FW: Ch 16 question from students

[Zellmer, Robert]

I received a couple of questions about how to calculate the pH for very small concentrations
of strong acid or base in water (conc. less than or around 10^-6 M).  I also received a couple
of questions about a problem dealing with calculating the pH of a solution made by mixing
a strong acid and strong base (neutralization reaction).  People said they couldn't find either
of these in the book or end-of-chapter exercises.  Here's what I want to say about both of
these questions.

I actually answered the first one, 16.106 (in the Additional Exercises in the EOC exercises)
a couple of weeks ago.  Granted you may not have read it if you were a little behind at that
time.  Still, when you don't read the e-mails right away file them away and try to remember
something was sent and take a quick look when you're working on the material.  You'll find
my explanation below.

The other question about the pH after mixing a strong acid and strong base is 16.100
from the Additional Exercises.  It may seem this one would make more sense in ch 17,
since it is a neutralization rxn, which is what happens in titrations.  However, you did SA-SB
neutralization rxns in Ch 4.

How many of you are trying the Additional Exercises or Integrative Exercises?  I know
some of you are as I've had questions about them during office hours and via e-mail.  But
that's coming from just a few people.  Hopefully, everyone is trying these at some point.
These questions aren't all in Mastering yet so I can't even assign them in the MC
homework.  These are viable questions to be asked on quizzes and exams.

Dr. Zellmer


From: Zellmer, Robert <zellmer.1 at osu.edu>
Sent: Friday, February 25, 2022 8:31 AM
To: cbc-chem1220 at groups.asc.ohio-state.edu
Cc: cbc-chem1220-ta at groups.asc.ohio-state.edu
Subject: Ch 16 question from students

I often get the following question from students.  How do I correctly calculate
the pH for something like a 1 x 10^-8 M NaOH or 1 x 10^-7 M HCl solution
There's actually a homework problem like this in the additional exercises
(16.106).

I'll give a little hint.  While NaOH and HCl are strong acids and will completely
dissociate or ionize, these conc. are very small and close to the conc. of OH-
or H+ from pure water.  Normally, with conc. of 10^-5 M or greater for the strong
acids and bases we can ignore the conc. of H+ or OH- coming from the water.
As a matter of fact, if you look at the autoionization rxn for H2O it goes back to
the left when H+ or OH- is added from an outside source (acid or base) so you
get even less H+ or OH- from the autoionization rxn.  This applies even for weak
acids or bases, as long as their conc. are relatively high and they're not too weak.
Thus, for an acid, the conc. of H+ we normally consider is coming just from the
acid and we ignore any H+ from the water itself (same for a base and OH-).

However, with very small conc. of acid or base (whether strong or weak) you
might not be able to ignore the H+ or OH- coming from the water.

So here's the hint.  Set this up as you would for the autoionization of H2O,

    H2O (l) <=>   H+ (aq) +  OH- (aq)

Normally, the first line in the ICE table would have zero for both the H+ and
OH- on the right.   When you add strong acid or base and their conc. are really
small (close to 10^-7), treat the above problem kind of like a what is referred to
as a common-ion problem (actually 17.1).  The initial conc. of the H+ or OH- in
the ICE table won't be zero (kind of like the 2nd step for a polyprotic acid).

Try this before reading the stuff below.





Lets say we have 1 x 10^-8 M HCl.  This means conc. of H+ from the strong
acid is [H+] = 1 x 10^-8 M.  Set up the autoionization equilibrium for H2O in
the following way,

    H2O (l) <=>   H+ (aq) +  OH- (aq)
                          10^-8          0
       -x                  +x            +x
   -----------------------------------------------
       ----             x + 10^-8      x

Plug these into Kw and calculate "x".  You will need to solve a quadratic eqn.
The "x" is the H+ and OH- coming from water.  It will be less than that from
pure water.  Then you add the "x" and the 10^-8 to get the total conc. of H+.
The total [H+] will be a little greater than 10^-7 but not by much.  You'll see
the pH will be close to 7 (a little less than 7), close to that of pure water.

Dr. Zellmer
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