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12.09.2019 •
Chemistry
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?
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Ответ:
You need to add 203 mL of 1,00M KH₂PO₄ and 147 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:
7,07 = 7,21 + log₁₀
0,7244 = (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 × = 0,203 L ≡ 203 mL of 1,00M KH₂PO₄
And:
0,4201 M × 0,350 L × = 0,203 L ≡ 147 mL of 1,00M K₂HPO₄
I hope it helps!
Ответ:
C. Physical Change.