# Question ce9be

Sep 7, 2015

The overall charge on the tetrapeptide is at $\text{pH 7.6}$ is $\text{-1}$.

#### Explanation:

Two principles to remember:

• If the $\text{pH}$ of the environment is less than the $\text{p} {K}_{a}$ of the weak acid ($\text{HA}$ or ${\text{BH}}^{+}$), the acid will be predominantly in the protonated form ($\text{HA}$ or ${\text{BH}}^{+}$).
• If the $\text{pH}$ of the environment is greater than the $\text{p} {K}_{a}$ of the weak acid, the acid will be predominantly in the deprotonated form (${\text{A}}^{-}$ or $\text{B}$).

Let's list the charges at various pH values.

$\text{ "" "" ""H"_3"N"^+"-gly-glu-cys-ala-COOH}$
$\textcolor{w h i t e}{1} \text{p"K_a" "color(white)(1)9.5" "" "color(white)(1)4.25" "" "" "" } 2.1$
stackrel(——————————————————————)("pH 0"" "" "+" "" "" "0" "" "" "" "color(white)(1)0)#
$\text{ ""pH 7.6"" "+" "" "color(white)(1)-" "" "" "" } -$
$\text{ ""pH 10"" "color(white)(1)0" "" "" "-" "" "" } \textcolor{w h i t e}{1} -$

At $\text{pH 7.6}$, you deprotonate both carboxyl groups but not the $N$-terminal ${\text{H"_3"N}}^{+}$ group.

You have two negative charges and a positive charge, for a net charge of $- 1$.

At $\text{pH > 9.5}$, you deprotonate all groups, so the charge on the tetrapeptide is $- 2$.