In: Chemistry
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?
Ka = 3.6 × 10-13
Hint: The answer is NOT 18.7 I already tried that.
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