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.

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

Explanation:

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

the answer is a and have a great day

im not exactly sure but i think its b!