Question

How many grams of Nah2PO4 would have to be added to 200 mL of 2.5 M Na3PO4 to produce a ph of 7.25 buffer? Assume no significacnt change in volume.

Phosphoric acid pka1=2.148, pka2=7.1499, pka3=12.15

Answer 1

Na3PO4 can act as a base and NaH2PO4 as an acid :

Na3PO4 + NaH2PO4 ----> 2 Na2HPO4

weak acid and its salt forms a buffer system

pH = pKa + log [salt] / [acid]

salt = Na2HPO4

Acid = NaH2PO4

since pH is nearer to pKa2 value ; we have to consider that pKa value

7.25 = 7.1499 + log [salt] / [acid]

[salt] / [acid] = 1.259

MOles of Na3pO4 = molarity*V in L = 200*2.5 / 1000 = 0.5

Na3PO4 + NaH2PO4 ----> 2 Na2HPO4

0.5 x 0

x-0.5 2*0.5

[salt] / [acid] = 1.259

2*0.5 / x-0.5 = 1.259

x = 1.2948

MOles of NaH2PO4 = 1.2948

Molar mass of NaH2PO4 = 120 gms/mol

Mass = moles*Molar mass = 1.2948*120 = 155.313 gms of NaH2PO4 is needed