Question

A buffer solution with a pH of 12.08 consists of Na₃PO₄ and Na₂HPO₄. The volume of solution is 200.0 mL. The concentration of Na₃PO₄ is 0.470 M.

What mass of Na₃PO₄ is required to change the pH to 12.33?

K_a = 3.6 × 10^-13

*************The answer is NOT 27.3g

Answer 1

Let us first calculate the initial concentration of Na2HPO4 in buffer,

pH = pKa + log(base/acid)

[base] = Na3PO4

[acid] = Na2HPO4

Ka = -log(Ka) = 12.44

so,

12.08 = 12.44 + log(0.470/x)

x = [Na2HPO4] = 1.077 M

When the pH was changed to 12.33 by addition of Na3PO4,

[base] concentration increases, [acid] concentration reduces

let x mmols of Na3PO4 has been added

new [Na2HPO4] = (1.077 M x 200 ml - x)

new [Na3PO4] = (0.470 M x 200 ml + x)

12.33 = 12.44 + log[(0.470 M x 200 ml + x)/(1.077 x 200 ml - x)]

167.204 - 0.776x = 94 + x

x = (167.204 - 94)/(1 + 0.776)

   = 41.22 mmol (Na3PO4)

Thus,

mass of Na3PO4 to be added = 0.04122 mol x 163.94 g/mol

                                                 = 6.76 g