Phosphoric acid is a triprotic acid with the following pKa values: pka1= 2.148 p
ID: 948774 • Letter: P
Question
Phosphoric acid is a triprotic acid with the following pKa values: pka1= 2.148 pka2= 7.198 pka3= 12.375
You wish to prepare 1.000 L of a 0.0300 M phosphate buffer at pH 7.680. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?
What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer?
(Hint: Use the Henderson-Hasselbalch equation to get the molar ratio of Na2HPO4 to NaH2PO4 required, then the fraction of each form from the ratio. The total moles needed will be 1.000 L × 0.0300 M = 0.0300 moles. Use the formula mass to calculate the mass needed. (FM NaH2PO4 = 119.98; FM Na2HPO4 = 141.96))
Explanation / Answer
We know that,
pH for this buffer = pKa + log ([Na2HPO4] / [NaH2PO4])
Given,
pKa for H2PO4- = pKa2 = 7.198
and [Na2HPO4] + [NaH2PO4] = 0.03 M ------------- (1)
Applying the equation we get,
7.68 = 7.198 + log ([Na2HPO4] / [NaH2PO4])
=> log ([Na2HPO4] / [NaH2PO4]) = 0.482
=> ([Na2HPO4] / [NaH2PO4])) = 3.034
=> [Na2HPO4] = 3.034 x [NaH2PO4]
From (1)
3.034 x [NaH2PO4] + [NaH2PO4] = 0.03 M
=> [NaH2PO4] = 0.00744 M
and [Na2HPO4] = 0.0226 M
Volume of solution = 1 L
We know that,
Moles = Molarity x Volume (L)
=> Moles of NaH2PO4 = 0.00744 x 1 = 0.00744 moles
=> Mass of NaH2PO4 = 0.00744 x 119.98 = 0.893 g
Moles of Na2HPO4 = 0.0226 x 1 = 0.0226 moles
=> Mass of Na2HPO4 = 0.0226 x 141.96 = 3.21 g