dealing with Ka's/acethedat=dirt

[H to the Izzy]

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This is my first post on this forum...I have checked it out real frequently lately cause my dat is approaching aug 19th. Unfortunately, I have to use my first post for this kind of question. 👎 I'd appreciate if you guys could give me some insight on this one...

A 0.02M solution of a monoprotic acid has a pH=5. What is the Ka of this acid? The answer is 5 X 10 -9

ps. that acethedat is a complete and toal hoax.....bastards

 

[ems5184]

I always get frustrated with these type of problems, but I'll give this one a try.

So, first, [H+] = antilog (5) = 1*10^-5 M which is also equal to [OH- or whatever the conjugate base of the acid is]

The equilibrium concentration of the acid left after dissociating would be 0.02 M - 1*10^-5 M which is equal to 0.01999 M.

In this case, Ka = [H+][conjugatebase] / [acid]
and this is equal to [1*10^-5][1*10^-5] / 0.02
0.02 is used to simplify the calculations

Simplifying this expression, we get 1*10^-10 / 0.02

Multiply the numerator by 50 (reciprocal of 0.02) and you get 5*10^-9

Hope this helps 🙂 , let me know if anything is not clear.

 

[sinned]

so a monoprotic acid gets deprotonated in the following fashion,

HA --> A- + H3O+
0.02 0 0
-x +x +x
0.02-x x x



so Ka = (x^2)/0.02-x

we know the pH which is 5, this means that -log[H30+]=5, where [H30+] is equal to "x."

the antilog of -5 equals, 1 X 10 -5, so you plug this into the Ka equation as x. the answer you get is 5 X 10 -9