Question

On the basis of the information above, a buffer with a pH = 9 can best be made by using

A pure NaH2PO4

B H3PO4 + H2PO4–

C H2PO4– + PO42–

D H2PO4– + HPO42–

E HPO42– + PO43

Answer 1

D H2PO4– + HPO42–

Step-by-step explanation:

The acid dissociation constant for
`\mathbf{H_3PO_4 , H_2PO^(-)_4 , HPO_4^(2-)}` are
`\mathbf{7* 10^(-3), \ \ 8* 10^(-8) ,\ \ 5* 10^(-13)}` respectively.

`\mathbf{pka (H_3PO_4) = -log (7* 10^(-3) )=2.2}`

`\mathbf{pka (H_2PO_4^-) = -log (8* 10^(-8) )=7.1}`

`\mathbf{pka (HPO_4^(2-)) = -log (5* 10^(-13) )=12.3}`

The reason while option D is the best answer is that, the value of pKa for both

`\mathbf{H_2PO^(-)_4 ,\ \& \ HPO_4^(2-)}` lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

`\mathbf{pH = pKa + log \Big( (HPO_4^(2-))/(H_2PO_4^-) \Big)}`