You wish to prepare 1.000 l of a 0.0600 m phosphate buffer at ph 7.640 . to do this, you choose to mix the two salt forms involved in the second ionization, nah2po4 and na2hpo4 , in a 1.000 l volumetric flask and add water to the mark. what mass of each salt will you add to the mixture?

We need 1.908 grams of NaH2PO4 and 6.1299 grams of Na2HPO4

Explanation:

Step 1: Given data

We want to prepare a buffer with pH = 7.64, a volume of 1L and a concentration of 0.06M

pKa1 = 2.148

pKa2 = 7.198

pKa3 = 12.375

Step 2: Calculate [H2PO4-]

For the equation H3PO4 ⇔ H+ + H2PO4-  we consider [H+] and [H2PO4-] as X and [H3PO4] = 0.06 - X

pKa1 = 2.148 = [H+][H2PO4 -] / [H3PO4]

2.148 = x² / 0.06 - x

x² + 2.148x - 0.12888 = 0

x = [H2PO4-] = 0.05841

Step 3: Plug into the pH formula

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

7.64 = 7.198 + log[Na2HPO4]/[NaH2PO4]

0.442 = log[Na2HPO4]/[NaH2PO4]

2.77 = [Na2HPO4]/[NaH2PO4]

This means we will need 2.77x the amount of Na2HPO4compared to NaH2PO4

x / y = 2.77

x + y = 0.06

2.77y + y = 0.06

3.77y = 0.06

y = 0.0159M = [NaH2PO4]

x = 0.0441M = [Na2HPO4]

Step 4: Calculate mass

0.0159 moles NaH2PO4 * 120g/mole = 1.908 g

0.0441 moles Na2HPO4 x 139g/mole = 6.1299 g

We need 1.908 grams of NaH2PO4 and 6.1299 grams of Na2HPO4

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