my average pH is 2.38 and the avg molarity is 0.07101... i'm supposed to find Ka using those... please help?
2007-02-21 21:09:51 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Philip K, pH=-log[H+] and not -logKa.
I don't understand what you mean by "average".
Assuming that we are talking about a monoprotic acid
Ka=[H+][A-]/[HA]
Since the reaction is HA <=> H+ +A-
[A-]=[H+]
and [HA]=[HA]total-[H+]= 0.07101-[H+]
so
Ka= [H+]^2/(0.07101-[H+])
[H+]=10^-pH= 10^-2.38 =0.004169
so
Ka= (0.004169^2) / (0.07101-0.004169) =2.6*10^-4
2007-02-21 21:29:20 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋
2
2016-08-23 01:07:12 · answer #2 · answered by ? 3 · 0⤊ 0⤋
Well pH is defined as -log Ka. while the equation of (an acid i assume you are working with) is Ka = [H3O+][A-]/[HA]. For example, a pH of 1 would be .1 Ka which would be .1 M acidic solution. Thus, if you have the Average Molarity, and you know how to derive pH from Average Molarity... well you know concentrations so just plug into the equation. Or even easier would be to use the equation -log (Ka)= pH and input pH already known to find the Ka value. By the way, by Log, it means base 10 if you were not familiar with the function.
2007-02-21 21:20:50 · answer #3 · answered by Philip K 1 · 0⤊ 0⤋
i think of the assumption is right, in spite of the undeniable fact that i could think of the acid dissociation equation ought to be HN3 + H2O --> H3O+ and N3- (which ought to be 0.848M, considering NaN3 is monobasic)
2016-12-18 08:27:43 · answer #4 · answered by ? 4 · 0⤊ 0⤋