Today was buffers day in class so we looked at 294 different ways to use the Henderson-Hasselbalch equation. Remember, the Henderson-Hasselbalch equation is just a rearrangement of the K

_{a}expression we use for understanding any acid... You can always plug directly into the K_{a}expression and get the same result you will get using Henderson-Hasselbalch. So the problems...
1. Combine 12.642g of chlorous acid
with 15.372g of lithium chlorite, dilute to 500.0mL. What is the
expected pH of this buffer? (Chlorous acid K

_{a}= 1.12x10^{-2})
Plug into the Henderson-Hasselbalch
equation...

And solve...

2. Prepare 500.0mL of a 0.650M HN

_{3}/N_{3}^{-1}buffer at pH = 5.10 from HN_{3}(s) and NaN_{3}(s). (Hydrazoic acid K_{a}= 1.93x10^{-5})
Start off by solving for the ratio of
conjugate acid to conjugate base using either the
Henderson-Hasselbalch equation or just an unmodified K

_{a}expression. I'll use H-H...
[N

_{3}^{-1}] / [HN_{3}] = 2.4297
[N

_{3}^{-1}] = 2.4297[HN_{3}]
Now that we know the ratio
of these concentrations, we can solve for the actual concentrations
by using the relationship...

0.650M = [HN

_{3}] + [N_{3}^{-1}]
0.650M = [HN

_{3}] + 2.4297[HN_{3}] = 3.4297[HN_{3}]
[HN

_{3}] = 0.1895M
[N

_{3}^{-1}] = 0.650 – [HN_{3}] = 0.650 – 0.1895 = 0.4605M
To make a solution that's 0.1895M HN

_{3}at 500.0mL, we need...
(0.5000L)(0.1895M) =
0.09475mols HN

_{3}
(0.09475mols
HN

_{3})(43.029g/mol) = 4.077g HN_{3}(s)
To make a solution that's 0.4605M N

_{3}^{-1}at 500.0mL, we need...
(0.5000L)(0.4605M) =
0.23025mols N

_{3}^{-1}
(0.23025mols
N

_{3}^{-1})(65.011g/mol) = 14.969g NaN_{3}(s)
So we should be able to make the target
buffer by combining 4.077grams of HN

_{3}and 14.969g of NaN_{3}in enough water to make 500.0mL of solution.
NOTE: If you're actually making a
buffer, be very careful about the order of addition of the
components. Whenever you're combining a solid or concentrated
solution with a solvent, it's usually a good practice to add the
solid or concentrated stock to the larger volume of solvent slowly
with very good mixing. Dissolving and/or mixing can liberate a LOT of
heat in some cases that could be dangerous if the order of addition
is reversed.

NOTE2: Hydrazoic acid is not a solid at
room temperature, and the pure liquid is a non-trivial safety risk...
We can talk about this type of a buffer on paper, but there's very
little chance you (or I) will ever prepare or use a hydrazoic
acid-based buffer.

3. What is K

_{a}of a weak acid, “HA”, if a solution made by dissolving 0.316mol HA and 0.327mol A^{-1}in water and diluting to 750.0mL has a pH of 9.374?
Plug in to Henderson-Hasselbalch or the
generic Ka expression...

9.374 = pK

_{a}+ log (0.327 / 0.316)
pK

_{a}= 9.359
K

_{a}= 4.374x10^{-10}
4. What is K

_{b}of a weak base, “B”, if a solution made by dissolving 0.143mol B and 0.158mol HB^{+1}in water and diluting to 400.0mL has a pH of 5.975?
Similar to the previous problem, plug
in to Henderson-Hasselbalch or the generic Ka expression...

5.975 = pK

_{a}+ log (0.143 / 0.158)
pK

_{a}= 6.018
pK

_{b}= 14 – 6.018 = 7.892
K

_{a}= 1.043x10^{-8}
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