Question

Prepare 1.0L of 0.1M phosphate buffer, pH 7.21. Assume the availability of H₃PO₄ (FW=98), KH₂PO₄ (FW=136), K₂HPO₄ (FW=174), K₃PO₄ (FW=212), and 1.0M KOH. pK_a1=2.12; pK_a2=7.21; pK_a3=12.66.

Answer 1

Since, we need pH value = 7.21, which is exactly equal to the pK2 = 7.21. We can use KH₂PO₄ and K₂HPO₄ to prepare the buffer solution.

the Henderson–Hasselbalch equation

pH = pKa + log [Acid]/[Base]

Since, pH = pKa,

log [Acid]/[Base] = 0

Therefore [Acid] = [Base]

Here, we have [Acid] = KH₂PO₄ and [Base] = K₂HPO₄

Finally we need to prepare a solution of 0.1 M phosphate buffer, that means =[ KH₂PO₄   + K₂HPO₄ ] = 0.1 M

We know [ KH₂PO₄] = [ K₂HPO₄]

Therefore, [ KH₂PO₄] = [ K₂HPO₄] = 0.05 M

Now, we have to find out how many grams of each KH₂PO₄ and K₂HPO₄ is needed to prepare 1L solution.

Grams of KH₂PO₄ required = Molarity x mol. Mass x volume in L

                                           = (0.05 mole/L)( 136 g/mol)(1L)

                                          = 6.8 g

Grams of K₂HPO₄ required = Molarity x mol. Mass x volume in L

                                           = (0.05 mole/L)( 174 g/mol)(1L)

                                          = 8.7 g

Therefore, we need to take 6.8 g of KH₂PO₄ and 8.7 g K₂HPO₄ and add water to make 1 L solution. The resulting solution will give you the pH 7.21.