Calculate values from the following ’s:

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

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

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:

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

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$\text{p}{K}_a$ .

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

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