Answer
I would put 4.2 mL of $12M$ HCl in a recipient, and then I would fill the recipient with water until it reaches 1600mL of solution.
Work Step by Step
1. Calculate the $[H_3O^+]$ molarity:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.5}$
$[H_3O^+] = 3.2 \times 10^{- 2}M$
2. Find the number of moles we would need to have this concentration in a 1600 mL solution:
$1600mL \times \frac{1L}{1000mL} \times \frac{3.2 \times 10^{-2}mol}{1L} = 0.051mol (H_3O^+)$
3. Since $HCl$ is a strong acid, the amount of produced $H_3O^+$ is equal to the inicial amount of $HCl$. Calculate the volume of $12M$ $HCl$ that would have 0.0512mol $(H_3O^+)$.
$0.051mol(H_3O^+) \times \frac{1mol(HCl)}{1mol(H_3O^+)} \times \frac{1L(solution)}{12mol(HCl)} = 4.2 \times 10^{-3}L = $ 4.2 mL.
