The combination used in the preparation of a buffer with a pH around 8.8 accounts for ka = 7 * 10 -3 for h 3po4
The ph of the 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.
Hence, the correct answer is option A
Learn more about buffering systems here,
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