The first step in making a buffer solution at pH = 7.4 and β = 0.1 is to choose the appropriate acid- conjugate base combination. The best choice is for the acid to have a pKa value close to the desired pH. Therefore, the acid is sodium dihydrogen phosphate ( NaH2PO4 , pKa2 = 7.21) and the conjugate base is sodium monohydrogen phosphate (Na2HPO4). The following chemical formula describes the dissociation in solution:
H2PO4- + H2O ↔ HPO4-2 + H3O+
The second step is to calculate the ratio [HPO4-2]/[H2PO4-] using the buffer equation as the following:
The third step is to calculate the total buffer concentration, C, using the buffer capacity equation:
The fourth step is to calculate the concentration of each component of the buffer (NaH2PO4 and Na2HPO4) according to the following:
[HPO4-2]/[H2PO4-] = 1.55
[HPO4-2]= 1.55 × [H2PO4-]
C = [HPO4-2] + [H2PO4-] = 0.18 M
1.55 × [H2PO4-] + [H2PO4-] = 0.18
2.55 × [H2PO4-] = 0.18
[H2PO4-] = 0.07 M
[HPO4-2]= 1.55 × 0.07 = 0.11 M
Finally, the mass of each salt is calculated. The molecular weight of NaH2PO4 is 119.98 g/ mole and that of Na2HPO4 is 141.96 g/ mole. The number of moles and mass of each salt is calculated as the following:
moles of NaH2PO4 = 0.07 (mole/ liter) × 1 liter = 0.07 moles.
moles of Na2HPO4 = 0.11 (mole/ liter) × 1 liter = 0.11 moles.
mass of NaH2PO4 = 0.07 (mole) × 119.98 (g/ mole) = 8.40 g.
mass of Na2HPO4 = 0.11 (mole) × 141.96 (g/ mole) = 15.62 g.
Feel free to write comments or questions.
How did you calculate the concentration of acid? [H3O+]
ReplyDeleteIt is calculated from pH = 7.4.
Delete[H3O+] = 10^(-pH) = 10^(-7.4) = 3.98 × 10^(-8).
Note: the symbol ( ^ ) used above stands for (to the power).
In IP 1996 Potassium dihydrogen phosphate and disodium hydrogen phosphate is given. why potassium used?
ReplyDeleteThe counter ion (sodium or potassium) affects the calculations in terms of the molecular weight of the corresponding salt. The dissociation of both salts (sodium salt or potassium salt) is complete. Therefore, it should not make a difference in the calculations as long as the correct molecular weight is used for each salt form.
DeleteWhy does 1.55 become 2.55?
ReplyDelete1.55 × [H2PO4-] + [H2PO4-] = 0.18
Delete[H2PO4-] × (1.55 + 1) = 0.18 Factorization using the common factor [H2PO4-]
2.55 × [H2PO4-] = 0.18
How can we prepare 1000ml of 7.4ph buffer
ReplyDelete1000ml is equal to 1L. The above example shows how to make calculations for a 1L (1000ml) phosphate buffer at pH = 7.4 So you can use the above numbers to prepare the buffer which is at buffer capacity of 0.1. Always measure the pH of the prepared buffer before using it.
Deletethank you sir
ReplyDeletehow can we prepare PBS from this buffer please (i mean the quantity of NaCl to add)
ReplyDeleteIs sodium phosphate monobasic dihydrate a combination of monosodium phosphate
ReplyDeletedisodium phosphate