First, we need to determine the number of moles of HCl and NaOH in the solution.moles of HCl = volume L concentration M moles of HCl = 0.025 L 0.50 mol/L = 0.0125 molmoles of NaOH = volume L concentration M moles of NaOH = 0.025 L 0.50 mol/L = 0.0125 molSince the moles of HCl and NaOH are equal, the reaction goes to completion and all the HCl and NaOH react to form water and NaCl.The heat of reaction H is given as -57.1 kJ/mol. The negative sign indicates that the reaction is exothermic releases heat . To find the heat released during the reaction, we can multiply the moles of the limiting reactant in this case, both HCl and NaOH have the same moles by the heat of reaction.q = moles Hq = 0.0125 mol -57.1 kJ/mol = -0.714 kJNow, we need to find the mass of the resulting solution. Since the density is 1.00 g/mL, we can multiply the total volume by the density.mass = volume densitymass = 50.0 mL 1.00 g/mL = 50.0 gNext, we can calculate the change in temperature T using the heat capacity Cp of the solution and the heat released q .q = mass Cp TT = q / mass Cp T = -0.714 kJ / 50.0 g 4.18 J/gC T = -0.714 10^3 J / 50.0 4.18 T -3.42 CSince the initial temperature was 25 C, the final temperature of the solution after mixing is:T_final = T_initial + TT_final = 25 C - 3.42 C 21.58 CThe heat of mixing when 25.0 mL of 0.50 M HCl solution is added to 25.0 mL of 0.50 M NaOH solution at 25 C is -0.714 kJ, and the final temperature of the solution is approximately 21.58 C.