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 involv

Question

4 years ago

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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 involv

ed 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?

Chemistry

1 answer:

umka21 [38] · Answer 1 · 2022-01-25T13:42:52+03:00

Answer:

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