To determine the concentration of the sodium hydroxide (NaOH) solution required to titrate a known volume and concentration of hydrochloric acid (HCl), we'll use the concept of titration, specifically employing the formula:
[tex]\[ M_1 V_1 = M_2 V_2 \][/tex]
Where:
- [tex]\( M_1 \)[/tex] is the molarity of the HCl solution.
- [tex]\( V_1 \)[/tex] is the volume of the HCl solution.
- [tex]\( M_2 \)[/tex] is the molarity of the NaOH solution.
- [tex]\( V_2 \)[/tex] is the volume of the NaOH solution.
We have the following given values:
- [tex]\( V_1 \)[/tex] (volume of HCl) = 50 mL
- [tex]\( M_1 \)[/tex] (molarity of HCl) = 1 M
- [tex]\( V_2 \)[/tex] (volume of NaOH) = 33 mL
We need to find [tex]\( M_2 \)[/tex], the molarity of the NaOH solution. Rearranging the formula to solve for [tex]\( M_2 \)[/tex] gives us:
[tex]\[ M_2 = \frac{M_1 V_1}{V_2} \][/tex]
Now, substituting the known values:
[tex]\[ M_2 = \frac{1 \, \text{M} \times 50 \, \text{mL}}{33 \, \text{mL}} \][/tex]
[tex]\[ M_2 = \frac{50 \, \text{M} \cdot \text{mL}}{33 \, \text{mL}} \][/tex]
[tex]\[ M_2 = 1.515 \, \text{M} \][/tex]
Therefore, the concentration of the NaOH solution must be approximately 1.515 M to titrate 50 mL of a 1 M HCl solution.
