Making buffer solutions
Definitions & Key takeaways
A buffer is a solution that resists changes in pH, and it generally consists of a weak acid and its conjugate base. There are two ways of making a buffer solution. The first way is to start with an aqueous solution of a weak acid and then add a soluble salt that contains the conjugate base. The second way to make a buffer solution is to start with an aqueous solution of a weak acid and then add a strong base to neutralize some of the weak acid. When some of the weak acid is neutralized, it turns into the conjugate base.
Buffer solutions contain relatively high concentrations of a weak acid and its conjugate base. When small amounts of acid or base are added to the solution.
Buffers react with the added acid or base and resist large changes in Ph. Many biochemical reactions require specific buffers to work properly.
Therefore, making buffer solutions is an important laboratory skill in biochemistry. OK.
So for a brief review of how buffers work, let's look at a generic buffer that consists of a weak acid H A and its conjugate base A minus.
If an acid H plus is added to the buffer solution, the conjugate base A minus will react with it. We can write the balanced equation for this acid base neutralization reaction as A minus plus H plus gives H A.
If a base oh minus is added to the buffer solution, the weak acid H A will react with it. We can write the balanced equation for this acid base neutralization reaction as oh minus plus H A gives H2O plus A minus.
Therefore, a buffer resists changes in Ph because it contains both an acid H A to neutralize added oh minus ions and a base A minus to neutralize added H plus ions.
There are two ways to make a buffer solution. The first way starts with an aqueous solution of a weak acid.
H since weak acids have a low percent ionization in aqueous solution, most of the weak acid will stay as H A and only a small percentage of H will turn into a minus.
So to represent an aqueous solution for our weak acid. In our particulate diagram, we start out with 4h molecules and zero A minus anions, water molecules are left out of the particulate diagram.
For clarity. Next, we add a solid salt.
NAA. When NAA dissolves in water, it turns into the NA A plus ion and the A minus anion.
So if we add four particles of NAA to the aqueous solution of H A, they will dissociate and form four N A plus ions and four A minus anions buffers work the best when the concentrations of the weak acid and its conjugate base are equal.
Since we have 4h molecules and four A minus an ions in solution, we have a good buffer solution. There are also four sodium ions present in the solution.
However, sodium ions don't react with water. Since the sodium ions don't react with water, they don't affect the ph.
The second way to make a buffer solution also starts with an aqueous solution of a weak acid H A. However, this time instead of adding a salt will add a strong base.
This time, let's start with eight H A molecules in the particulate diagram. Our next step is to add an aqueous solution of sodium hydroxide.
Sodium hydroxide is a strong base that dissociates into sodium ions. N A plus and hydroxide ions O minus when dissolved in water oh minus reacts with H A in an acid base neutralization reaction to form H2O and A minus.
So in the particulate diagram, if we pour in a solution that contains four particles of oh minus, the 40 minus ions will react with the 4h A molecules to form four A minus ions.
So after the acid base neutralization reaction is complete, we're left with 4h molecules and four A minus anions. Since we have equal amounts of H and A minus, we have a good buffer solution.
Since we added four particles of oh minus, we would also have four sodium ions in the solution. Notice how the buffer solution that we made the second way is identical to the buffer solution that we made.
The first way they both have 4h A molecules, four A minus anions and four N A plus ions. Now that we know how to make buffers, let's look at why buffers are the most effective over a certain PH range.
Since buffers work the best when the concentration of H A and A minus are equal. Let's find the optimal Ph for a buffer to find the PH of a buffer solution.
We use the Henderson Hasselbach equation which says that the PH is equal to the PK A of the weak acid plus the log of the concentration of A minus divided by the concentration of H A.
OK. So that sounds pretty complex, but we can break it down since the concentrations of A minus and H A are equal, the ratio of A minus to H A is 1 to 1.
Since the log of one is equal to zero, the PH of the buffer solution is equal to the PK A of the weak acid. If we add enough acid or base, eventually, a buffer stops being effective.
This is often called breaking a buffer. So let's start with a buffer solution that has equal concentrations of H and A minus.
And if we add H plus to the buffer solution, the A minus reacts with the H plus to form H. Therefore, adding acid causes the concentration of A minus to decrease and the concentration of H to increase when the ratio of A minus to H A becomes less than 1 to 10 buffer solutions break and lose their buffering action.
