A buffer solution with a pH of 12.05 consists of Na3PO4 and Na2HPO4. The volume of solution is 200.0 mL. The concentration of Na3PO4 is 0.370 M.
What mass of Na3PO4 is required to change the pH to 12.30?
Ka = 3.6 × 10-13
_____ g
Ka = 3.6 × 10-13
pKa = -log (3.6 × 10-13)
= 12.44
pH = pKa + log [salt / acid]
12.05 = 12.44 + log [Na3PO4 / Na2HPO4]
[Na3PO4 / Na2HPO4] = 0.4074
0.200 x 0.370 / Na2HPO4 = 0.4074
Na2HPO4 = 0.1816
moles of Na2HPO4 = 0.1816
pH = pKa + log [salt / acid]
12.30 = 12.44 + log [Na3PO4 / Na2HPO4]
[Na3PO4 / Na2HPO4] = 0.7244
Na3PO4 / 0.1816 = 0.7244
moles of Na3PO4 = 0.132
mass of Na3PO4 = 0.132 x 163.94 = 21.57 g
mass of Na3PO4 is required = 21.57 g
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