To determine the original concentration of the hydrochloric acid (HCl) solution, we need to analyze the reaction between HCl and NaOH, which proceeds according to the following balanced chemical equation:
[tex]\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
Here are the detailed steps to find the concentration of the HCl solution:
1. Identify the volumes and concentration provided:
- Volume of HCl solution, \( V_{\text{HCl}} \) = 150 mL
- Volume of NaOH solution, \( V_{\text{NaOH}} \) = 60.0 mL
- Concentration of NaOH solution, \( C_{\text{NaOH}} \) = 0.100 M
2. Convert volumes from mL to L for consistency with molarity units:
- Volume of HCl in liters: \( V_{\text{HCl (L)}} = \frac{150 \text{ mL}}{1000} = 0.150 \text{ L} \)
- Volume of NaOH in liters: \( V_{\text{NaOH (L)}} = \frac{60.0 \text{ mL}}{1000} = 0.060 \text{ L} \)
3. Calculate the moles of NaOH used:
- Moles of NaOH, \( n_{\text{NaOH}} \) = \( C_{\text{NaOH}} \times V_{\text{NaOH (L)}} \)
- \( n_{\text{NaOH}} = 0.100 \text{ M} \times 0.060 \text{ L} = 0.006 \text{ moles} \)
4. Use the stoichiometry of the reaction to find moles of HCl reacted:
- According to the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the moles of HCl reacted will be equal to the moles of NaOH.
- Moles of HCl, \( n_{\text{HCl}} \) = 0.006 moles
5. Calculate the concentration of the HCl solution:
- Concentration of HCl, \( C_{\text{HCl}} \) = \( \frac{n_{\text{HCl}}}{V_{\text{HCl (L)}}} \)
- \( C_{\text{HCl}} = \frac{0.006 \text{ moles}}{0.150 \text{ L}} = 0.040 \text{ M} \)
Thus, the original concentration of the HCl solution is:
[tex]\[ \boxed{0.040 \text{ M}} \][/tex]
