You are asked to prepare 2 L of 50 mM sodium phosphate pH 7.0. The

pK2 of phosphate is 6.8. What are the amounts of sodium phosphate mono
basic monohydrate (NaH2PO4 • 1H2O, MW=137.99 Da) and anhydrous sodium
phosphate dibasic (Na2HPO4, MW=141.96 Da) that must be weighed out?

See your other question about tris and tris-HCl.

In this problem determine the (Base) and (Acid), then convert to mols you will need, then grams = mols x molar mass = ?
Post your work if you get stuck.

To determine the amounts of NaH2PO4 • 1H2O and Na2HPO4 required to prepare 2 L of a 50 mM solution of sodium phosphate at pH 7.0, you need to follow these steps:

Step 1: Calculate the moles of sodium phosphate required
- The concentration of the solution is given as 50 mM, which means there are 50 millimoles of sodium phosphate per liter.
- To find the total moles of sodium phosphate required, multiply the concentration by the volume in liters: 50 mM * 2 L = 100 millimoles (0.1 moles) of sodium phosphate.

Step 2: Determine the molar ratio of NaH2PO4 • 1H2O to Na2HPO4
- The desired ratio will depend on the pK2 value of phosphate (6.8).
- At pH 7.0, the pH is slightly above the pK2 value, indicating that phosphate is in its dihydrogen phosphate form (HPO4^2-).
- Considering the pK2 and the desired pH, the molar ratio of NaH2PO4 • 1H2O to Na2HPO4 is 1:1.

Step 3: Calculate the masses of NaH2PO4 • 1H2O and Na2HPO4
- The molecular weights of NaH2PO4 • 1H2O and Na2HPO4 are given as 137.99 Da and 141.96 Da, respectively.
- Since the molar ratio is 1:1, the masses of NaH2PO4 • 1H2O and Na2HPO4 will also be equal.

To calculate the mass of each compound, divide the total moles of sodium phosphate required by 2 (as there are two compounds in equal amounts):

Mass of NaH2PO4 • 1H2O = 0.1 moles / 2 = 0.05 moles
Mass of Na2HPO4 = 0.1 moles / 2 = 0.05 moles

Next, convert moles to grams by multiplying the number of moles by the molecular weight:

Mass of NaH2PO4 • 1H2O = 0.05 moles * 137.99 g/mol = 6.90 grams
Mass of Na2HPO4 = 0.05 moles * 141.96 g/mol = 7.10 grams

Therefore, to prepare 2 L of a 50 mM sodium phosphate solution at pH 7.0, you would need to weigh out approximately 6.90 grams of NaH2PO4 • 1H2O and 7.10 grams of Na2HPO4.