Question

7. Calculate the volume of 1 M solution of HCl that we need to add to 100mL of 0.1 M of phosphate buffer (0.1 M H3PO4 and 0.1 M Na2HPO4) to get pH =4. Is it a buffer and why?

Answer 1

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

I hope the calculations are clear.

Wish you luck.

The reason of adjusting is coming from

1) it is very hard to calibrate the weights of the salts added into the solution. You cannot be 0.00 perfect.

2) the theoretical calculations are for perfect occasions, where nothing is on the road of solution formation. This is not the case in practical one

The reason of choosing those two salts is because of the pKa value, as I have written above. If the needed pH was 12, we would use K3PO4 and K2HPO4, and if it was 2, then they would have been of H3PO4 and KH2PO4