To determine the enthalpy change of precipitation for the reaction between NaOH and CuSO4, we first need to write the balanced chemical equation for the reaction:NaOH aq + CuSO4 aq Na2SO4 aq + Cu OH 2 s Next, we need to determine the limiting reactant. Since both solutions have the same concentration and volume, they will react in a 1:1 ratio, and neither is in excess.Now, we need to calculate the moles of the reactants:moles of NaOH = 0.1 M 0.05 L = 0.005 molmoles of CuSO4 = 0.1 M 0.05 L = 0.005 molSince the reaction is exothermic, the heat released during the reaction will cause a temperature increase in the solution. We can use the specific heat capacity formula to calculate the heat released:q = mcTwhere q is the heat released, m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.First, we need to find the mass of the solution. Assuming the density of the solution is approximately the same as water 1 g/mL , the mass of the 100 mL solution is:mass = 100 mL 1 g/mL = 100 gNext, we need to find the temperature change. Since the final temperature is 25C, and we can assume the initial temperature of both solutions was also 25C, the temperature change is:T = 0CSince there is no temperature change, the heat released q is:q = 100 g 4.18 J/gC 0C = 0 JTherefore, the enthalpy change of precipitation for this reaction is 0 J. However, this result is not realistic, as there should be some heat released during the reaction. The assumption that the initial and final temperatures are the same might not be accurate. To obtain a more accurate enthalpy change, the initial temperature of the solutions should be measured before mixing, and the final temperature should be measured after the reaction has occurred.