Question

You need to prepare 0.2 liters of a 0.1 M solution of a phosphate buffer with a pH of 7.2. If you have solid K2HPO4 (FW=136.09) and 6M HCl or 6 M NaOH describe how you would prepare this buffer. Note: pKa values for phosphoric acid are 2.15, 6.86, and 12.35

Answer 1

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 $\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}$

7,2 = 6,86 + log₁₀ $\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}$

2,18776 =  $\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}$ (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× $\frac{136,09 g}{1 mole}$ = 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.