Certainly! Let's go through the steps to determine the molarity of the HCl solution given the data from your titration experiment:
1. Volume of the titrant (NaOH) used in the titration:
- Volume [tex]\( V_{\text{NaOH}} = 12.33 \)[/tex] mL
2. Molarity of the titrant (NaOH):
- Molarity [tex]\( M_{\text{NaOH}} = 0.5 \)[/tex] mol/L
3. Volume of the analyte (HCl):
- Volume [tex]\( V_{\text{HCl}} = 19.54 \)[/tex] mL
4. Determine the moles of NaOH used in the titration:
- To find the moles of NaOH, use the formula:
[tex]\[
\text{moles of NaOH} = M_{\text{NaOH}} \times \frac{V_{\text{NaOH}}}{1000}
\][/tex]
Substituting the given values:
[tex]\[
\text{moles of NaOH} = 0.5 \, \text{mol/L} \times \frac{12.33 \, \text{mL}}{1000} = 0.006165 \, \text{mol}
\][/tex]
5. Since HCl and NaOH react in a 1:1 molar ratio:
- The moles of HCl will be equal to the moles of NaOH.
- Hence, the moles of HCl [tex]\( = 0.006165 \, \text{mol} \)[/tex]
6. Calculate the molarity of the HCl solution:
- Molarity [tex]\( M_{\text{HCl}} \)[/tex] is calculated using the formula:
[tex]\[
M_{\text{HCl}} = \frac{\text{moles of HCl}}{\frac{V_{\text{HCl}}}{1000}}
\][/tex]
Substituting the known values:
[tex]\[
M_{\text{HCl}} = \frac{0.006165 \, \text{mol}}{\frac{19.54 \, \text{mL}}{1000}} = 0.31550665301944736 \, \text{mol/L}
\][/tex]
So, the molarity of the HCl solution is approximately [tex]\( 0.3155 \, \text{mol/L} \)[/tex].
