Buffer Calculations Simplified Into 3 Easy Methods!

For this topic, I m assuming you already understand the Acids Bases topic. If not, do not fret, I will be releasing an article covering it soon! What are Buffers?

Buffer solutions are made from a weak acid mixed with a solid salt or a strong base. These reactants form an equilibrium, which responds to the addition of an acid or base by counteracting the change in [H+] (remember Le Chatelier s Principle?). Simply put, buffer solutions can maintain a constant pH, and are very useful in lab experiments for replicating biological conditions. Types of Buffer Questions To break down the questions into manageable chunks, I m going to group them into three categories: Types 1, 2 and 3. The reactants and given data may change (asking for different answers), but the general method is still the same. Type 1 and Type 2 are questions about the pH of an already made buffer. Buffer solutions can be made in two main ways, for the two types of questions respectively:Weak acid mixed with its complementary, solid salt e.g. ethanoic acid and sodium ethanoate.Excess weak acid mixed with a strong, liquid basee.g. ethanoic acid and sodium hydroxide.Type 3 questions concern the addition of a strong acid or base to an already made buffer resulting in a new pH
Type 1: Weak acid mixed with its complementary, solid salt
For a buffer solution, the equilibrium expression looks like this:
HA_(aq) & H⁺_(aq) + A^-_(aq)
( & is the symbol I will be using for equilibriums. In fact, it is 2 half arrows painting in opposite directions, so don t use my symbol in your actual work! Also, sometimes I or X are used instead of A, but they are all just used to represent the negative ion.)
In a buffer made from a weak acid and its complementary salt, [HA] is the concentration of the acid used to make the buffer, and [A^-] is equal to the concentration of the salt. So, Type 1 questions aren t too bad. Once you ve recognised that you re dealing with this type of question, this is the method were going to want to use:
1. Find K_a, [HA] and [A^-], using the information given in the question (remember to convert number of moles into concentration.)
2. By rearranging (having to make do with typed out formula)
K_a = ([H⁺][A^-])/[HA]
Into
[H⁺] = (K_a[HA])/[A^-]
We can substitute in our numbers to find [H⁺]

3. Finally, we can calculate the pH using
pH = -log[H⁺]

Type 2: Excess weak acid mixed with a strong, liquid base
Normally for this type of question, the weak acid will be mixed with a metal hydroxide (remember that group 2 hydroxides will be dibasic, so you will need to multiply the concentration of them by 2). The equation for the reaction of the weak acid with a group 1 hydroxide base is as follows. M represents the group 1 metal.
HA_(aq) + MOH & MA_(aq) + H₂O
The unreacted weak acid (HA) provides the HA in our buffer equation, and the [A^-] ions are provided by the (MA), or the reacted (A^-) from the weak acid. As the molar ratio is 1:1, this concentration will also be equal to the concentration of reacted OH^- ions from the strong base. Once we have calculated these amounts, our buffer equilibrium is the same.
HA_(aq) & H⁺_(aq) + A^-_(aq)

Therefore, to calculate the pH of this solution:
1. From the data, find initial moles of HA and OH^- (don t forget to multiply OH^- by 2 if dibasic)

2. Using the molar equation, calculate the amount of HA that reacts and subtract this from the initial to find the unreacted HA.

3. Using the total volume, calculate [HA] and [OH^-] in the buffer solution
[OH^-]will be equal to the produced [A^-] in the buffer equilibrium

4. Using the unreacted [HA] and the produced [A^-], we can now use our rearranged expression for K_a to find [H⁺]:
[H⁺] = (K_a[HA])/[A^-]
&And therefore, the pH:
pH = -log[H⁺]

Type 3: Adding a strong acid or base to an already made buffer solution
For this type of question, you will have been given values for the moles of the acid or base in the buffer initially, and the amount of strong acid or base to be added.
When a strong acid is added, H⁺ ions are added to the buffer equilibrium. Following Le Chatelier s principle, the equilibrium shifts to the left, so the amount of HA increases, and amount of A^- decreases. As a result, pH slightly increases.

&HA &H⁺ + A^-

When a strong base is added, OH^- ions are added. These ions react with the H⁺ ions in the solution, causing a decrease in the amount of H⁺. In response to this change, the equilibrium shifts to the right, so the amount of HA decreases, and amount of A^- increases. As a result, pH slightly decreases.

&HA & & H⁺ + A^-

If we want to calculate the new pH:
1. Find the initial moles of HA and A^- in the buffer solution

2. Find the moles of H⁺ added or removed (if OH^- is added)

3. Using an ICE table, we can find the moles of HA and A^- at equilibrium, e.g.:
& & & & & & & & & & & & & HA & & & & & H⁺ & & & & & A^-
Initial & & & & & & & 0.108 & & & & & & & & & & & 0.087
Change & & & & & & +x & & & & & +x & & & & & & -x
Equilibrium & & 0.108 & & & +x & & & & & & 0.087-x
NOTE: Initial and Equilibrium amounts of H⁺ are not important for this table, we will calculate the final [H⁺] next.

4. Now that we have found the equilibrium amounts of HA and A^- in moles, we can calculate [HA] and [A^-] by dividing by the total volume of the solution. NOTE: If the added acid or base is solid, volume will be unchanged, but if it is a liquid with a given volume, you will need to add it to the original volume of the buffer solution!

5. Once we have [HA] and [A^-], things start to look familiar. We can now calculate [H⁺] using [HA], [A^-] and K_a (given for the acid in the buffer).

[H⁺] = (K_a[HA])/[A^-]
6. And finally, the pH
pH = -log[H⁺]

There we have it. All buffer questions for A-Level Chemistry simplified into 3 methods. Once you ve learnt the methods, the best thing is to do plenty of questions as practice. I recommend finding question packs on Physics and Maths Tutor. I hope you found this article helpful, and if you have any further questions, feel free to contact me on my Tutor Hunt page.

Until Next Time,
James