Consider how best to prepare one liter of a buffer solution with pH = 6.55 using

Question

Consider how best to prepare one liter of a buffer solution with pH = 6.55 using one of the weak acid/conjugate base systems shown here.

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 produce1.00 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 =

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

Explanation / Answer

H2PO4- + HPO4-2 mixture can act as best buffer . because its pKa value nearer to its pH value.

moles of buffer :

NaH2PO4 + Na2HPO4 = 1 x 1 = 1

acid + base = 1 ------------------>1

pH = pKa + log [salt /acid]

6.55 = 7.21 + log (salt /acid)

0.210 = salt / acid -------------->2

salt + 0.210 salt = 1

salt = 0.820 moles

acid moles = 0.180

acid NaH2PO4 mass = 0.180 x 120 = 21.6 g

salt Na2HPO4 mass = 0.820 x 142 = 116.4 g

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