$$HCl + KOH → KCl + H_2O$$
From the equation, we can see that 1 mole of HCl reacts with 1 mole of KOH. Therefore, the number of moles of HCl required to neutralize 25.0 mL of 1.00 M KOH is:
$$25.0 \text{ mL KOH} \times \frac{1 \text{ mmol KOH}}{1 \text{ mL KOH}} \times \frac{1 \text{ mmol HCl}}{1 \text{ mmol KOH}} = 25.0 \text{ mmol HCl}$$
The volume of 0.45 M HCl required to provide 25.0 mmol HCl is:
$$25.0 \text{ mmol HCl} \times \frac{1 \text{ mL HCl}}{0.45 \text{ mmol HCl}} = 55.6 \text{ mL HCl}$$
Therefore, 55.6 mL of 0.45 M HCl is required to neutralize 25.0 mL of 1.00 M KOH.
