Given the data for the following separate titrations, calculate the molarity of the \(\mathrm{HCl}\) : $$ \begin{array}{|c|c|c|c|c|} \hline & \begin{array}{c} \text { mL } \\ \text { HCl } \end{array} & \begin{array}{c} \text { Molarity } \\ \text { HCl } \end{array} & \begin{array}{c} \mathbf{m L} \\ \mathbf{N a O H} \end{array} & \begin{array}{c} \text { Molarity } \\ \mathbf{N a O H} \end{array} \\ \hline \text { (a) } & 40.13 & M & 37.70 & 0.728 \\ \text { (b) } & 19.00 & M & 33.66 & 0.306 \\ \text { (c) } & 27.25 & M & 18.00 & 0.555 \\ \hline \end{array} $$

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It’s like a recipe that tells us how much of each ingredient we need. In our titration exercise, we use stoichiometry to relate the amount of acid (HCl) to the amount of base (NaOH). First, understand that in a 1:1 acid-base titration, one mole of HCl reacts with one mole of NaOH. This relationship is captured in the formula: \[ n_{\text{HCl}} = n_{\text{NaOH}} \], where \( n \) represents moles. To find the moles, we use \( n = M \times V \), where \( M \) is the molarity (concentration) and \( V \) is the volume in liters. By using these stoichiometric relationships, we can determine how much of each reactant we need.

Acid-base titration is a technique used to determine the concentration of an unknown acid or base. During titration, we slowly add a titrant (a solution of known concentration) to the analyte (solution of unknown concentration) until the reaction reaches an endpoint. The endpoint is when the acid completely neutralizes the base. In our exercise, we titrate HCl with NaOH. The volume of NaOH needed to reach the endpoint helps us calculate the molarity of HCl. We use a chemical indicator or pH meter to detect the endpoint, showing us the exact point at which the titration is complete. By plotting the volume of titrant added versus pH, we can visually see the titration curve.

Solution concentration refers to the amount of solute (substance dissolved) in a given volume of solvent. In our context, concentration is expressed as molarity (M), which is moles of solute per liter of solution. For example, if you have 1 mole of HCl in 1 liter of solution, the molarity is 1 M. Concentration helps us understand how much of a substance is present. During titration, knowing the concentration of our titrant (NaOH) is crucial to determining the concentration of our analyte (HCl). We can follow these steps to calculate solution concentration:

• Measure the volumes of both solutions.
• Use stoichiometry to relate the volumes and concentrations.
• Apply the molarity formula \( M = \frac{n}{V} \).

Molarity calculations involve finding the number of moles of solute in a solution and then using that to determine the molarity. Here’s how we calculate the molarity of HCl in our exercise:
1. Convert all volumes to liters (1 mL = 0.001 L).
2. Use the titration formula: \( n_{\text{HCl}} = M_{\text{HCl}} \times V_{\text{HCl}} \) and \( n_{\text{NaOH}} = M_{\text{NaOH}} \times V_{\text{NaOH}} \).
3. Set the moles of HCl equal to the moles of NaOH (since we have a 1:1 ratio).
4. Rearrange the formula to solve for molarity of HCl (\( M_{\text{HCl}} \)).
5. Plug in the numbers and solve:

• Example: \( M_{\text{HCl}} = \frac{0.728 \times 37.70}{40.13} \) for titration (a).

By following these steps, we find that the molarities for titrations (a), (b), and (c) are 0.684 M, 0.542 M, and 0.367 M, respectively.