Answer :
To solve the problem, we first determine the number of moles of hydrochloric acid ([tex]$\text{HCl}$[/tex]) that were used in the neutralization reaction, and then use the 1:1 molar ratio between [tex]$\text{HCl}$[/tex] and sodium hydroxide ([tex]$\text{NaOH}$[/tex]) to find the moles of [tex]$\text{NaOH}$[/tex] present. Finally, we calculate the concentration of the [tex]$\text{NaOH}$[/tex] solution.
Step 1. Convert the volume of [tex]$\text{HCl}$[/tex] solution to liters
The problem gives the volume of the [tex]$\text{HCl}$[/tex] solution as
[tex]$$37.9\,\text{mL}.$$[/tex]
Convert this to liters:
[tex]$$
V_{\text{HCl}} = \frac{37.9\,\text{mL}}{1000} = 0.0379\,\text{L}.
$$[/tex]
Step 2. Calculate the moles of [tex]$\text{HCl}$[/tex]
Given the molarity ([tex]$M$[/tex]) of [tex]$\text{HCl}$[/tex] is
[tex]$$4.21 \times 10^{-2}\,\text{M},$$[/tex]
the number of moles is calculated by:
[tex]$$
n_{\text{HCl}} = M_{\text{HCl}} \times V_{\text{HCl}} = \left(4.21 \times 10^{-2}\,\text{M}\right) \times 0.0379\,\text{L}.
$$[/tex]
Evaluating the product gives:
[tex]$$
n_{\text{HCl}} \approx 0.00159559\,\text{mol}.
$$[/tex]
Step 3. Use the stoichiometry of the neutralization reaction
The neutralization reaction between [tex]$\text{HCl}$[/tex] and [tex]$\text{NaOH}$[/tex] is:
[tex]$$
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}.
$$[/tex]
Since the mole ratio between [tex]$\text{HCl}$[/tex] and [tex]$\text{NaOH}$[/tex] is 1:1, the number of moles of [tex]$\text{NaOH}$[/tex] is:
[tex]$$
n_{\text{NaOH}} = n_{\text{HCl}} \approx 0.00159559\,\text{mol}.
$$[/tex]
Step 4. Convert the volume of [tex]$\text{NaOH}$[/tex] solution to liters
The volume of the [tex]$\text{NaOH}$[/tex] solution is given as
[tex]$$65.0\,\text{mL},$$[/tex]
which in liters is:
[tex]$$
V_{\text{NaOH}} = \frac{65.0\,\text{mL}}{1000} = 0.0650\,\text{L}.
$$[/tex]
Step 5. Calculate the concentration of the [tex]$\text{NaOH}$[/tex] solution
The molarity of the [tex]$\text{NaOH}$[/tex] solution is the number of moles of [tex]$\text{NaOH}$[/tex] divided by its volume in liters:
[tex]$$
M_{\text{NaOH}} = \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}} = \frac{0.00159559\,\text{mol}}{0.0650\,\text{L}}.
$$[/tex]
Evaluating the division, we get:
[tex]$$
M_{\text{NaOH}} \approx 0.02455\,\text{M}.
$$[/tex]
Final Answer
The concentration of the [tex]$\text{NaOH}$[/tex] solution is approximately
[tex]$$\boxed{0.02455\,\text{M}}.$$[/tex]
Step 1. Convert the volume of [tex]$\text{HCl}$[/tex] solution to liters
The problem gives the volume of the [tex]$\text{HCl}$[/tex] solution as
[tex]$$37.9\,\text{mL}.$$[/tex]
Convert this to liters:
[tex]$$
V_{\text{HCl}} = \frac{37.9\,\text{mL}}{1000} = 0.0379\,\text{L}.
$$[/tex]
Step 2. Calculate the moles of [tex]$\text{HCl}$[/tex]
Given the molarity ([tex]$M$[/tex]) of [tex]$\text{HCl}$[/tex] is
[tex]$$4.21 \times 10^{-2}\,\text{M},$$[/tex]
the number of moles is calculated by:
[tex]$$
n_{\text{HCl}} = M_{\text{HCl}} \times V_{\text{HCl}} = \left(4.21 \times 10^{-2}\,\text{M}\right) \times 0.0379\,\text{L}.
$$[/tex]
Evaluating the product gives:
[tex]$$
n_{\text{HCl}} \approx 0.00159559\,\text{mol}.
$$[/tex]
Step 3. Use the stoichiometry of the neutralization reaction
The neutralization reaction between [tex]$\text{HCl}$[/tex] and [tex]$\text{NaOH}$[/tex] is:
[tex]$$
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}.
$$[/tex]
Since the mole ratio between [tex]$\text{HCl}$[/tex] and [tex]$\text{NaOH}$[/tex] is 1:1, the number of moles of [tex]$\text{NaOH}$[/tex] is:
[tex]$$
n_{\text{NaOH}} = n_{\text{HCl}} \approx 0.00159559\,\text{mol}.
$$[/tex]
Step 4. Convert the volume of [tex]$\text{NaOH}$[/tex] solution to liters
The volume of the [tex]$\text{NaOH}$[/tex] solution is given as
[tex]$$65.0\,\text{mL},$$[/tex]
which in liters is:
[tex]$$
V_{\text{NaOH}} = \frac{65.0\,\text{mL}}{1000} = 0.0650\,\text{L}.
$$[/tex]
Step 5. Calculate the concentration of the [tex]$\text{NaOH}$[/tex] solution
The molarity of the [tex]$\text{NaOH}$[/tex] solution is the number of moles of [tex]$\text{NaOH}$[/tex] divided by its volume in liters:
[tex]$$
M_{\text{NaOH}} = \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}} = \frac{0.00159559\,\text{mol}}{0.0650\,\text{L}}.
$$[/tex]
Evaluating the division, we get:
[tex]$$
M_{\text{NaOH}} \approx 0.02455\,\text{M}.
$$[/tex]
Final Answer
The concentration of the [tex]$\text{NaOH}$[/tex] solution is approximately
[tex]$$\boxed{0.02455\,\text{M}}.$$[/tex]
