Calculate the pH using the relationship between pH and pOH.

$$\text{pH} + \text{pOH} = 14$$ $$\text{pH} = 14 - \text{pOH}$$ $$\text{pH} = 14 - 11.40$$ $$\text{pH} = 2.60$$

$$\text{pH} = -\log_{10}[\text{H}^+]$$ $$[\text{H}^+] = 10^{-\text{pH}}$$ $$[\text{H}^+] = 10^{-2.60}$$ $$[\text{H}^+] = 2.51 \times 10^{-3} \text{ mol/dm}^3$$

The concentration of hydrogen ion is $\boxed{\text{}2.51 \times 10^{-3} \text{ mol/dm}^3\text{}}$.

2. Find the hydrogen ion and hydroxide ion concentration in a $0.02 \text{ M}$ solution of $\text{H}_2\text{SO}_4$.

Sulfuric acid is a strong acid and dissociates completely.

$$\text{H}_2\text{SO}_4(aq) \longrightarrow 2\text{H}^+(aq) + \text{SO}_4^{2-}(aq)$$

Since each mole of $\text{H}_2\text{SO}_4$ produces 2 moles of $\text{H}^+$ ions:

$$[\text{H}^+] = 2 \times [\text{H}_2\text{SO}_4]$$ $$[\text{H}^+] = 2 \times 0.02 \text{ M}$$ $$[\text{H}^+] = 0.04 \text{ M}$$

At $25^\circ\text{C}$, $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ M}^2$.

$$[\text{OH}^-] = \frac{K_w}{[\text{H}^+]}$$ $$[\text{OH}^-] = \frac{1.0 \times 10^{-14} \text{ M}^2}{0.04 \text{ M}}$$ $$[\text{OH}^-] = 2.5 \times 10^{-13} \text{ M}$$

The hydrogen ion concentration is $\boxed{\text{}0.04 \text{ M}\text{}}$, and the hydroxide ion concentration is $\boxed{\text{}2.5 \times 10^{-13} \text{ M}\text{}}$.

3. Find the pH in which the hydrogen ion concentration is $6.38 \times 10^{-6} \text{ mol/dm}^3$.

$$\text{pH} = -\log_{10}(6.38 \times 10^{-6})$$ $$\text{pH} = -(-5.195)$$ $$\text{pH} = 5.195$$

The pH is $\boxed{\text{}5.195\text{}}$.