To calculate the enthalpy change for the precipitation reaction, we can use the following formula:H_reaction = H_f products - H_f reactants First, we need to determine the moles of each reactant involved in the reaction. Since we have 50 mL of 0.1 M lead nitrate solution and 50 mL of 0.1 M sodium chloride solution, we can calculate the moles as follows:moles of Pb NO3 2 = volume concentration = 0.050 L 0.1 mol/L = 0.005 molmoles of NaCl = volume concentration = 0.050 L 0.1 mol/L = 0.005 molNow, we can use the stoichiometry of the balanced chemical equation to determine the moles of products formed:1 mol Pb NO3 2 reacts with 2 mol NaCl to form 1 mol PbCl2 and 2 mol NaNO3.Since we have equal moles of Pb NO3 2 and NaCl, the reaction will go to completion, and we will have 0.005 mol of PbCl2 and 0.005 mol of NaNO3 formed.Now we can calculate the enthalpy change for the reaction using the standard enthalpies of formation:H_reaction = [ 1 mol PbCl2 -359.57 kJ/mol + 2 mol NaNO3 -467.93 kJ/mol ] - [ 1 mol Pb NO3 2 -460.56 kJ/mol + 2 mol NaCl -407.33 kJ/mol ]H_reaction = [ -359.57 kJ + -935.86 kJ ] - [ -460.56 kJ + -814.66 kJ ]H_reaction = -1295.43 kJ - -1275.22 kJ H_reaction = -20.21 kJ/molSince we have 0.005 mol of reactants, the enthalpy change for this specific reaction is:H = -20.21 kJ/mol 0.005 mol = -0.10105 kJThe enthalpy change involved in the precipitation reaction between 50 mL of 0.1 M lead nitrate solution and 50 mL of 0.1 M sodium chloride solution at 25C is approximately -0.101 kJ.