Question

A buffer solution with a pH of 12.19 consists of Na₃PO₄ and Na₂HPO₄. The volume of solution is 200.0 mL. The concentration of Na₃PO₄ is 0.320 M.

What mass of Na₃PO₄ is required to change the pH to 12.44?

K_a = 3.6 × 10^-13

^{Hint: The answer is NOT 18.7 I already tried that.}

Answer 1

Ka = 3.6 x 10^-13

pKa = -log Ka = - log (3.6 x 10^-13) = 12.44

Now, using henderson-hesselbalach equation

pH = pKa + log {[salt] / [acid]}

12.19 = 12.44 + log {[Na3PO4] / [Na2HPO4]}

- 0.25 = log {[0.320] / [Na2HPO4]}

0.320 / [Na2HPO4] = 10^-0.25 = 0.56

[Na2HPO4] = 0.320 / 0.560 = 0.57 M

So, moles of Na2HPO4 = 0.57 M x 0.2 L =0.114 moles

Molar mass of Na2HPO4 = 142 g/mol

1 mole of Na2HPO4 = 142 g

0.114 moles of Na2HPO4 = 0.114 x 142 g = 16.19 g