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

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

7,2 = 6,86 + log₁₀

2,18776 =   (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× = 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.

3 1/4 %

explanation:

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