Cheni help needed?
Answer:
AcOH = H+ + OAc-
[H+][OAc-] / [HOAc] = 1.75 E-05
But [H+] = [OAc-]
So [H+]^2 / [HOAc] = 1.75 E-05
Plug in [HOAc] = 0.1 and solve for [H+]
pH = -log(10)([H+])
Note that we are assuming the concentration of HOAc does not change significantly because of the low ionization constant.
I deliberately left a few of the easy things for you to do so that you can "get into" the problem a little bit which will help you learn it a little more. I hope it helps you on the test.
Im going to give it a shot here. I assume K4= Ka?.
So a general formula for Acid dissociation is
|HA|= |H+| + |A-|
Ka = [H+][A-]/ [HA]
When you put acid in the solution, since its a weak acid, only a certain amount of acid dissociate. But one thing is for sure...that is...the amount of [H+] and [A-] is the same. And [HA] after dissociation is the original concentration - the concentration of either [H+] or [A-].
Set [H+] = [A-] = x
*[HA] = 0.10M - x <---0.10M is given.
* but usually x is negligible so you can just assume that it doesn't affect the concentration of [HA]
You know Ka.
Plug into the equation and solve for x.
1.75* 10^-5 = (x * x)/0.1
Solve for x. = 0.001323M
Therefore [H+] = [A-] = 0.001323 M. <---hydronium concentration.
pH = - log [H+]
so plug that in.
pH= - log (0.001323) =2.8784.
Hope this helps and if you have any question or if you find any mistakes in my answer...please let me know.
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