Ch 16 question from students (EOCE 16.105)

[Zellmer, Robert]

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
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