A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$ \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?

The given reaction is a neutralization reaction between H+ from HCl and OH- from Ba(OH)2 to form water. The HCl solution is 0.862 M with a volume of 200 mL (or 0.2 L) which gives us \(0.862 mol/L * 0.2 L = 0.1724 mol\) of H+. The Ba(OH)2 solution is 0.431 M with a volume of 200 mL (or 0.2 L) which gives us \(0.431 mol/L * 0.2 L = 0.0862 mol\) of Ba(OH)2. Because each Ba(OH)2 provides two OH-, we have \(0.0862 mol * 2 = 0.1724 mol\) of OH-. This makes H+ and OH- react in a 1:1 ratio, so none of them are in excess.

In neutralization reactions, the heat of reaction can be calculated as heat = moles of limiting reactant * heat of neutralization. Here, we have 0.1724 mol as the number of moles (from both H+ and OH-) and -56.2 kJ/mol as the heat of neutralization. Hence, heat transferred = \(0.1724 mol * -56.2 kJ/mol = -9.68 kJ\). The negative sign indicates that the heat is released in the reaction.

We can now calculate the change in temperature using the formula q = mcΔT where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution and ΔT is the change in temperature. We can assume that the specific heat capacity of the solution is the same as that of water (4.18 J/g°C). Also, the mass of the solution is approximately equal to the total volume in grams as the density of water is approximately 1g/mL. So, m = 200 mL + 200 mL = 400 g. Re-arranging the formula for ΔT gives ΔT = q/(mc). Substituting, we get ΔT = \(-9.68 kJ * 1000 / (400 g * 4.18 J/g°C) = -5.79°C\). The final temperature would be the initial temperature plus the change, \(20.48°C - 5.79°C = 14.69°C\).

Given data has three significant figures. Therefore, rounding off the final temperature to the appropriate significant figures gives 14.7°C.