# How to prepare 100 ml of 50 mM Phosphate buffer of pH 6.8?

(Last Updated On: October 23, 2017)

Preparation of the 50 mM phosphate buffer

To prepare a 50 mM phosphate buffer of pH 6.8, you may use either sodium salt or potassium salt of the orthophosphoric acid. There are two components of the phosphate buffer; acid and salt (conjugate base). For sodium phosphate buffer, acid is sodium dihydrogen phosphate and salt is disodium hydrogen phosphate. In the same way, potassium dihydrogen phosphate is an acid and dipotassium hydrogen phosphate is a salt for potassium phosphate buffer. Read more: What is a buffer system and its importance?

For the preparation of phosphate buffer, you need to consider three factors. These factors are; concentration, pH, and final volume of the buffer. In this case, the pH, concentration, and the volume are 6.8, 50 mM and 100 ml.

Henderson-Hasselbalch equation; $\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;pH=pKa+\log_{10}\tfrac{[Salt]}{[Acid]}$ is most often used in the lab for the prepare different types of buffers such as phosphate buffer. You may use this formula to calculate the concentration of the individual components of the phosphate buffer or any buffer. However, you need to know the pKa of the buffer that you are going to prepare.

## Molar concentration of individual components

pKa value of a buffer system is the negative logarithm of the acid dissociation constant the key compound of the buffer. For the phosphate buffer, the pKa value that we use is the negative logarithm of the acid dissociation constant of the second ionization of the phosphoric acid.  The pKa value (specifically pKa2) of the phosphate buffer is 7.21. Now, using pH and pKa values you can calculate molar concentration of individual components of the phosphate buffer. Read more: How to calculate pKa of phosphate buffer?

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;pH=pKa+&space;\log_{10}\tfrac{\left&space;[&space;Salt&space;\right&space;]}{\left&space;[&space;Acid&space;\right&space;]}$

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;6.8=7.21+\log_{10}\tfrac{\left&space;[&space;Salt&space;\right&space;]}{\left&space;[&space;Acid&space;\right&space;]}$

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;\log_{10}\tfrac{\left&space;[&space;Salt&space;\right&space;]}{\left&space;[&space;Acid&space;\right&space;]}=-0.41$

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;\tfrac{\left&space;[&space;Salt&space;\right&space;]}{\left&space;[&space;Acid&space;\right&space;]}=0.389$

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;\left&space;[Salt&space;\right&space;]=0.389\left&space;[Acid&space;\right&space;]\cdots&space;equation-1$

Since, $\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;\left&space;[&space;Salt&space;\right&space;]+\left&space;[Acid&space;\right&space;]=50&space;mM\cdots&space;equation-2$

Now, solving equation-1 and equation-2, you will get molar concentration of individual components; [Salt]= 14.004 mM and [Acid]=35.996 mM. After that, it’s easy for you to calculate the required amount of both components using the molarity formula.

## Weights of the individual components

Here, I have illustrated the calculation of required weight for the NaH2PO4. Now, we have the following information regarding the NaH2PO4; molarity: 35.996 mM (0.035996 M), molecular weight: 137.993 g/mol, total volume: 100 ml (0.1 L). Using the molarity formula, the required weight of the NaH2PO4 can be calculated which is 0.4967g.

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;Molarity&space;(M)=\tfrac{Given&space;Weight}{Molecular&space;Weight}\times&space;\frac{1000}{Volume(ml)}$

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;0.035996=\frac{Given&space;Weight}{137.992}\times&space;\frac{1000}{100}$

$\inline&space;\dpi{80}&space;\dpi{120}&space;\large&space;GivenWeight=\tfrac{0.035996\times&space;137.992}{10}=0.4967g$

In the same way, you can also calculate the required amount of Na2HPO4 which is 0.1988 g. Now, put these components in a 100 ml volumetric flask and add 50 ml of distilled water. Dissolve the components completely and make the final volume to 100 ml. In this way, you can prepare 100 ml of 50 mM phosphate buffer of pH 6.8. This is a manual calculation for the preparation of phosphate buffer. However, if you want to save your time in calculating these things, you can use our online phosphate buffer calculator. I have recently developed an online calculator especially for the preparation of phosphate buffer. You will find it more useful than any other online calculators available for the preparation of phosphate buffer.