Buffers Preparation

Did you know that baby lotion contains a solution that acts as a buffer to keep the pH of the cream around 6, thus preventing bacteria from multiplying? Buffers play significant roles in our lives. Some of them are even responsible for maintaining our blood pH at 7.4.

So, if you are interested in learning more about buffers, let's dive into the world of preparing buffers!

• This article is about buffers preparation.
• First, we will talk about how to prepare a buffer and also look at titration.
• Then, we will learn the procedure involving buffer preparation.
• After, we will look at buffer preparation and standardization.
• Lastly, we will learn about the preparation of neutral buffered formalin.

Definition of a Buffer

First, let's define what a buffer solution is.

A buffer solution usually consists of a mixture of a weak acid and its salt, or a weak base and its salt. Acidic buffers are those have a pH that is less than 7, whereas a basic buffer has a pH that surpasses 7.

Buffers work by removing any hydrogen (H⁺) or hydroxide (OH^-) ions that are added to it, preventing the pH from changing! But, how exactly do buffers accomplish this? Let's take a look.

Let's say that you have a CH₃COOH/CH₃COO^- buffer solution. Acetic acid is a weak acid, so this is a type of acidic buffer.

$$CH_{3}COOH_{(aq)}+H_{2}O_{(l)}\rightleftharpoons CH_{3}COO^{-}+H^{+}_{(aq)}$$

If you add a small amount of strong acid to this buffer, the acetate ions (CH₃COO^-) in the buffer will combine with the recently added hydrogen ions (H⁺) and create more acetic acid. When this happens, the buffer solution removes most of the new hydrogen ions, preventing the pH from undergoing significant changes.

Now, you add a small amount of a strong base to this buffer, the buffer will remove the majority of the added OH^- ions by reacting with H⁺ to form water!

Basic buffers work a similar way. A common example of a basic buffer is a NH₃/NH₄⁺ buffer solution.

$$NH_{(3\ aq)}+H_{2}O_{(l)}\rightleftharpoons NH_{(4\ aq)}^{+}+OH^{-}_{(aq)}$$

pKa is the negative log of the acid dissociation constant, known as K_a. This acid dissociation constant is used by chemists to determine the strength of an acid, as it measures its ability to dissociate in water. The higher the K_a of an acid, the stronger it will be!

$$Ka=\frac{[products]}{[reactants]}=\frac{[H^{+}]\cdot [A^{-}]}{[HA]}$$

$$pKa=-logKa$$

Buffer Preparation and Titration

Now that we know how buffers work, let's look at buffer preparation. There are three ways to prepare a buffer solution:

The first way is by mixing an acid with its conjugate base while keeping the concentrations of both components equal. For example, a buffer containing 0.10 M HClO and 0.10 M NaClO (aq).

The second way involves mixing a weak base with a strong acid. This buffer should have a higher amount of the weak species compared to the strong species. For example, a buffer made up of 1.0 M HCl mixed with 1.25 M NH₃.

Another way to prepare a buffer solution is by mixing a weak acid with a strong base. Similarly, this buffer should also have a higher amount of the weak species compared to the strong species. For example, a buffer containing 1.50 moles of the weak oxyacid HNO₂ and 1.25 moles of the strong acid NaOH.

When deciding on a buffer to use, the best choice is to prepare a buffer that has the pKa closest to the desired pH. Let's put this into practice.

Buffer solutions are also seen in titrations. During a titration, a solution of known concentration is added to the solution of unknown concentration until the endpoint of the titration is reached. You can learn more about titrations by reading "Types of Acid-Base Titrations".

As an example, let's look at the titration of 50.0 mL of a 0.1 M acetic acid, CH₃COOH (aq) with 0.1 M NaOH (aq). Notice that this is an example of a titration of a weak acid with a strong base.

1. At the start of the titration, the solution only contains CH₃COOH (aq) and has a pH of 2.89.
2. Between the initial pH and the equivalence point, NaOH is added dropwise to the solution, and the added hydroxide ions (OH^-) convert, CH₃COOH (aq) into CH₃COO^- (aq), forming a buffer solution!
3. At the equivalence point, the acid gets completely neutralized by the added base (moles of acid = moles of base), resulting in the formation of the salt solution CH₃COONa (aq).
4. After the equivalence point, there is no more acid left to react with excess OH^-.

