Let's solve each part step by step and explain at the end of each step. --- ## **A. Calculate the initial concentration of NaOH** ### **Step 1: Find the moles of NaOH** - Volume of NaOH solution = 25. mL = .025 L - Concentration of NaOH = .5 M \[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (L)} = .5 \times .025 = .0125 \text{ mol} \] **Explanation:** We multiplied the given molarity by the volume (converted to liters) to get the moles of NaOH present in the solution. --- ### **Step 2: Find the final concentration after mixing** - Total final volume after mixing = 25. mL NaOH + 25. mL ethyl acetate = 50. mL = .050 L \[ \text{Final concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Total Volume (L)}} = \frac{.0125}{.050} = .25 \text{ M} \] **Explanation:** After mixing, the total volume doubles, so the concentration is halved (since both solutions have same volume and concentration). --- **Final Answer for A:** \[ \boxed{.25 \text{ M}} \] --- ## **B. Calculate the rate constant of the reaction at 25°C** ### **Step 1: Identify the rate law and use the slope** - From Table 2 at 25°C, the slope = .101 min\(^{-1}\). - For a second-order reaction (since both reactants are mixed in equal concentrations), the integrated rate law is: \[ \frac{K_-K_t}{K_\infty-K_t} = k t \] or, as given, the slope of the graph gives the rate constant \(k\). **Explanation:** The rate law for a second-order reaction (with equal initial concentrations) gives a straight line when plotting \(\frac{K_-K_t}{K_\infty-K_t}\) vs time, with the slope equal to \(k\). --- ### **Step 2: State the value** - At 25°C, \(k = .101 \text{ min}^{-1}\) **Explanation:** We use the slope given in Table 2 for 25°C. --- **Final Answer for B:** \[ \boxed{.101 \text{ min}^{-1}} \] --- Would you like to proceed with **C** and **D/E** in the same detailed stepwise manner?