1. What is a buffer? What is needed to create a buffer (i.e. what should you look for to
identify a buffer?
A buffer is a solution that is able to resist changes in pH when additional amounts of
acid and base are added to the solution. Look for solutions that contain about equal
concentrations of weak acid and its conjugate base or of weak base and its conjugate
acid.
2. Give the Henderson-Hasselbalch equation. When is it used and what is it used for?
pH = pKa + log ([base]/[acid])
This equation is used ONLY FOR BUFFERS. It can be used to calculate the pH or
the pOH of a buffer or the concentration of acid or base present in a buffer.
3. What is the pH of a solution that is 0.20 M NH3 and 0.30 M NH4Cl? The Kb of NH3 is 1.8
x 10-5.
This solution is a buffer – has about equal amounts of weak base and conjugate acid
present  can use Henderson-Hasselbach to calculate the pH.
pH = pKa + log ([base]/[acid])
We have Kb and need Ka  Kw = Ka x Kb  Ka = Kw / Kb = (1.0 x 10-14)/(1.8 x 10-5) =
5.56 x 10-10
pH = -log(5.56 x 10-10) + log(.2 M / .3 M) = 9.08
4. A solution is initially 0.15 M in HOCl and 0.25 M in NaOCl. What is the pH of this
solution after 0.050 mol of HCl has been bubbled into the solution?
Ka of HOCl is 3.5 x 10-8.
Our initial solution is a buffer – has about equal amounts of weak acid and
conjugate base present. We add strong acid to the buffer, so we will need to use a
limiting reagent ICE table to see how much of each species is remaining after the
addition. The hardest part of this calculation is writing a reaction to go along with
the ICE table. Remember that acids always react with bases, and this will become a
little easier.
Since HCl is a strong acid, we are really just adding H3O+ to the buffer. This H3O+
will react with the base in the buffer (OCl-) to be neutralized.
We can assume this solution is in 1 L, making it easier to convert between moles and
molarity. We can also assume that the bubbling of the strong acid DID NOT change
the volume of solution, so we don’t have to be worried about a total volume change.
Use your ICE table to completely get rid of the limiting reagent, subtract it from
both reactants and add to the products side. This is different than ICE tables we
have done in the past – we know the value of the change row in these cases. H3O+ is
our limiting reagent here, since there is the least amount of it, so we will subtract out
all of the H3O+ from the reactants side and add that same value to the products.
Initial
Change
Equilibrium
(Final)
H 3 O+
0.050
- 0.050
0
OCl0.25
- 0.050
0.2

HOCl
0.15
+ 0.050
0.2
H2 O
n/a
n/a
n/a
Now, look at the type of solution present once the addition of strong acid has been
accounted for. We have exactly the same amount of acid and conjugate base present
in solution, making our final solution a BUFFER. So, we can use HendersonHasselbach to calculate the pH.
pH = pKa + log([base]/[acid])
pH = -log(3.5 x 10-8) + log (0.2/0.2) = 7.46
5. Draw and label the titration curve for the titration of a strong acid with a strong base.
6. Draw and label the titration curve for the titration of a strong base with a strong acid.
7. Draw and label the titration curve for a weak acid titrated with a strong base.
8. Draw and label the titration curve for a weak base titrated with a strong acid.
9. Calculate the molar solubility and solubility in g/L of silver phosphate (Ksp = 8.9 x 10-17).
Use ICE chart to calculate molar solubility (x corresponds to the molar solubility of
the solid).
Ag3PO4 (s)
3 Ag+
PO43
Initial
n/a
0
0
Change
n/a
+3x
+x
Equilibrium
n/a
3x
x
Ksp = 8.9 x 10-17 = (3x)3 * x = 27x4  x = 4.26 x 10-5 M
In grams/L  4.26 x 10-5 mol/L * molar mass of Ag3PO4 (418.6 g/mol) = .018 g/L