The titration involves a weak base, triethylamine (B), with a strong acid, HCl.

Here are the solutions for parts (a), (b), and (c) of Question 2.

The titration involves a weak base, triethylamine ($B$), with a strong acid, $HCl$.
The reaction is: $B(aq) + H^+(aq) \to BH^+(aq)$
Given:
Initial volume of weak base = 20.00 mL = 0.02000 L
Initial concentration of weak base = 0.1000 M
Initial moles of weak base = $0.02000 \text{ L} \times 0.1000 \text{ M} = 0.002000 \text{ mol}$
Concentration of titrant ($HCl$) = 0.1000 M
$K_b$ of base = $5.2 \times 10^{-4}$
$pK_b = -\log(5.2 \times 10^{-4}) \approx 3.284$

Part (a): Find the pH during the titration.

i. 0.00 mL HCl added:
This is a solution of a weak base.
The equilibrium is: $B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)$
Let $[OH^-] = x$.

$$K_b = \frac{[BH^+][OH^-]}{[B]}$$ $$5.2 \times 10^{-4} = \frac{x^2}{0.1000 - x}$$

Using the quadratic formula ($x^2 + (5.2 \times 10^{-4})x - (5.2 \times 10^{-5}) = 0$):

$$x = \frac{-5.2 \times 10^{-4} + \sqrt{(5.2 \times 10^{-4})^2 - 4(1)(-5.2 \times 10^{-5})}}{2}$$ $$x = \frac{-5.2 \times 10^{-4} + \sqrt{2.704 \times 10^{-7} + 2.08 \times 10^{-4}}}{2}$$ $$x = \frac{-5.2 \times 10^{-4} + \sqrt{2.082704 \times 10^{-4}}}{2}$$ $$x = \frac{-5.2 \times 10^{-4} + 0.01443}{2} = 0.006955 \text{ M}$$

So, $[OH^-] = 0.006955$ M.

$$pOH = -\log(0.006955) \approx 2.1578$$ $$pH = 14 - pOH = 14 - 2.1578 = 11.8422$$

The pH is $\boxed{11.