Q18P What percent of imidazole is pr... [FREE SOLUTION]

Chapter 19: Q18P (page 941)

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

Short Answer

Expert verified

The percentage of imidazole protonated at physiological pH (7.4) is 20 %.

Step by step solution

Step-by-Step Solution Step 1: Acid dissociation constant

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 ( $K_{a}$ )for the acid-dissociation equilibrium.

Example: $\begin{array}{l} {HA+H}_{2} O 鈬� A^{鈭�} {+H3O}^{鈭�} \\ K_{a} \underset{}{} = \underset{}{} \frac{{[H3O}^{+} {][A}^{鈭�}]}{[HA]} \end{array}$

Acid strengths are normally determined using $p K_{a}$ values rather than $K_{a}$ values, where the $p K_{a}$ is the negative common logarithm of $K_{a}$ :

$p K_{a} =-log K_{a}$

The stronger acid has a smallervalue of $p K_{a}$ , and a weaker acid has a larger $p K_{a}$ value.

pH scale

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⁺].

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

Calculate percentage of imidazole

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

Fraction of imidazole in acidic form $= \frac{[H^{+}]}{K_{a} + [H^{+}]}$

We have given pH = 7.4 and pK_a= 7.8

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

${[H}^{+}] = 4.0 脳 10^{鈭� 8}$

And

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

$K_{a} \underset{}{} = \underset{}{} 1.58 脳 10^{鈭� 7}$

Put these values in equation 1 and we have,

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