Question

As a technician in a large pharmaceutical research firm, you need to produce 350. mL of 1.00 M potassium phosphate buffer solution of pH = 7.07. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution?

Answer 1

You need to add 203 mL of 1,00M KH₂PO₄ and 147 mL of 1,00M K₂HPO₄.

Step-by-step explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for 7,07 phosphate buffer is:

7,07 = 7,21 + log₁₀
`([HPO4^(2-)] )/([H2PO4^(-)])`

0,7244 =
`([HPO4^(2-)] )/([H2PO4^(-)])` (1)

As the buffer concentration must be 1,00 M:

1,00 = [H₂PO₄⁻] + [HPO4²] (2)

Replacing (2) in (1):

[H₂PO₄⁻] = 0,5799 M

Thus:

[HPO4²] = 0,4201 M

To obtain these concentrations you need to add:

0,5799 M × 0,350 L ×
`(1L)/(1mol)` = 0,203 L ≡ 203 mL of 1,00M KH₂PO₄

And:

0,4201 M × 0,350 L ×
`(1L)/(1mol)` = 0,203 L ≡ 147 mL of 1,00M K₂HPO₄

I hope it helps!