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In this module we're going to look at polyprotic acids.
Our objective for this module is to be able to identify polyprotic acids
as well as write the ionization reactions for each step.
What distinguishes polyprotic acids from monoprotic acids is in the proton.
In this case we have multiple protons
that are ionizable from this particular molecule.
Here we have _2CO_3 that has two protons
and therefore can undergo 2 ionization steps.
The first step will remove one proton and produce HCO_3-
Then HCO_3-
will lose its proton and form H+
and CO_3^2-. Notice that each step has its own K_a
value and one thing we notice is that K_a2,
for the second reaction, is much smaller than K_a1.
This is a trend that will continue whenever we look at polyprotic.
Here are some examples are some common
polyprotic acids. We've already seen H_2SO_4 and what we notice is that out K_a
value is much greater than one.
Remember that the first ionization step of H_2SO_4
is actually representative of a strong acid. The second step is not.
The second step is a weak acid and as such has a K_a value
even though it's considerably higher than many other K_a then we see for a
second step.
Sulfurous acid has 2 ionizable protons,
so it has 2 K_a values. Carbonic acid also has 2 K_a values
for each of it's ionizable protons. Phosphoric acid actually has three
ionizable protons.
And for that it has three K_a values.
Notice that the trend we saw with the carbonic acid that K_a2 is less than K_a1
continues as we go to phosphoric. K_a3 is the smallest of the three values.
K_a2 is a little bit bigger and K_a1 is somewhat larger.
We will see this trend continue for all polyprotic acids.
Let's look at an example...
Which of the following best describes the pH 0.03 molar
H_2SO_4 solution. Will the pH be ess than 1.5
or greater than 1.5?
When we look at H_2SO_4 we need to remember that it ionizes: in two steps.
The first step H_2SO_4
is a complete reaction and complete ionization to H+
plus HSO_4-. Remember this first step is a strong acid step.
Then we have HSO_4-
in equilibrium with H+
and SO_4^2-. If I go back to my problem
I know that the H_2SO_4 concentration initially
is 0.03. Therefore after the first ionization
the H+ concentration for my first step
will be equal to 0.03
molar. The same as the initial concentration of H_2SO_4.
If I were to find the pH only from that
amount of H+, I would take the pH because the negative
log have 0.03
and I would find that pH equals
1.5. However,
I do have a second step that's producing a little bit more H+.
Now we don't know the exact amount so we can't quantify with the pH is
but what we do know is that with the second step
the H+ concentration is increasing if only slightly.
It is increasing and therefore
the pH must be decreasing, because we have a solution that is more acidic more H+
therefore the pH value will decrease. If you're not sure about why that's the case
then go ahead and put in a different number that is greater than 0.03.
For example what if we said H+ was equal to 0.04 then we could say that pH
equals the negative log of 0.04
and what we find is that pH
equals 1.4. So as we have
increased our H+ concentration we have decreased the pH of the solution.
In the next module we will look at different types acids
and the periodic trends related to their strength.