Let's solve this problem step by step.

Given:

1. Mass of lactic acid ($HC_3H_5O_3$): $0.222 \, \text{g}$
2. Molar mass of lactic acid ($HC_3H_5O_3$): $90.08 \, \text{g/mol}$
3. Volume of $NaOH$ solution: To be calculated.
4. Concentration of $NaOH$: $0.1200 \, \text{M}$ (Molarity = moles/L).
5. Reaction between $HC_3H_5O_3$ and $NaOH$ is 1:1.

Step 1: Calculate moles of $HC_3H_5O_3$

$\text{Moles of } HC_3H_5O_3 = \frac{\text{Mass}}{\text{Molar mass}}$ $\text{Moles of } HC_3H_5O_3 = \frac{0.222}{90.08} = 0.002465 \, \text{mol}$

Step 2: Use the 1:1 molar ratio to find moles of $NaOH$

Since $HC_3H_5O_3$ and $NaOH$ react in a 1:1 ratio: $\text{Moles of } NaOH = \text{Moles of } HC_3H_5O_3 = 0.002465 \, \text{mol}$

Step 3: Calculate volume of $NaOH$ solution required

Using the formula for molarity: $M = \frac{\text{moles}}{\text{volume (L)}}$ Rearranging for volume: $\text{Volume (L)} = \frac{\text{moles}}{M}$ $\text{Volume (L)} = \frac{0.002465}{0.1200} = 0.02054 \, \text{L}$ Convert to milliliters: $\text{Volume (mL)} = 0.02054 \times 1000 = 20.54 \, \text{mL}$

Final Answer:

The student will need 20.5 mL of $NaOH$ solution (rounded to 3 significant digits).

Let me know if you'd like further clarification or additional details!

Follow-up Questions:

1. What is the significance of using the 1:1 molar ratio in this problem?
2. How does the molar mass of $HC_3H_5O_3$ affect the calculation of moles?
3. Why do we convert liters to milliliters in the final step?
4. What would happen if the concentration of $NaOH$ was different?
5. How does rounding to significant digits affect accuracy?

Tip: Always double-check units during stoichiometry calculations to avoid errors.