# What is the percent dissociation of glycine if the solution has a pH = 8.60 and pKa = 9.60?

Nov 22, 2015

It's worth mentioning that Glycine has TWO pKas, not just one. One for the amine portion and one for the carboxylic acid portion. This is called a zwitterion.

The relevant pKa is the $\setminus m a t h b f \left(9.60\right)$, which is of the amine group.

At this pH, glycine is deprotonated on the carboxyl and protonated on the amine group since the pH > pKa1 (~2.2), and pH < pKa2 (9.60). Since the reference pKa is 9.60, we are considering the acid glycine and its conjugate base in which the amine group is deprotonated.

Using the Henderson-Hasselbalch equation, we get:

$p H = p K a + \log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

$- 1.00 = \log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

$0.100 = \frac{\left[{A}^{-}\right]}{\left[H A\right]}$

Since the number of $\text{mol}$s of base created is equal to the number of $\text{mol}$s of acid deprotonated, we can normalize the concentrations to \mathbf(100%) and then rewrite this as:

$0.100 = \frac{x}{1 - x}$

$0.100 - 0.100 x = x$

$0.100 = 1.100 x$

$\frac{0.100}{1.100} = x$

color(blue)(x = 9.09%)

That means there is 90.91% left of Glycine as the not-yet-deprotonated form, and there is 9.09% dissociated.