In a coffee-cup calorimeter, 150.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) HCl is added to 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NaOH}\) to make 200.0 \(\mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\) . If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) , calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.184 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assume no heat loss to the surroundings.

Given the volumes (V) and concentrations (C) of both HCl and NaOH, we can calculate the moles using the formula moles = C * V. Moles of HCl: \( moles = C_{HCl} * V_{HCl} \) \( moles = 0.50 \frac{mol}{L} * 150.0 mL = 0.50 \frac{mol}{L} * 0.150 L = 0.075 mol \) Moles of NaOH: \( moles = C_{NaOH} * V_{NaOH} \) \( moles = 1.00 \frac{mol}{L} * 50.0 mL = 1.00 \frac{mol}{L} * 0.050 L = 0.050 mol \)

In this case, we see that there are fewer moles of NaOH (0.050 mol) than HCl (0.075 mol), making NaOH the limiting reactant. To find the heat released during the reaction, we multiply the moles of the limiting reactant by the enthalpy of neutralization. Heat released: \( q = moles_{NaOH} * \Delta H \) \( q = 0.050 mol * (-56 \, kJ/mol) = -2.80 \, kJ \)

We will now use the heat released, the specific heat capacity of the solution (C), and the initial temperature to calculate the final temperature. The formula we will use is: \( q = mc \Delta T \) Here, m is the mass of the solution (200 g) and \(\Delta T\) is the change in temperature. First, we need to convert the heat released from kJ to J: \( q = -2.80 \, kJ * 1000 \frac{J}{kJ} = -2800 \, J \) Now, rearrange the formula to find \(\Delta T\): \( \Delta T = \frac{q}{mc} = \frac{-2800 \, J}{(200 \, g)(4.184 \frac{J}{^\circ C \cdot g})} = -3.35 ^\circ C \) Finally, add the change in temperature to the initial temperature to find the final temperature: \( T_{final} = T_{initial} + \Delta T = 48.2 ^\circ C - 3.35 ^\circ C = 44.85 ^\circ C \) Thus, the final temperature of the calorimeter contents is approximately \(44.85^{\circ}C\)..