Question

Question: Histidine is a triprotic amino acid:

Find the value of the equilibrium constant for the reaction

Short answer

The equilibrium constant for the reaction is 1.17×10^-8

Step-by-step solution

Given data

Histidine is one type of amino acid. It follows the following equilibrium constant values for protonation

$$\begin{gathered} K_1=3\times {10}^{-2} \\ {K}_2=8.5\times {10}^{-7} \\ {K}_3=4.6\times {10}^{-10} \end{gathered}$$

We need to determine equilibrium constant for the following reaction

Determine the equilibrium constant

be denoted by A and the equilibrium constant for the reaction be K.

The equilibrium constant for the reaction will be

$\begin{array}{rcl}\left[H^{+}\right]\left[O{H}^{-}\right]& =& K_w\\ K_w& =& {10}^{-14}\\ \frac{\left[A{H}^{+}\right]\left[O{H}^{-}\right]}{\left[AH\right]}& =& K\\ \frac{\left[AH\right]}{\left[A\right]}& =& K_2\\ \frac{\left[H^{+}\right]\left[A\right]\left[O{H}^{-}\right]}{\left[AH\right]}& =& \frac{K_w}{K_2}\\ \frac{\left[A{H}^{+}\right]\left[O{H}^{-}\right]}{\left[AH\right]}& =& \frac{K_w}{K_2}\\ K& =& \frac{K_w}{K_2}\\ & =& \frac{{10}^{-14}}{8.5\times {10}^{-7}}\\ & =& 1.17\times {10}^{-8}\end{array}$

Therefore, the equilibrium constant for the reaction is 1.17×10^-8