Sure! Let's walk through the solution to this problem step-by-step:
1. Understand the Neutralization Reaction:
The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be written as:
[tex]\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\][/tex]
From the equation, we see that one mole of HCl reacts with one mole of NaOH.
2. Calculate the Moles of NaOH:
First, we need to calculate the number of moles of NaOH present in the solution:
[tex]\[
\text{Moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH (in liters)}
\][/tex]
Given:
- Molarity of NaOH = 0.8000 M
- Volume of NaOH = 50.00 mL
Convert the volume from milliliters to liters:
[tex]\[
\text{Volume of NaOH (L)} = \frac{\text{Volume of NaOH (mL)}}{1000} = \frac{50.00}{1000} = 0.05000 \, \text{L}
\][/tex]
Now, calculate the moles of NaOH:
[tex]\[
\text{Moles of NaOH} = 0.8000 \, \text{M} \times 0.05000 \, \text{L} = 0.04000 \, \text{mol}
\][/tex]
3. Determine the Moles of HCl Required:
Since the reaction is a 1:1 molar ratio, the moles of HCl required to neutralize the NaOH will be equal to the moles of NaOH:
[tex]\[
\text{Moles of HCl} = \text{Moles of NaOH} = 0.04000 \, \text{mol}
\][/tex]
4. Calculate the Volume of HCl Required:
Using the molarity of HCl and the moles of HCl needed, we can find the volume of the HCl solution:
[tex]\[
\text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl (in liters)}
\][/tex]
Rearranging to solve for the volume:
[tex]\[
\text{Volume of HCl (L)} = \frac{\text{Moles of HCl}}{\text{Molarity of HCl}} = \frac{0.04000 \, \text{mol}}{0.2000 \, \text{M}} = 0.200 \, \text{L}
\][/tex]
Convert the volume from liters to milliliters:
[tex]\[
\text{Volume of HCl (mL)} = 0.200 \, \text{L} \times 1000 = 200.0 \, \text{mL}
\][/tex]
So, the volume of 0.2000 M HCl required to neutralize 50.00 mL of 0.8000 M NaOH is 200.0 mL.
