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-?

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.