If K_a of weak acid HA is 1 × 10^ -7 , calculate the [H_3O^+] of...

For a weak acid HA, the dissociation reaction is:

$\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$

The dissociation constant is given by $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$.

Since $K_a$ is very small, the degree of dissociation $\alpha$ is much less than $1$, and the concentration of hydronium ions can be approximated as:

$[\text{H}_3\text{O}^+] = \sqrt{K_a \times C}$

Substituting the given values $K_a = 1 \times 10^{-7}$ and $C = 0.1\ \text{M}$:

$[\text{H}_3\text{O}^+] = \sqrt{1 \times 10^{-7} \times 0.1}$

$[\text{H}_3\text{O}^+] = \sqrt{10^{-8}}$

$[\text{H}_3\text{O}^+] = 1 \times 10^{-4}\ \text{M}$

Answer: $1 \times 10^{-4}\ \text{M}$