You have at your lab bench the following chemicals: NaH2PO4(s), Na2HPO4(s), Na3PO4(s) and deionized water. You also have standard glassware available. Describe how you would make 1.00 L of buffer with a pH of 7.00 using only solid materials and deionized water so that the concentration of the weak acid in the buffer is 0.150 M.

A buffer is made with the mixture of a weak acid and its conjugate base. We have as first to find the correct mixture based on the pKa's of the acids:

pKa H3PO4- NaH2PO4 = 2.148

pKa NaH2PO4 - Na2HPO4 = 7.198

pKa Na2HPO4 - Na3PO4 = 12.375

A buffer works in a range of pH's of pKa±1. Thus, we choice NaH2PO4 - Na2HPO4.

The first equation that we can write is:

NaH2PO4 = 0.150 moles -Because the concentration of the weak acid is 0.150M, in 1L = 0.150moles

Using H-H equation for this mixture:

pH = pKa + log [Na2HPO4] / [NaH2PO4]

pH is desire pH = 7.00

pKa is 7.198

[Na2HPO4] and [NaH2PO4] could be taken as the moles of each salt.

Replacing:

7.00 = 7.198 + log [Na2HPO4] / [0.150moles]

-0.198 = log [Na2HPO4] / [0.150moles]

0.6339 = [Na2HPO4] / [0.150moles]

0.095 moles = [Na2HPO4]

The mass of each salt that we must weigh is:

NaH2PO4 -Molar mass: 119.98g/mol-

0.150mol * (119.98g/mol) = 17.997g Na2HPO4

Na2HPO4 -Molar mass: 141.96g/mol-

0.095 moles * (141.96g / mol) = 13.486g of NaH2PO4

Thus, to prepare the buffer we have to:

Weigh 17.997g Na2HPO4 and 13.486g of NaH2PO4 dissolving those salts in 1.00L of deionized water