What is the molarity of an HCl solution if 11.0 mL HCI solution is titrated with 27.6 mL of 0.160 M NaOH solution? HCl(aq) + NaOH(aq)→¬N2CI(aq) + H;0(1)

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### Determining the Molarity of an HCl Solution through Titration

**Problem Statement:**
What is the molarity of an HCl solution if 11.0 mL of HCl solution is titrated with 27.6 mL of 0.160 M NaOH solution?

**Chemical Reaction:**
\[ \text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \]

**Solution Method:**

1. **Identify the known quantities:**
   - Volume of HCl solution (V_HCl) = 11.0 mL
   - Volume of NaOH solution (V_NaOH) = 27.6 mL
   - Molarity of NaOH solution (M_NaOH) = 0.160 M

2. **Write the expression for the titration reaction:**
   - HCl and NaOH react in a 1:1 molar ratio.

3. **Calculate the moles of NaOH used:**
   \[ \text{moles of NaOH} = M_{\text{NaOH}} \times V_{\text{NaOH}} \]
   \[ \text{moles of NaOH} = 0.160 \, \text{M} \times \frac{27.6 \, \text{mL}}{1000} \]
   \[ \text{moles of NaOH} = 0.004416 \, \text{mol} \]

4. **Since the reaction ratio is 1:1:**
   - Moles of HCl = Moles of NaOH = 0.004416 mol

5. **Calculate the molarity of HCl:**
   \[ M_{\text{HCl}} = \frac{\text{moles of HCl}}{V_{\text{HCl}}} \]
   \[ M_{\text{HCl}} = \frac{0.004416 \, \text{mol}}{\frac{11.0 \, \text{mL}}{1000}} \]
   \[ M_{\text{HCl}} = 0.401 \, \text{M} \]

**Express Your Answer Numerically in Molarity:**
- Molar
Transcribed Image Text:### Determining the Molarity of an HCl Solution through Titration **Problem Statement:** What is the molarity of an HCl solution if 11.0 mL of HCl solution is titrated with 27.6 mL of 0.160 M NaOH solution? **Chemical Reaction:** \[ \text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \] **Solution Method:** 1. **Identify the known quantities:** - Volume of HCl solution (V_HCl) = 11.0 mL - Volume of NaOH solution (V_NaOH) = 27.6 mL - Molarity of NaOH solution (M_NaOH) = 0.160 M 2. **Write the expression for the titration reaction:** - HCl and NaOH react in a 1:1 molar ratio. 3. **Calculate the moles of NaOH used:** \[ \text{moles of NaOH} = M_{\text{NaOH}} \times V_{\text{NaOH}} \] \[ \text{moles of NaOH} = 0.160 \, \text{M} \times \frac{27.6 \, \text{mL}}{1000} \] \[ \text{moles of NaOH} = 0.004416 \, \text{mol} \] 4. **Since the reaction ratio is 1:1:** - Moles of HCl = Moles of NaOH = 0.004416 mol 5. **Calculate the molarity of HCl:** \[ M_{\text{HCl}} = \frac{\text{moles of HCl}}{V_{\text{HCl}}} \] \[ M_{\text{HCl}} = \frac{0.004416 \, \text{mol}}{\frac{11.0 \, \text{mL}}{1000}} \] \[ M_{\text{HCl}} = 0.401 \, \text{M} \] **Express Your Answer Numerically in Molarity:** - Molar
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