Question

Soft drinks contain phosphate buffers. calculate the ph of 8 oz drink containing 4.4g of sodium dihydrogen phosphate. (120g/mol) and 5.4g of sodium hydrogen phosphate (142g/mol)

Answer 1

H₂PO₄⁻ = 4.4 g x 1 mol / 120 g = 0.037 mol
0.037 mol / 0.24 L = 0.15

HPO₄⁻² = 5.4 g x 1 mol / 142 g = 0.038 mol
0.038 mol / 0.24 L = 0.16

pH = pKa + log $\frac{[HPO_4^{-2} ]}{[H_2PO_4^{-}]}$
Since pKa = 7.2

pH = 7.2 + log (0.16/0.15)
      = 7.2 + 0.028 = 7.23

Answer 2

Answer:

7.21

Explanation:

A buffer is a solution in which a weak acid is in equilibrium with its conjugate base (or a weak base with its conjugate acid), and because of the equilibrium, the pH remains constant.

The sodium hydrogen phosphate dissociates and forms the same number of moles of sodium and hydrogen phosphate because it's a 1:1:1 stoichiometric reaction. Thus, the buffer presented is:

H₂PO₄⁻ ⇄ H⁺ + HPO₄⁻²

The equilibrium is characterized by a constant (Ka), and the value of pKa of it is 7.2. The number of moles of each substance is:

H₂PO₄⁻ = 4.4/120 = 0.0367

HPO₄⁻² = 5.4/142 = 0.0380

According to the Handerson-Halsebach equation, the pH of the buffer is:

pH = pKa + log[HPO₄⁻²]/[H₂PO₄⁻]

Because the volume is the same (8 oz), the concentration can be replaced by the number of moles:

pH = 7.2 + log(0.0380/0.0367)

pH = 7.21