Question

A buffer is made such that it is 0.04 M in Na2HPo4 and 0.08 in KH2PO4 .what is the pH of the buffer .the second ionization constant of phosphoric acid is 6.2 × 10^-8​

Answer 1

Answer:

Firstly you need to do the calculations via Handerson Hasselbach equation:

pKa's of phosphoric acid are 2.3, 7.21 and 12.35. If required pH is 6, then, 7.21 will be used. This means, monopotassium dihydrogen phosphate and dipotassium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO4--) potassium salts) will be used.

pH = pKa + log ([A-]/[HA])

Now, for A- you put K2HPO4 concentration, and for HA you put KH2PO4 concentration:

6 = 7.21 + log(A/HA)

log(A/HA)=-1.2

A/HA = 0.063

So, now you know the fold difference between these salts, and the Molarity of the solution would be given to you, and you can calculate the required amount:

HA+A=0.05

A/HA=0.063

HA=0.047M

A=0.003M

This means, you need to put 0.047 moles of KH2PO4 and 0.003 moles of K2HPO4 salts into 1 liters of solution.

for molecular weights: KH2PO4= 39+97=136amu; K2HPO4= 39*2+96=174amu

Thus, 0.047*136=6.392g of KH2PO4 and 0.003*174= 0.522g of K2HPO4 should be added.

But, you should also note that, if |pH-pKa| >1 , then buffer capacity of solution decreases. In this case, it is equal to 1.2