Question

Calculate values from the following ’s:

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

Short 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.

Step-by-step solution

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 Ka and pKa

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 Ka 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 Ka 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.