Calculate the pH of a 1.0 × 10−8 M solution of HNO3
In this problem we’re adding a very small amount of acid to water. The pH should be less
than 7, since it must be now an acidic solution, but the pH should be only slightly less than
7 given how little HNO3 was added.
1. Many people answered this question incorrectly by saying
pH = − log10 [H+ ]
= − log10 1 × 10−8
=8
This answer would imply that the solution is basic, which is contrary to our intuition.
Adding acid to water should not make the solution basic.
If you think about it more deeply, this is incorrect because it doesn’t take into account
the autoionization of water. Think about how the mathematics of your answer should
work as the concentration of HNO3 approaches 0. Clearly the pH should approach 7,
since if you add no HNO3 , the pH would be exactly 7. But mathematically, work like
what is shown above which doesn’t take into account the autoionization of water will
return a pH that is approaching infinity as the amount of added HNO3 goes to zero,
since it will be taking the negative log of a number very close to zero.
2. Many other people answered the question in the following way
−
−−
*
First, write an ICE table for the “equilibrium” of HNO3 )
−
− H+ + NO3 , and used the
pKa given in the lecture slides of something like -1.3.
This yields
Ka = 101.3
=
x2
1 × 10−8 − x
If you solve this exactly (the approximation that x is small relative to 1 × 10−8 is not
valid since the equilibrium constant is large and 1 × 10−8 tiny), you get
x = 0.99999999949 × 10−8
x ≈ 1 × 10−8
− log10 x = 8.0
So again an answer of pH = 8.0 was reported.
Remember that HNO3 is a strong acid. It dissociates completely. As a calculation like
this shows, there is no use doing an ICE table for a reaction that you know
goes to completion. You just get an x that represents a complete conversion of reactants into products.
Thus, this answer doesn’t add anything over the previous answer. It does not take into
account the autoionization of water.
3. Another answer was as follows. This one is close, but not all the way there.
Remember that the autoionization of neutral water produces [H+ ] = 1 × 10−7 M. The
addition of HNO3 produces a concentration of [H+ ] = 1.0 × 10−8 . Therefore one might
argue that the total concentration of H+ is simply 1.0 × 10−7 + 1.0 × 10−8 = 1.1 × 10−8 .
Taking the -log of this proposed [H+ ] gives a pH 6.96.
This answer has the right asymptotics. The mathematics would return the right pH,
7, in the limit that zero moles of HNO3 were added, so thats good. The problem with
this approach is that it disregards Le Chatlier’s principle.
Consider the auto-ionization equilibrium:
−
+
−
*
H2 O −
)
−
− H + OH
By adding HNO3 , since it dissociates completely, we’re upping the concentration of H+ .
Le Chatlier’s principle tells us that this perturbation will cause the reaction above to
shift to the left. Because of the shift, the equilibrium concentration of H+ will not
simply be equal to the sum of 1.0 × 10−8 and 1.0 × 10−7 . We need to take this into
account in our answer.
4. The correct approach is as follows. This is a slightly different presentation than the
way it was done in the answer key posted on coursework.
Remember that we have the auto-ionization of water going on
−
+
−
*
H2 O −
)
−
− H + OH
In pure water, we have [H+ ] = [OH− ] = 1.0 × 10−7 M.
Imagine that we add the HNO3 , thus upping the concentration of H+ , but that the
equilibrium above hasn’t yet had time to relax. By upping the concentration of H+ ,
we’ve shifted the reaction out of equilibrium, and set up a new initial condition, and
the equilibrium is going to shift left to respond.
H2 O
I
C
E
-
−
−
*
)
−
−
H+
1.0 × 10 + 1.0 × 10−8
−x
1.1 × 10−7 − x
−7
OH−
1.0 × 10−7
−x
1.0 × 10−7 − x
K = Kw = 1.0 × 10−14
K = [H+ ][OH− ]
1 × 10−14 = (1.1 × 10−7 − x) · (1.0 × 10−7 − x)
x = 0.488 × 10−8
[H+ ] = 1.1 × 10−7 − x
= 1.05 × 10−7
pH = − log10 [H+ ]
= − log10 1.05 × 10−7
= 6.98
So we get a pH that is very close to 7. Note that compared to answer 3, we have a
lower H+ here of 1.05 × 10−7 as opposed to 1.1 × 10−7 . As you can see from the ICE
table, this is because we’re allowing the H2 O equilibrium to respond via Le Chatlier’s
principle and shift to the left.
The principle of a problem like this is that you’ve got to extend your ICE table a
bit. Whenever you have a perturbed equilibrium; what you’ve got is your perturbation,
added on the concentrations of your equilibrium system, sets up a new initial condition
that your equilibrium is going to shift to respond to.
Just like your “E” line in an ICE table is the sum of your initial plus a change, your
“I” line in an ICE table might be the sum of the former equilibrium concentration plus
a perturbation that pulls it out of equilibrium.
I hope this makes sense to you. If it doesn’t please come and visit me in office hours or send
me an email. I’m here to help!