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.

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

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

0,3467 =  (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 × = 0,0743 L ≡ 74,3 mL of 1,00M KH₂PO₄

And:

0,2574 M × 0,100 L ×  = 0,0257 L ≡ 25,7 mL of 1,00M K₂HPO₄

I hope it helps!

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