What percent of imidazole is protonated at physiological pH (7.4)?

Acids differ in their ability to donate H⁺. Stronger acids, such as HCl, react almost completely with water, whereas weaker acids, such as acetic acid (CH₃COOH), react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant (${\mathrm{K}}_{\mathrm{a}}$)for the acid-dissociation equilibrium.

Example: $\begin{array}{l}{\text{HA+H}}_2\text{O}\rightleftarrows {\text{A}}^{-}{\text{+H3O}}^{-}\\ {\mathrm{K}}_{\mathrm{a}}\underset{}{}\text{=}\underset{}{}\frac{{\text{[H3O}}^{+}{\text{][A}}^{-}\text{]}}{\left[\mathrm{HA}\right]}\end{array}$

Acid strengths are normally determined using $\text{p}{\mathrm{K}}_{\mathrm{a}}$values rather than $K_a$values, where the $\text{p}{K}_a$is the negative common logarithm of $K_a$:

$\text{p}{K}_{\text{a}}\text{=-log}{K}_a$

The stronger acid has a smallervalue of $\text{p}{K}_a$, and a weaker acid has a larger $\text{p}{K}_a$value.

The pH scale is used to describe the acidity of a solution, whereas the pKa indicates the tendency of a compound to lose its proton.Thus, the pK_ais characteristic of a particular compound, much like a melting point or a boiling point.

The concentration of protons in a solution is indicated by pH. This concentration is written as either [H⁺]or [H₃O⁺].

${\text{pH=-log[H}}^{+}]$

We know, fraction of imidazole in acidic medium is given by:

Fraction of imidazole in acidic form $=\frac{\left[{\mathrm{H}}^{+}\right]}{{\mathrm{K}}_{\mathrm{a}}+\left[{\mathrm{H}}^{+}\right]}$

We have given pH = 7.4 and pK_a= 7.8

So, $\begin{array}{l}\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ 7.4=-\log \left[{\mathrm{H}}^{+}\right]\\ {\mathrm{e}}^{-7.4}=\left[{\mathrm{H}}^{+}\right]\end{array}$

${\text{[H}}^{+}]=4.0\times {10}^{-8}$

And

$\begin{array}{l}{\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}\\ 6.8=-{\mathrm{logK}}_{\mathrm{a}}\\ {\mathrm{e}}^{-6.8}={\mathrm{K}}_{\mathrm{a}}\end{array}$

${\mathrm{K}}_{\mathrm{a}}\underset{}{}\text{=}\underset{}{}1.58\times {10}^{-7}$

Put these values in equation 1 and we have,

$\begin{array}{l}\mathrm{Fraction}\mathrm{of}\mathrm{imidazole}\mathrm{in}\mathrm{acidic}\mathrm{form}=\frac{4.0\times {10}^{-8}}{1.58\times {10}^{-7}+4.0\times {10}^{-8}}\\ =\frac{4.0\times {10}^{-8}}{1.98\times {10}^{-7}}\\ =0.20\end{array}$

Hence, 20 % of imidazole is protonated at physiological pH (7.4) i.e. in acidic form.