Given the data for the following separate titrations, calculate the molarity of the \(\mathrm{NaOH}\) : $$ \begin{array}{|c|c|c|c|c|} \hline & \begin{array}{c} \text { mL } \\ \text { HCl } \end{array} & \begin{array}{c} \text { Molarity } \\ \text { HCl } \end{array} & \begin{array}{c} \mathbf{m L} \\ \mathbf{N a O H} \end{array} & \begin{array}{c} \text { Molarity } \\ \mathbf{N a O H} \end{array} \\ \hline \text { (a) } & 37.19 & 0.126 & 31.91 & M \\ \text { (b) } & 48.04 & 0.482 & 24.02 & M \\ \text { (c) } & 13.13 & 1.425 & 39.39 & M \\ \hline \end{array} $$

In a neutralization reaction, an acid reacts with a base to produce water and a salt. This reaction is essential in titration calculations because it allows us to determine the concentration (molarity) of an unknown solution. For this problem, we are working with hydrochloric acid (HCl) and sodium hydroxide (NaOH), which react in a 1:1 molar ratio:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
Here, one mole of HCl neutralizes one mole of NaOH. Knowing this ratio helps us set up the calculations correctly.

Molarity (M) is a measure of the concentration of a solution, defined as the number of moles of solute per liter of solution. It's a fundamental concept in titration because it allows us to quantify how concentrated our solutions are. The formula for molarity is:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In titrations, we often use the volumes of solutions in milliliters (mL), but we convert these to liters (L) by dividing by 1000. For example, if we have 50 mL of solution, that equals 0.050 L.

When performing titration calculations, we use the titration formula:
\[ M_1V_1 = M_2V_2 \]
Here:

• \(M_1\) is the molarity of the acid (HCl in this case)
• \(V_1\) is the volume of the acid used
• \(M_2\) is the molarity of the base (NaOH in this case)
• \(V_2\) is the volume of the base used

This equation arises from the stoichiometry of the neutralization reaction, assuming a 1:1 molar ratio. Rearranging the formula allows us to solve for the unknown molarity. For instance, to find the molarity of NaOH (\(M_2\)), we rearrange as:
\[ M_2 = \frac{M_1V_1}{V_2} \].

Stoichiometry involves the calculation of reactants and products in chemical reactions. In titration, stoichiometry is crucial because it ensures the correct ratios of chemicals are used. For the neutralization between HCl and NaOH, we know the following:
\[ 1 \text{ mole of } \text{HCl} + 1 \text{ mole of }\text{NaOH} \rightarrow \text{salt and water} \]
Given this 1:1 ratio, if we know the molarity and volume of HCl, we can determine the moles of HCl, and thus the moles of NaOH required for neutralization. For example, if we have 0.126 M HCl solution and we use 37.19 mL of it, the moles of HCl would be:
\[ \text{moles of HCl} = M_1 \times V_1 = 0.126 \times 0.03719 = 0.00469 \text{ moles} \]
These same moles of NaOH will be required, and from here, using the titration formula, you can find the unknown molarity of NaOH.