What is the molarity of an HCl solution if 125 mL is neutralized in a titration by 76.0 mL of 1.22 M KOH?

To determine the molarity of the HCl solution, we start by using the neutralization reaction between HCl and KOH, which follows a 1:1 stoichiometry. This means that one mole of HCl reacts with one mole of KOH.

First, calculate the number of moles of KOH used in the titration. This is done using the molarity of the KOH solution and the volume used in liters:

1. Convert the volume of KOH from milliliters to liters:

[

76.0 \text{ mL} = 0.076 \text{ L}

]

2. Calculate the moles of KOH:

[

\text{moles KOH} = \text{Molarity} \times \text{Volume} = 1.22 , \text{M} \times 0.076 , \text{L} = 0.09272 , \text{mol}

]

Since the stoichiometry of the reaction is 1:1, this means the moles of HCl will also be 0.09272 mol.

Next, we can calculate the molarity of