Answer:
To prepare 0,2 L of a 0,1M solution of a phosphate buffer to a pH of 7,2 you need to add 1,05 mL of 6M HCl, 2,722 g of K₂HPO₄ and complete 0,2 L with water.
Explanation:
The acid equilibrium of phosphate buffer for a pH of 7,2 is:
H₂PO₄⁻ ⇄ HPO₄²⁻+ H⁺ pka = 6,86
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [tex]\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}[/tex]
7,2 = 6,86 + log₁₀ [tex]\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}[/tex]
2,18776 = [tex]\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}[/tex] (1)
You need to add 0,2L× 0,1M = 0,02 moles of phosphate buffer, that means:
0,02 moles = H₂PO₄⁻ + HPO₄²⁻ (2)
Replacing (2) in (1):
H₂PO₄⁻: 6,2739x10⁻³ moles
Thus,
HPO₄²⁻: 0,013726 moles
K₂HPO₄ reacts with HCl thus:
K₂HPO₄ + HCl → KH₂PO₄ + KCl
Thus, you need to add 0,02 moles of K₂HPO₄ that will react with 6,2739x10⁻³ moles of HCl To produce the necessary moles for the buffer:
0,02 moles of K₂HPO₄× [tex]\frac{136,09 g}{1 mole}[/tex] = 2,722 g
6,2739x10⁻³ moles of HCl÷ 6 M = 1,05x10⁻³ L ≡ 1,05 mL of 6M HCl
Thus, to prepare 0,2 L of a 0,1M solution of a phosphate buffer to a pH of 7,2 you need to add 1,05 mL of 6M HCl, 2,722 g of K₂HPO₄ and complete 0,2 L with water.
