Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: kct on May 21, 2005, 12:41:33 AM

Title: help on Calculating pH
Post by: kct on May 21, 2005, 12:41:33 AM
Heres the problem:
Calcaulte hte pH of a .20M aqueous solution of (CH3)2NH2C2H3O2, dimethylamine acetate.

Here's what I did:

(CH3)2NH2C2H3O2 --> (CH3)2NH2+ + C2H3O2-.

Since acetate doesn't do anything to the pH I take it out of the equation so I have

(CH3)2NH2C2H3O2 --> (CH3)2NH2 + H+

I   .20                            0               0
C -x                              +x           +x
F .20 - x                         x             x

x2/(.20-x) = 1.0 x 10-14/5.9x10-4 = 1.7 x 10-11

x=1.8 x 10-6
.log(x) = 5.74

What am I doing wrong?? Thanks for your help


Title: Re:help on Calculating pH
Post by: Garneck on May 21, 2005, 02:54:06 AM
Quote from: kct on May 21, 2005, 12:41:33 AM
(CH3)2NH2C2H3O2 --> (CH3)2NH2+ + C2H3O2-.

It doesn't dissociate like that.
Title: Re:help on Calculating pH
Post by: Borek on May 21, 2005, 04:20:47 AM
Quote from: kct on May 21, 2005, 12:41:33 AM
Since acetate doesn't do anything to the pH I take it out of the equation

There are two hydrolysis processes here. One of acetic ion, one of dimethylamine. The dissociation constants are too close for any of the equlibriums to be neglected, so it is not possible to do any simple approximations. I will take a look at the problem, right now all I know is that pH is 7.72 (but it was calculated using BATE, so no fancy thinking on my side).
Title: Re:help on Calculating pH
Post by: Borek on May 21, 2005, 04:49:50 AM
No better answer so far and no time to delve deeper at the moment. From what I see the best solution I can get to requires third degree equation solving. That's just to symmetrical problem. Hmm... perhaps that symetry is a key to success? Perhaps later.

As for now download BATE, I wrote it to not do such calculations by hand ;)
Title: Re:help on Calculating pH
Post by: Borek on May 21, 2005, 05:17:27 AM
 :Lighten:

Symmetry does the trick!

We have a base and an acid - lets call them HA and BOH.

Ka = [A][H]/[HA]
Kb = [ B][OH]/[BOH]

so

[H] = Ka[HA]/0.2
[OH]=Kb[BOH]/0.2

Why 0.2? Because the hydrolysis doesn't go too far and we can safely assume that the dissociated form concentration is almost the same as the concentration of salt. Besides, that's only assumption - we can later check if it was OK, and if not, start again.

Now, first fancy move:

[H][OH] = Ka[HA]/0.2 * Kb[BOH]/0.2 = Kw

after rearranging:

[HA][BOH] = 0.2^2*Kw/(Ka*Kb)

And now... SYMMETRY DOES THE TRICK!

Ka and Kb have very similar values, so lets assume that [HA] = [BOH]. If so

[HA]^2 = 0.2^2*Kw/(Ka*Kb)

and [HA] = 0.00022

Getting back to [H] and acid dissociation constant:

[H] = Ka [HA] / 0.2 = 10^(-4.75) * 0.00022 / 0.2 = 1.96E-8

and pH = 7.71

That's the same value as the one calculated with BATE. To be OK we now should check if our assumption was right, putting known values of [H], [OH], [HA] and [BOH] into all equations and checking if they hold - but I can assure you they do.

OK, I am going before my wife filles for divorce....
Title: Re:help on Calculating pH
Post by: kct on May 21, 2005, 12:34:45 PM
Ok,  I think i got it! Thanks Borek!