Question

Which may be used to prepare a buffer having a ph of 8.8? ka= 7 × 10–3 for h3po4; 8 × 10–8 for h2po4–; 5 × 10–13 for hpo42–?

Answer 1

h3po4 and h2po4– can be used to prepare buffer of pH = 8.8

The ph of buffer can be shown as:

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

[Salt] /[ Acid ] = x

For h3po4 with ka= 7 × 10–3

8.8 = - log (7 × 10^–3) + log x

8.8 = 2.21 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For h2po4- with ka= 8 × 10–8

8.8 = - log (8 × 10^–8) + log x

8.8 = 7.14 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For hpo42– with ka= 5 × 10–13

8.8 = - log (5 × 10–13) + log x

8.8 = 12.31 + log x

Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.