Questions: By titration, it is found that 44.7 mL of 0.158 M NaOH(aq) is needed to neutralize 25.0 mL of HCl. Calculate the concentration of the HCl solution. [HCl]=

Solution Steps

The neutralization reaction between NaOH and HCl is:

\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

This reaction shows a 1:1 molar ratio between NaOH and HCl.

The moles of NaOH can be calculated using its concentration and volume:

\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.158 \, \text{M} \times 44.7 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \]

\[ = 0.158 \times 0.0447 = 0.0070566 \, \text{mol} \]

Since the reaction is 1:1, the moles of HCl are equal to the moles of NaOH:

\[ \text{Moles of HCl} = 0.0070566 \, \text{mol} \]

The concentration of HCl is calculated using its moles and volume:

\[ [\text{HCl}] = \frac{\text{Moles of HCl}}{\text{Volume of HCl in L}} = \frac{0.0070566 \, \text{mol}}{25.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}} \]

\[ = \frac{0.0070566}{0.0250} = 0.2823 \, \text{M} \]

Final Answer