To calculate the standard enthalpy change for the reaction, we first need to write the balanced chemical equation for the reaction between HCl and NaOH:HCl aq + NaOH aq NaCl aq + H2O l Since both solutions have the same concentration 1.0 M and the same volume 50 mL , the reaction will go to completion, and all of the HCl and NaOH will react with each other.Next, we need to calculate the heat released during the reaction. The heat released q can be calculated using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.Since the density of the solution is 1.00 g/mL, the mass of the solution can be calculated as follows:mass = volume densitymass = 50 mL HCl + 50 mL NaOH 1.00 g/mLmass = 100 gThe temperature change T can be calculated using the heat released during the reaction and the specific heat capacity of the solution:T = q / mc We know that the specific heat capacity c is 4.18 J g K, but we don't have the heat released q yet. To find the heat released, we can use the enthalpy change of the reaction H . The enthalpy change for the reaction between HCl and NaOH can be found in literature or a reference table, and it is approximately -57.1 kJ/mol.Since both solutions have the same concentration 1.0 M and the same volume 50 mL , the number of moles of HCl and NaOH can be calculated as follows:moles = concentration volumemoles = 1.0 mol/L 0.050 Lmoles = 0.050 molNow we can calculate the heat released q during the reaction:q = H molesq = -57.1 kJ/mol 0.050 molq = -2.855 kJNow we can calculate the temperature change T :T = q / mc T = -2855 J / 100 g 4.18 J g K T = -6.82 KSince the reaction is exothermic H is negative , the temperature of the solution will increase. Therefore, the final temperature of the solution will be:T_final = T_initial + TT_final = 25C + 6.82CT_final 31.82CThe standard enthalpy change for the reaction is -57.1 kJ/mol.