Question

Phosphoric acid has a pka of 2.1. at what ph will 75% of phosphoric acid be in the conjugate base form?

Answer 1

`from henderson hasselbalch equation\\</p><p>pH=pka+log[(conjugate)/(acid)]\\</p><p>given[(conjugate base)/(acid)]\\</p><p>=(75)/(25)=3\\</p><p>PH=2.1+log3\\</p><p>pH=2.1+0.5=2.6`

Answer 2

pH is 2.58

Step-by-step 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