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Topic: Acid  (Read 2950 times)

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Offline KungKemi

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Acid
« on: January 15, 2017, 07:24:49 AM »
So basically there is this Acid-Base problem that precedes as follows: "An aqueous solution contains a mixture of 0.0500 M HCOOH (Ka = 1.77×10-4) and 0.150 M CH3CH2COOH (Ka = 1.34×10-5). Calculate the pH of this solution."

Okay, so firstly I considered the dissociation equilibrium established by the weaker of the two acids (Propionic acid) which, after equilibrium has been established, yields a [H+] of 1.42×10-3 M. I then preceded to consider the following, concerning the dissociation of Formic acid:

         HCOOH   ::equil::   H+  +  COOH-

Initial:    0.15   ::equil::    0.00142     +  0
Change:  -x      ::equil::       +x         +  +x
Equili.: 0.15-x  ::equil:: (0.00142+x) +   x

0.000177 = (x)(0.00142+x)/(0.15-x)

Eliminate (+x and -x; assuming that its loss and addition is negligible in comparison to itself as an isolated variable)...

0.000177 = 0.00142x/0.15

 x = 6.24×10-3 M

pH = -log[0.00624] = 2.20

I was just wondering if my methology seemed accurate, and if the following is the correct answer?

p.s.: I'm not the strictest on significant figures.

Thank you,
KungKemi

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Re: Acid
« Reply #1 on: January 15, 2017, 07:52:31 AM »
In general it is better to start with the dissociation of the stronger acid, as it will be less changed after the weaker acid powers the pH. However here the concentration of the weaker acid is much higher, so it is not that easy.

The best approach is to make several iterations.

Have you checked if the assumption that +x and -x can be ignored is a correct one? When can you make such an assumption?

Your final result is quite off.
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Offline AWK

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Re: Acid
« Reply #2 on: January 15, 2017, 09:07:19 AM »
AWK

Offline KungKemi

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Re: Acid
« Reply #3 on: January 15, 2017, 09:34:58 PM »
Right, okay so after looking at what AWK suggested I decided upon the following two systems:

         HCOOH   ::equil::   H+  +  HCOO-  K = 1.77×10-4

Initial:    0.05   ::equil::  0   +  0
Change:  -x      ::equil:: +x  +  +x
Equili.: 0.05-x  ::equil::     x + x

CH3CH2COOH   ::equil::   H+  +  CH3CH2COO-  K = 1.34×10-5

Initial:    0.15   ::equil::  0   +  0
Change:  -y      ::equil:: +y  +  +y
Equili.: 0.15-y  ::equil::     y + y

[H+] = x + y

1.77×10-4 = (x+y)x/(0.05-x)

1.34×10-5 = (x+y)y/(0.15-y)

I then continued to solve the following using appropriate assumptions, and I ended up with a final solution pH of 2.50.

I was just wondering if this would be the more correct method to solving this question?
KungKemi

Offline AWK

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Re: Acid
« Reply #4 on: January 16, 2017, 12:10:43 AM »
Almost exact solution will be quintic equation with ionic strength correction. Sufficient approximation is
[H3O+]=SQRT(K1c1+K2c2)
AWK

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