If we plug in a ratio of 1 to 10 in the Henderson Hasselbach equation, the log of one divided by 10 is negative one. So the PH of the solution is equal to the PK A minus one.
Let's return to a buffer that has equal concentrations of H A and A minus this time. If we add oh minus to the buffer solution, the H A reacts with the oh minus to form a minus.
Therefore, adding a base causes the concentration of H A to decrease. And the concentration of A minus to increase when the ratio of A minus to H A becomes greater than 10 to 1 buffer solutions break and lose their buffering action.
So if we plug in a ratio of 10 to 1 in the Henderson Hasselbach equation, the log of 10 divided by one is one. So the PH of the solution is equal to the PK A plus one.
These calculations show us that the effective PH range of A buffer is equal to plus or minus one of the PK A value. For example, we can consider a buffer system consisting of acetic acid and its conjugate base.
The acetate anion, the PK A value for acetic acid is 4.75. Therefore, the range of effectiveness of the acetic acid acetate buffer solution is plus or minus one from this PK A value corresponding to a PH range from 3.75 to 5.75.
All right now that we understand that it's possible to break a buffer. Let's look at the buffering capacity of two buffers using particulate diagrams.
The buffer on the left has 4h molecules and four A minus ions. The buffer on the right has six H molecules and six A minus ions.
Since both buffer solutions have equal concentrations of H A and A minus. Both buffer solutions have a PH equal to the PK A.
Even though their PH S are the same, their buffering capacities are different. Since the buffer on the left has a smaller number of H A and A minus.
If we added enough acid or base, it would take less acid or base to break this buffer. Since the buffer on the right has a greater number of H A and A minus, it would take more acid or base to break this buffer.
Therefore, the buffer on the right has the greater buffering capacity. Now, let's use what we've learned about buffers to make a specific buffer solution with a PH of nine.
The first step is to choose a good weak acid conjugate base pair. An acetic acid acetate buffer would not work because the PK A of acetic acid is 4.75.
So the effective ph range of the buffer is like we learned plus or minus one from 4.75. So 3.75 to 5.75 a better buffer would consist of the ammonium ion NH four plus and its conjugate base ammonia.
NH three, the ammonium ion has AP K A value of 9.25. So the buffer would have an effective range plus or minus one from 9.25.
So 8.25 to 10.25. Since a ph of nine falls within this range.
The ammonium ion ammonia buffer is a great choice. Now that we've chosen our buffer.
Suppose we just so happen to have some solid ammonium chloride and 1 L of a 0.10 molar aqueous solution of ammonia just sitting around the laboratory.
Since ammonium chloride is a source of ammonium ions, we can dissolve a certain number of moles of ammonium chloride in the ammonia solution to make our buffer.
To figure out just how many moles of the ammonium ion. We need, we start again with the Henderson Hasselbach equation which says that the PH is equal to the PK A plus the log of the concentration of the conjugate base over the concentration of the weak acid for the buffer that we chose the conjugate base is ammonia.
So we plug the concentration of NH three in the numerator. The weak acid is the ammonium ion.
So we plug the concentration of NH four plus into the denominator. We can also plug in the desired ph of nine.
And the PK A of the ammonium ion which is 9.25. When we subtract 9.25 from nine, we get negative 0.25 is equal to the log of the concentration of NH three divided by the concentration of NH four plus.
Since the concentration of ammonia is 0.10 we can plug that in and solve for the concentration of the ammonium ions to get rid of the log we take 10 to the power of each side of the equation.
All right, it's a quick recap. A buffer consists of a weak acid and its conjugate base.
There are two ways of making a buffer solution. The first way is to start with an aqueous solution of a weak acid and then add a soluble salt that contains the conjugate base.
The second way to make a buffer solution is to start with an aqueous solution of a weak acid and then add a strong base to neutralize some of the weak acid.
When some of the weak acid is neutralized, it turns into the conjugate base. The effective PH range of a buffer is equal to plus or minus one of the PK A value.
Therefore, when choosing a buffer, the PK a value of the weak acid must be relatively close to the desired PH of the buffer solution.
Once the identity of the buffer has been chosen, the Henderson Hasselbach equation allows us to calculate the exact concentrations necessary to achieve the desired
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