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.