CH 117 Spring 2015Worksheet 13
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. Describe how a buffer works.
The purpose of a buffer is to resist changes in pH, so it contains both acid and base. If extra acid is added to a buffer, that acid will react with the base that is already present in the buffer and be neutralized. If extra base is added to a buffer, that base will react with the acid that is already present in the buffer and be neutralized. Basically, a buffer makes use of basic acid-base chemistry, in which bases react with acids and vice versa.
4. What are the limits for the range of a buffer? If the pKa of benzoic acid is 4.19, what is the possible range of a buffer created from this acid?
Buffer range is limited to about one pH unit above or below the pKa of the acid present in the buffer system. The possible range of a buffer made from benzoic acid is 3.19-5.19.
5. Define buffer capacity.
The amount of acid or base that can be added to a buffer without causing a significant change in pH
6. What are the two types of problems we will work concerning buffers?
1. Simple calculations using Henderson-Hasselbach, solving for the pH of a buffer solution or the amount of acid/base present in the buffer
2. Calculating the final pH or pOH after the addition of strong acid or strong base to a buffer solution (involves the use of a limiting reagent ICE table and Henderson-Hasselbach)
7. 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
8. 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.
H3O+ / OCl- / ® / HOCl / H2OInitial / 0.050 / 0.25 / 0.15 / n/a
Change / - 0.050 / - 0.050 / + 0.050 / n/a
Equilibrium (Final) / 0 / 0.2 / 0.2 / 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 Henderson-Hasselbach to calculate the pH.
pH = pKa + log([base]/[acid])
pH = -log(3.5 x 10-8) + log (0.2/0.2) = 7.46