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.

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.