Ch 16 questions from students, 16.100 and 16.105 (again)

[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 answered the first one, 16.105 (in the Additional Exercises in the EOC exercises) just
last Thur.  This was also a problem in Mastering.  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.  This is one reason you wind up seeing multiple e-mails from me to the
class, people don't read the first one (or more) I send.  You'll find the e-mail with the
explanation below.  This question has been on past midterms.

The other question is 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?  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.  Not all these questions
are in Mastering yet so I can't even assign them in the MC homework.  These are viable
questions for quizzes and exams.

Dr. Zellmer


From: Zellmer, Robert <zellmer.1 at osu.edu>
Sent: Thursday, February 16, 2023 5:39 PM
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 (EOCE 16.105)

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^-8 M HCl solution
There's a homework problem like this in the additional exercises (16.105).
I got a question about this problem this week and after class today.

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 like you're starting with one of
the products before the water has a chance to ionize.  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 for 16.105 before reading the stuff below.


Let's 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 slightly less than 7 since you added a small amount of strong
acid, but very close to that of pure water.

Dr. Zellmer
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.osu.edu/pipermail/cbc-chem1220/attachments/20230221/5f8549d3/attachment-0001.html>