49E - Organic Chemistry (Mcmurry)

Question

Calculate  values from the following ’s:

(a) Acetone,  $p{K}_a$=19.3 (b) Formic acid,  $p{K}_a$=3.75

Step-by-Step Solution

Answer

(a) The $K_a$ of Acetone is $5\times {10}^{-20}$if its  $p{K}_a$ = 19.3.

(b) The  $K_a$ of Formic acid is $1.8\times {10}^{-4}$if its  $p{K}_a$= 3.75.

Acid and base strength

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:$\begin{array}{l}{\text{HA + H}}_2\text{O }\rightleftarrows {\text{ A}}^{-}{\text{ + H}}_{\text{3}}{\text{O}}^{-}\\ K_a\underset{}{}\text{=}\underset{}{}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{+}{\text{][A}}^{-}\text{]}}{\left[HA\right]}\end{array}$

Relation between and K a and p K a

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

 $\text{p}K\text{a = -log }{K}_a$

The stronger acid has a smaller  $p{K}_a$ , and the weaker acid has a larger $p{K}_a$ .

 

Calculate K a of compound (a) Acetone

As we know from the above step,

  $\text{p}K\text{a = -log }{K}_a$   -------eq(1)

 

Taking antilog on both sides of eq(1)

 $Anti\log \left(\text{p}K\text{a}\right)\text{ = - }{K}_a$

 $Anti\log \left(\text{-p}K\text{a}\right)\text{ = }{K}_a$

 Hence

 $\begin{array}{l}Anti\log \left(-19.3\right)\text{ = }{K}_a\\ {10}^{-{19.3}_{}}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}=\stackrel{}{}{K}_a\\ 5.0\times {10}^{-20}\underset{}{\underset{}{}}\underset{}{}\underset{}{\underset{}{}}=\underset{}{}{K}_a\end{array}$

The  $K_a$of Acetone is $5\times {10}^{-20}$if its $p{K}_a$ = 19.3.

Calculate K a of compound (b) Formic acid

As we know from the above step

    $\text{p}K\text{a = -log }{K}_a$ -------eq(1)

 

Taking antilog on both sides of eq(1)

 

 $Anti\log \left(\text{p}K\text{a}\right)\text{ = - }{K}_a$

$Anti\log \left(\text{-p}K\text{a}\right)\text{ = }{K}_a$

 Hence

 $\begin{array}{l}Anti\log \left(-3.75\right)\text{ = }{K}_a\\ {10}^{-{3.75}_{}}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}=\stackrel{}{}{K}_a\\ 1.8\times {10}^{-4}\underset{}{\underset{}{}}\underset{}{}\underset{}{\underset{}{}}=\underset{}{}{K}_a\end{array}$

The  $K_a$of Formic acid is $1.8\times {10}^{-4}$if its $p{K}_a$ = 3.75.