Ch 17 EOC 17.18(b) question and error in solutions manual

[Zellmer, Robert]

I've been asked the following about 17.18(b), a common-ion (CI) problem.
It's followed by my answer.

"I just had a question regarding one of the EOC problems for chapter 17, specifically
18b. The question asks to calculate the percent ionization of 0.125M Lactic acid solution
in a solution containing 0.0075M Sodium Lactate. When setting up the Acid-Dissociation
constant equation, I noticed that the answer key ignores the x's in (0.0075+x) and (0.125-x).
This makes the equation =
(0.0075)(x)/(0.125) = 1.4*10^-4, instead of (0.0075+x)(x)/(0.125-x) = 1.4*10^-4.

Wouldn't it be incorrect to omit the x's in this equation? By taking the acid disassociation
constant divided by the concentration of Lactic acid or Lactate, you get a value well above
the general rule you described of <1.0*10^-3. "

The following was my response:

Before I go into the stuff below about the question, I gave a process in class to check
whether ignoring "x" in a weak acid problem would be okay before even going through
the ICE table.  Take the Ka and divide it by the original conc. of acid. If that ratio is less
than about 10^-3 it will okay to ignore "x" in the resulting expression for Ka obtained
from the ICE table.  The same applies to a weak base problem (using Kb and the original
conc. of base).   Strictly speaking, this was for when you start with only weak acid or only
weak base, not a common-ion (CI) problem.

  Ka
-------- <  ~ 10^-3                     Okay to ignore "x" in the expression for Ka.
[HA]o

(You can do the same thing for weak bases, Kb/[base]o < 10^-3).

How about for a CI problem?

Let's start by looking at something simpler,  just 0.125 M lactic acid (HL for short).  If you
set up the ICE table for 0.125 M HL you get,

              x^2
Ka = ----------------- = 1.4 x 10^-4
          (0.125 - x)

You can ignore the "x" in the denominator in this case

              x^2
Ka = ----------------- = 1.4 x 10^-4
           (0.125)

and get only a 3.3% error.  Note, in this case Ka/[HL]_o = 1.4 x 10^-3/0.125 = 1.12 x 10^-3
(which is close to what I have above) and ignoring "x" with respect to the 0.125 M is okay.

For the CI problem you have

            HL  <==>  H^+  +  L^-
         0.125            0       0.0075

           y (0.0075 + y)
Ka = ---------------------- = 1.4 x 10^-4
             (0.125 - y)

Since you're starting with one of the products this rxn won't go as far to the right as
it did with just the lactic acid.  That means this "y" will be smaller than the "x" above
with just 0.125 M HL (lactic acid).  That means ignoring "y" with respect to 0.125 is
a better approx. than above when ignoring the "x" w/rsp to 0.125 when you start with
only the lactic acid.  Thus, you only need to worry about the 0.0075.  Ignoring the
"y" in both the numerator and denominator gives the following for "y",

y = 2.33 x 10^-3 M

The % error with respect to the 0.125 is only 1.87%, less than 3.3% from above
when starting with only 0.125 M HL (lactic acid).  However, the % error w/rsp to the
0.0075 is 31%.  That means you really shouldn't ignore the "y" w/rsp to the 0.0075,
as is done in the solutions manual.

You can see the error by using this "y" value in the expression for Ka and see if
you get the Ka.

           (2.33 x 10^-3) (0.0075 + 2.33 x 10^-3)
Ka = ------------------------------------------------------ = 1.87 x 10^-4
                   (0.125 - 2.33 x 10^-3)

Comparing this to the actual Ka of 1.4 x 10^-4 gives a 34% error.  Not good!


You can solve a quadratic or do the following, just ignore "y" w/rsp to the 0.125
but not the 0.0075.

           y (0.0075 + y)
Ka = ---------------------- = 1.4 x 10^-4
               (0.125)

This is still going to be a quadratic equation but one that's easier to set up.
You could also solve it by using the method of successive approximations
as I did in class.  Doing this I get the following for 'y',

y = 1.868 x 10^-3

This is a better value for the [H+] that comes from this CI problem.  Using
this value for "y" to calculate Ka you get the following:

           (1. 868 x 10^-3) (0.0075 + 1. 868 x 10^-3)
Ka = ----------------------------------------------------------- = 1.42 x 10^-4
                        (0.125 - 1. 868 x 10^-3)

Comparing this to the actual Ka of 1.4 x 10^-4 gives a 1.5% error.  This
is less than 5% so this is fine.

It gives the following for % ionization,

% ionization = (1.868 x 10^-3)/0.125 * 100 = 1.49 %

This is a better answer compared to 1.87% ionization you get by ignoring
the "y" w/rsp to both 0.125 and 0.0075 as was done in the solutions manual.
The 1.87% is about a 26% error.

Dr. Zellmer
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