To calculate the standard enthalpy change for the neutralization of HCl with NaOH, we can use the following equation:q = mcTwhere q is the heat released or absorbed, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, we need to determine the moles of HCl and NaOH in the solution:moles of HCl = volume concentration = 0.050 L 0.10 mol/L = 0.005 molmoles of NaOH = volume concentration = 0.050 L 0.10 mol/L = 0.005 molSince the moles of HCl and NaOH are equal, the reaction will go to completion, and the neutralization reaction is as follows:HCl aq + NaOH aq NaCl aq + HO l Now, we need to find the mass of the solution:mass = volume density = 0.050 L + 0.050 L 1000 mL/L 1.00 g/mL = 100 gNext, we need to find the change in temperature T . Since the reaction is exothermic, the temperature of the solution will increase. The standard enthalpy change for the neutralization of a strong acid and a strong base is approximately -57.1 kJ/mol. We can use this value to find the change in temperature:q = -57.1 kJ/mol 0.005 mol = -0.2855 kJ = -285.5 JNow, we can use the equation q = mcT to find the change in temperature:-285.5 J = 100 g 4.18 J/ g C TSolving for T:T = -285.5 J / 100 g 4.18 J/ g C = -0.682 CSince the reaction is exothermic, the temperature of the solution will increase by 0.682 C. The standard enthalpy change for the neutralization of 50 mL of 0.10 M HCl with 50 mL of 0.10 M NaOH at 25C is -285.5 J or -0.2855 kJ.