Calculate the target hydrogen ion concentration ([H^+]) from the desired pH.

Since HCl is a strong acid, it dissociates completely, so $[H^+] = [HCl]$.

$$[H^+] = 10^{-pH}$$ $$[H^+] = 10^{-2.10}$$ $$[H^+] = 7.943 \times 10^{-3} \text{ M}$$

Therefore, the target concentration of HCl in the final solution is $7.943 \times 10^{-3}$ M.

The final volume is 12.5 L.

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of HCl} = (7.943 \times 10^{-3} \text{ mol/L}) \times (12.5 \text{ L})$$ $$\text{Moles of HCl} = 0.0992875 \text{ mol}$$

The concentrated HCl is 36.0% HCl by mass and has a density of 1.18 g/mL.
Assume we have 100 g of the concentrated solution.
Mass of HCl in 100 g solution = 36.0 g
Volume of 100 g solution = $\frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.18 \text{ g/mL}} = 84.746 \text{ mL} = 0.084746 \text{ L}$
Molar mass of HCl = 1.008 g/mol (H) + 35.45 g/mol (Cl) = 36.458 g/mol
Moles of HCl in 36.0 g = $\frac{36.0 \text{ g}}{36.458 \text{ g/mol}} = 0.9874 \text{ mol}$
Molarity of concentrated HCl = $\frac{\text{Moles of HCl}}{\text{Volume of solution (L)}}$

$$\text{Molarity of concentrated HCl} = \frac{0.9874 \text{ mol}}{0.084746 \text{ L}} = 11.651 \text{ M}$$

We need 0.0992875 mol of HCl (from Step 2) and the concentrated solution is 11.651 M (from Step 3).

$$\text{Volume of concentrated HCl} = \frac{\text{Moles of HCl}}{\text{Molarity of concentrated HCl}}$$ $$\text{Volume of concentrated HCl} = \frac{0.0992875 \text{ mol}}{11.651 \text{ mol/L}}$$ $$\text{Volume of concentrated HCl} = 0.008521 \text{ L}$$

Convert the volume to milliliters:

$$\text{Volume of concentrated HCl} = 0.008521 \text{ L} \times 1000 \text{ mL/L} = 8.521 \text{ mL}$$

Rounding to three significant figures (due to 36.0% and pH 2.10):
The volume of concentrated HCl required is $\text{8.52 mL}$.