solution:
[tex]from henderson hasselbalch equation\\
pH=pka+log[\frac{conjugate}{acid}]\\
given[\frac{conjugate base}{acid}]\\
=\frac{75}{25}=3\\
PH=2.1+log3\\
pH=2.1+0.5=2.6[/tex]
Answer:
pH is 2.58
Explanation:
The pH of the buffer made from the mixture of phosphoric acid (H₃PO₄) and its conjugate base form (H₂PO₄⁻) follows the Henderson-Hasselbalch equation:
pH = pka + log [H₂PO₄⁻] /[H₃PO₄]
If 75% of the buffer is in the conjugate base form (H₂PO₄⁻), 25% will be as H₃PO₄. Replacing to find pH:
pH = 2.1 + log [75%] /[25%]
pH = 2.58
pH is 2.58
