# Ratio of uncharged to charged side chains [closed]

The $$\mathrm{p}K_\mathrm{a}$$ of the side chain imidazole group of histidine is $$6.0.$$ What is the ratio of uncharged to charged side chains at $$\mathrm{pH}~7?$$

Here's my attempt to the solution:

$$7 = 6 + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

$$1 = \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

$$10[\ce{HA}] = [\ce{A-}]$$

$$\underset{\text{uncharged}}{1} : \underset{\text{charged}}{10}$$

I've used the Henderson-Hasselbalch equation to the solve this, considering imidazole group to be the acid $$\ce{HA}$$. However, the correct answer appears to be 10:1 uncharged to charged ratio. If I use $$\ce{HA^+}$$ to be the acid, the answer turns out correct.

Can anyone tell me what have I done wrong? If I was wrong in assuming $$\ce{HA}$$ please tell me why. Let me know if some clarification is needed.

## closed as off-topic by Mithoron, Mathew Mahindaratne, Jan, Buck Thorn, Todd MinehardtOct 3 at 21:15

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• It's not acidic, but basic - it's pKa of conjugated acid, not neutral base. – Mithoron Oct 2 at 23:57
• What Mith said. You have the $\mathrm pK_\mathrm a$ of histidinium, not histidine. A spoken and written inacurracy that is way too common. – Jan Oct 3 at 8:08

I think you are simply misunderstanding the "uncharged to charged" part of the question. Imidazole is a base, so it gains a proton to become charged when the pH < pKa. That's why you got the right answer when you used $$\ce{HA+}$$ as the acid, as you put it. In this case, unlike with most general chemistry problems of this nature, the acid is charged while the conjugate base is uncharged.
Below pH 6, imidazole exists largely as a protonated imidazolium ($$\ce{HA+}$$) species. Above pH 6, it is primarily in the neutral ($$\ce{A}$$) imidazole form. When imidazole is protonated (i.e., in acid or $$\ce{HA+}$$ form) it is charged. When it is deprotonated (i.e., in base or $$\ce{A}$$) form, it is uncharged.
So the proper expression is $$7 = 6 + \log\frac{[\ce{A}]}{[\ce{HA+}]}$$ which should lead you to the correct answer:
$$1 = \log\frac{[\ce{A}]}{[\ce{HA+}]}$$
$$10 = \frac{[\ce{A}]}{[\ce{HA+}]}$$
So, the ratio of uncharged:charged species is $$\frac{[\ce{A}]}{[\ce{HA+}]}$$, or 10:1.