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))

Answer 1

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