Question

Provide a detailed description of how you would prepare one liter of a pH 6.50 buffered solution with a total phosphate concentration of 0.50 M ( [Phosphate]T = formal concentration = 0.50 M) using potassium phosphate monobasic anhydrous and sodium phosphate dibasic heptahydrate.

Answer 1

Here we need to prepare 1L of pH 6.5 buffered solution of phosphate concentration 0.5 M using potassium phosphate KH₂PO₄ and sodium phosphate Na₂HPO₄

using henderson Hasselbalch equation pH = pKa + log (A/HA) ----(1)

So pKa of phosphoric acid are 2.3, 7.21 and 12.67.

If pH required is 6.5 then we will use pKa as 7.21.

Now [A] will be Na₂HPO₄ concentration & [HA] will be KH₂PO₄ concentration _.

Therefore equation (!) becomes 6.5= 7.21 + log [A] / [HA]

log [A] / [HA]= - 0.71

[A] / [HA]= 0.1949

Given molarity of the solution is 0.5

To find out required quantity:

HA+ A=0.5

A/HA = 0.1949

By using the above , We get HA=0.4194

A=0.0806

We need to put 0.4194moles of KH₂PO₄ and 0.0806 moles of  Na₂HPO₄   salts in to 1 litre of the solution.

For molecular weight of KH₂PO₄ =136g/mol &

molecular weight of Na₂HPO₄ =268.07g/mol.

Thus 0.4194 x 136 = 57.03 g of KH₂PO₄ and 0.0806 x 268.07 = 21.60g of Na₂HPO_{4 .}

To prepare 1 L of pH 6.5 buffer solution 57.03g of KH₂PO₄ and 21.60g of Na₂HPO₄ should be added.