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!

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.