To find the molarity of the HCl solution, we can use the stoichiometry of the titration reaction. The balanced chemical equation for the reaction between HCl and NaOH is:
\[ \text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \]
From the equation, we can see that the reaction occurs in a 1:1 molar ratio between HCl and NaOH.
Step 1: Calculate the moles of NaOH used
First, we calculate the number of moles of NaOH used in the titration:
\[ \text{Moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH in L} \]
Given:
- Molarity of NaOH = 0.250 M
- Volume of NaOH = 35.23 mL = 0.03523 L (conversion from mL to L by dividing by 1000)
Now, substitute the values:
\[ \text{Moles of NaOH} = 0.250 , \text{mol/L} \times 0.03523 , \text{L} = 0.0088075 , \text{mol} \]
Step 2: Calculate the moles of HCl
Since the reaction ratio is 1:1, the moles of HCl will be equal to the moles of NaOH:
\[ \text{Moles of HCl} = \text{Moles of NaOH} = 0.0088075 , \text{mol} \]
Step 3: Calculate the molarity of HCl
Now, we need to find the molarity of the HCl solution. Molarity is defined as moles of solute divided by the volume of solution in liters.
\[ \text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl in L}} \]
Given volume of HCl = 50.00 mL = 0.05000 L.
Substituting the values:
\[ \text{Molarity of HCl} = \frac{0.0088075 , \text{mol}}{0.05000 , \text{L}} = 0.17615 , \text{M} \]
Final Answer
Rounding this to three significant figures:
\[ \text{Molarity of HCl} \approx 0.176 , \text{M} \]
Thus, the molarity of the HCl solution is 0.176 M.