To find the concentration of the HCl solution, we can use the concept of molarity (M), which is defined as the number of moles of solute per liter of solution.
First, let's find the number of moles of NaOH used:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in liters)} \]
Given that the volume of NaOH used is \( 54 \, \text{mL} \) and the molarity of NaOH is \( 0.1 \, \text{M} \):
\[ \text{Volume (in liters)} = \frac{54 \, \text{mL}}{1000 \, \text{mL/L}} = 0.054 \, \text{L} \]
\[ \text{Moles of NaOH} = 0.1 \, \text{M} \times 0.054 \, \text{L} = 0.0054 \, \text{moles} \]
Since HCl and NaOH react in a 1:1 ratio, the moles of HCl present in the solution are also \( 0.0054 \, \text{moles} \).
Now, let's find the concentration of HCl:
\[ \text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume (in liters)}} \]
Given that the volume of the HCl solution is \( 125 \, \text{mL} \):
\[ \text{Volume (in liters)} = \frac{125 \, \text{mL}}{1000 \, \text{mL/L}} = 0.125 \, \text{L} \]
\[ \text{Molarity of HCl} = \frac{0.0054 \, \text{moles}}{0.125 \, \text{L}} \]
\[ \text{Molarity of HCl} = \frac{0.0054}{0.125} \, \text{M} \]
\[ \text{Molarity of HCl} = 0.0432 \, \text{M} \]
So, the concentration of the HCl solution is \( 0.0432 \, \text{M} \).
