Question

As a technician in a large pharmaceutical research firm, you need to produce 100. mL of 1.00 M potassium phosphate buffer solution of pH = 6.75. 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? Express your answer to three significant digits with the appropriate units.

Answer 1

Answer:

You need to add 74,3 mL of 1,00M KH₂PO₄ and 25,7 mL of 1,00M K₂HPO₄.

Explanation:

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

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

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:

6,75 = 7,21 + log₁₀ $\frac{[HPO_4^{2-}]}{[H_2PO_4^-}$

0,3467 =  $\frac{[HPO_4^{2-}]}{[H_2PO_4^-}$ (1)

As the buffer concentration must be 1,00 M:

1,00 = [H₂PO₄⁻] + [HPO4²⁻] (2)

Replacing (2) in (1):

[H₂PO₄⁻] = 0,7426 M

Thus:

[HPO4²] = 0,2574 M

To obtain these concentrations you need to add:

0,7426 M × 0,100 L × $\frac{[1 L]}{[1mol]}$ = 0,0743 L ≡ 74,3 mL of 1,00M KH₂PO₄

And:

0,2574 M × 0,100 L × $\frac{[1 L]}{[1mol]}$  = 0,0257 L ≡ 25,7 mL of 1,00M K₂HPO₄

I hope it helps!