Buffer Preparation Procedures

Now, let's look at the procedure to prepare a buffer solution of known pH.

Step 1. The first step in the preparation of a buffer solution is to choose the conjugate acid-base pair that best works for your needs. For example, if you needed to prepare a buffer solution of pH 6.3, you would choose an acid with a pKa that is closest to that of the desired pH. So, we do this by calculating K_a and then finding an acid with a K_a value close to the calculated K_a, and a pK_a value close to 6.3.

$$pKa=-logKa$$

$$Ka=10^{-pKa}$$

$$Ka=10^{-6.3}=5.01\cdot 10^{-7}$$

Carbonic acid (H₂CO₃) has a K_a value of 4.4 x 10^-7 and a pK_a1 of 6.37, so it would be a great choice of weak acid for our buffer. To supply the hydrogen carbonate ions, we can choose a salt such as sodium hydrogen carbonate (NaHCO₃).

Step 2. After choosing the components for our buffer solution (H₂CO₃/NaHCO₃), we need to calculate the ratio of buffer components by rearranging the Henderson-Hasselbalch equation.

$$pH=pKa+log\frac{[HCO_{3^{-}}]}{H_{2}CO_{3}}$$

$$6.30=6.37+log\frac{[HCO_{3^{-}}]}{H_{2}CO_{3}}$$

$$6.30-6.37=log\frac{[HCO_{3^{-}}]}{H_{2}CO_{3}}$$

$$10^{-0.07}=log\frac{[HCO_{3^{-}}]}{H_{2}CO_{3}}$$

$$\frac{[HCO_{3^{-}}]}{H_{2}CO_{3}}=\frac{0.85}{1}$$

So, for every 1 mole of H₂CO₃ in 1 L of solution, we will need 0.85 moles of NaHCO₃.

Step 3. Now that have the ratio of components, we can use it to calculate the buffer solution concentration. Let's say that in your laboratory you have a 0.5 M H₂CO₃ in stock. We can use it to calculate the amount of NaHCO₃ required to mix with the weak acid to create a buffer with a pH of 6.3.

First, calculate the moles of carbonic acid.

$$Mol\ H_{2}CO_{3}=1.0L\cdot \frac{0.5mol\ H_{2}CO_{3}}{1.0L}=0.5\ moles\ H_{2}CO_{3}$$

Then, use the buffer ratio from step 2 to find moles of NaHCO₃ from the moles of H₂CO₃.

$$Mol\ NaHCO_{3}=0.5\ moles\ H_{2}CO_{3}\cdot \frac{0.85\ mol\ NaHCO_{3}}{1.0\ mol\ H_{2}CO_{3}}=0.43NaHCO_{3}$$

After, convert moles of NaHCO₃ to mass (in grams).

$$g\ NaHCO_{3}=0.43\ mol\ NaHCO_{3}\cdot \frac{84.006g\ NaHCO_{3}}{1.0mol\ NaHCO_{3}}=36.12g\ NaHCO_{3}$$

Now, we know exactly how much NaHCO₃ we need to add to our H₂CO₃ to make a 1-L buffer solution!

Step 4. The final step in the buffer preparation procedure is to mix the components! First, we have to weigh 36.12g NaHCO₃ and dissolve it in enough 0.5 M H₂CO₃. After it dissolves, we have to add more 0.5 M H₂CO₃ solution until we get a volume of 1.0 L.

To make sure that the pH is correct, we can use a pH meter. If the pH is slightly higher, add a couple of drops of a strong acid until the desired pH is reached. If the pH is a bit lower than expected, add a few drops of a strong base until the desired pH is reached.

Congratulations! You just made an H₂CO₃/HCO₃^- buffer solution with a pH of 6.3!

Neutral Buffered Formalin Preparation

Have you heard of formalin before? Formalin is a solution that is used as a fixative for the preservation of biological specimens. At low concentrations, formalin is bacteriostatic, meaning that it inhibits the growth of microorganisms, but does not kill them.

One way of preparing a 10% Neutral Buffered Formalin from stock solutions is by mixing 100 .0mL of a 37% formaldehyde stock solution, 900.0 mL of distilled water, 4.0 g of NaH₂PO_{4 ,} and 6.5 g of Na₂HPO₄.

Now, I hope that you feel more confident when it comes to preparing buffers!