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 Ka expression we use for understanding any acid... You can always plug directly into the Ka 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 Ka = 1.12x10-2)
Plug into the Henderson-Hasselbalch
equation...
And solve...
2. Prepare 500.0mL of a 0.650M HN3/N3-1
buffer at pH = 5.10 from HN3(s) and NaN3(s).
(Hydrazoic acid Ka = 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 Ka
expression. I'll use H-H...
[N3-1]
/ [HN3] = 2.4297
[N3-1]
= 2.4297[HN3]
Now that we know the ratio
of these concentrations, we can solve for the actual concentrations
by using the relationship...
0.650M = [HN3]
+ [N3-1]
0.650M = [HN3]
+ 2.4297[HN3] = 3.4297[HN3]
[HN3] = 0.1895M
[N3-1]
= 0.650 – [HN3] = 0.650 – 0.1895 = 0.4605M
To make a solution that's 0.1895M HN3
at 500.0mL, we need...
(0.5000L)(0.1895M) =
0.09475mols HN3
(0.09475mols
HN3)(43.029g/mol) = 4.077g HN3(s)
To make a solution that's 0.4605M N3-1
at 500.0mL, we need...
(0.5000L)(0.4605M) =
0.23025mols N3-1
(0.23025mols
N3-1)(65.011g/mol) = 14.969g NaN3(s)
So we should be able to make the target
buffer by combining 4.077grams of HN3 and 14.969g of NaN3
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 Ka 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 = pKa +
log (0.327 / 0.316)
pKa = 9.359
Ka =
4.374x10-10
4. What is Kb 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 = pKa +
log (0.143 / 0.158)
pKa = 6.018
pKb = 14 –
6.018 = 7.892
Ka = 1.043x10-8
No comments:
Post a Comment