Answer
$[H_3O^+] = 1.5 \times 10^{-7}$
$pH = 6.81$
Work Step by Step
1. Write the $K_w$ expression:
$$K_w = [H_3O^+][OH^-]$$
2. In pure water: $[H_3O^+] = [OH^-]$.
$$K_w = [H_3O^+][H_3O^+]$$
$$\sqrt {K_w} = [H_3O^+]$$
$$[H_3O^+] = \sqrt{2.4 \times 10^{-14}} = 1.5 \times 10^{-7}$$
3. Calculate the pH:
$$pH = -log(\sqrt {2.4 \times 10^{-14}}) = 6.81$$
