To find the pH of the final solution, follow these steps:

Calculate the moles of HCl and NaOH before mixing:

HCl: Moles = $ 0.5 \text{ M} \times 0.1 \text{ L} = 0.05 \text{ mol} $

NaOH: Moles = $ 0.4 \text{ M} \times 0.1 \text{ L} = 0.04 \text{ mol} $

Determine the reaction outcome:

HCl and NaOH react in a 1:1 molar ratio as shown in the equation:

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Thus, 0.04 mol of HCl will completely react with 0.04 mol of NaOH. The remaining moles of HCl are:

$0.05 \text{ mol} - 0.04 \text{ mol} = 0.01 \text{ mol}$

Calculate the final volume after dilution:

By adding 800 mL of distilled water to 200 mL of the resultant solution, the total volume becomes:

$200 \text{ mL} + 800 \text{ mL} = 1000 \text{ mL} = 1 \text{ L}$

Determine the concentration of HCl after dilution:

$\text{Concentration of HCl} = \frac{0.01 \text{ mol}}{1 \text{ L}} = 0.01 \text{ M}$

Since HCl fully dissociates:

$\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$

Therefore, the concentration of $\text{H}^+$ ions is also 0.01 M.

Calculate the pH of the solution:

$ \text{pH} = -\log [\text{H}^+] \\ \text{pH} = -\log [0.01] = 2.0 $

Thus, the pH of the final solution is 2.0.

Physical Chemistry

Inorganic Chemistry