Question

use the Henderson-Hasselbalch equation to determine how to prepare 100 mL of 0.10 M sodium phosphate pH 6.80 and another solution of pH 7.70. The apparent pKa of phosphate is 6.77.

Answer 1

this is 2nd ionization of the phosphate

pKa = 677

so

pH = pKa2 + log(HPO4-2 / H2PO4-)

for pH = 6.80

6.80 = 6.77 + log(HPO4-2 / H2PO4-)

(HPO4-2 / H2PO4-) = 10^(6.80-6.77) = 1.07151

Na2HPO4 = 1.07151* NaH2PO4

we also know total mmol --> MV = 100*0.1 = 10 mmol

Na2HPO4 + NaH2PO4 = 10

then; substitte from Na2HPO4 = 1.07151* NaH2PO4

1.07151* NaH2PO4 + NaH2PO4 = 10

NaH2PO4 = 10/(1.07151+1) = 4.827 mmol

Na2HPO4 = 1.07151*4.827 = 5.172

for th epH = 6.80 buffer;

add:

mass of NaH2PO4 = 4.827 *119.9770 = 579.12 mg

mass of Na2HPO4 = 5.172*141.9588 = 734.210 mg

Now...

for the pH = 7.70 buffer:

7.70 = 6.77 + log(HPO4-2 / H2PO4-)

(HPO4-2 / H2PO4-) = 10^(7.70 -6.77) = 8.51

Na2HPO4 = 8.51* NaH2PO4

we also know total mmol --> MV = 100*0.1 = 10 mmol

Na2HPO4 + NaH2PO4 = 10

then; substitte from Na2HPO4 = 1.07151* NaH2PO4

8.51* NaH2PO4 + NaH2PO4 = 10

NaH2PO4 = 10/(8.51+1) = 1.05 mmol

Na2HPO4 = 8.51*1.05 = 8.9355

for the pH = 7.70 buffer;

add:

mass of NaH2PO4 = 1.05 *119.9770 = 125.97 mg

mass of Na2HPO4 = 8.9355 *141.9588 = 1268.46 mg