50E - Organic Chemistry (Mcmurry)

Question

Calculate  values from the following  ’s:

(a) Nitromethane,  $K_a$= $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{-11}$

(b) Acrylic acid, $K_a$ = $\mathbf{5}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{-5}$

Step-by-Step Solution

Answer

(a) The  $\text{p}{K}_a$value of Nitromethane is 10.30

(b) The  $\text{p}{K}_a$value of acrylic acid is 4.25

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 K a and p K a

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

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

 

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

Calculate K a of compound (a) Nitromethane

As we know from the above step, using equation (1)

 $\begin{array}{l}p{K}_a=-log{K}_a\\ p{K}_a=-log\left(5.0\times {10}^{-10}\right)\\ p{K}_a=-\left(-10.30\right)\\ Hence\\ p{K}_a=10.30\end{array}$

(a)  $\text{p}{K}_a$value of Nitromethane = 10.30

Calculate of compound (b) acrylic acid

As we know from the above step

$\begin{array}{l}p{K}_a=-log{K}_a\\ p{K}_a=-log\left(5.6\times {10}^{-5}\right)\\ p{K}_a=-\left(-4.25\right)\\ Hence\\ p{K}_a=4.25\end{array}$ 

 

(b)  $\text{p}{K}_a$value of acrylic acid = 4.25