerator? It seems like we could have done just as fine using the definition "Ka=[A-]/[HA]" or "Ka=[H+]/[HA]" instead of "Ka=[H+][A-]/[HA]". Is this conventional or am I missing something?
 
Last edited:
As both H+ and A- are products of the reaction changing concentration of either one shifts the equilibrium, so you need both.
 
I want to test a humidity sensor with one or more saturated salt solutions. The table salt that I have on hand contains one of two anticaking agents, calcium silicate or sodium aluminosilicate. Will the presence of either of these additives (or iodine for that matter) significantly affect the equilibrium humidity? I searched and all the how-to-do-it guides did not address this question. One research paper I found reported that at 1.5% w/w calcium silicate increased the deliquescent point by...
I'm trying to find a cheap DIY method to etch holes of various shapes through 0.3mm Aluminium sheet using 5-10% Sodium Hydroxide. The idea is to apply a resist to the Aluminium then selectively ablate it off using a diode laser cutter and then dissolve away the Aluminium using Sodium Hydroxide. By cheap I mean resists costing say £20 in small quantities. The Internet has suggested various resists to try including... Enamel paint (only survived seconds in the NaOH!) Acrylic paint (only...
Back
Top