Question

To make a phosphate buffer at pH 6.82 starting with one liter of 20 mM phosphoric acid (H₃PO₄; pKs are of 2.15, 6.82, and 12.38), you could add

a) 20 mM NaOH
b) 20 mM HCl
c) 20 mM Na₂HPO₄
d) 20 mM NaH₂PO₄

Answer 1

To prepare a phosphate buffer at pH 6.82, 20 mM Na₂HPO₄ should be added to the phosphoric acid, creating a buffer with equal concentrations of HPO₄²⁻ and H₂PO⁴⁻ ions, in line with the Henderson-Hasselbalch equation. Therefore, correct option is c.

Step-by-step explanation:

To make a phosphate buffer at pH 6.82 starting with one liter of 20 mM phosphoric acid (H₃PO₄), the best choice would be to add 20 mM Na₂HPO₄.

When Na₂HPO₄ (which is a weaker base) comes into contact with phosphoric acid, there will be a reaction that forms a buffer solution consisting of HPO₄²⁻ and H₂PO⁴⁻ ions, which are the conjugate base and acid pair necessary for a buffer solution near their pKa value of 6.82.

Using the Henderson-Hasselbalch equation, we can see that to make an effective buffer at the pKa value, there should be roughly equal amounts of the weak acid and its conjugate base.