Question

Consider how best to prepare one liter of a buffer solution with pH = 8.09 using one of the weak acid/conjugate base systems shown here.

Weak Acid	Conjugate Base	Ka	pKa
HC2O4-	C2O42-	6.4 x 10-5	4.19
H2PO4-	HPO42-	6.2 x 10-8	7.21
HCO3-	CO32-	4.8 x 10-11	10.32

How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams sodium salt of weak acid =

grams sodium salt of conjugate base =

Answer 1

Q1.

nearest pKa value to the pH is that of H2PO4- and HPO4-2

H2PO4- HPO42- with KA = 6.2 x 10-8 or pKa = 7.21

so the buffer equation

pH = pKa + log(HPO4-2 / H2PO4-)

Q2.

V = 1 L, M = 1 weak base

mmol of HPO4-2 = MV = 1 mol

pH = pKa + log(HPO4-2 / H2PO4-)

8.09= 7.21+ log(1 / H2PO4-)

1/ mol of H2PO4- = 10^(8.09-7.21) = 7.585

mol of H2PO4- = 1/7.585 = 0.13183 mol

now

we can get this from:

NaH2PO4 and N2HPO4

mol of NaH2PO4 = 0.13183

mass = mol*MW = 0.13183 *119.98 = 15.816 g

mol of Na2HPO4 = 1

mass = mol*MW = 1*141.96 = 141.96 g