Question

Calculate the heat of mixing when 50.0 mL of 1.00 M HCl is added to 50.0 mL of 1.00 M NaOH at 25°C. The heat capacity of the solution is 4.18 J/(g·°C) and the density of the final solution is 1.00 g/mL. Assume that the heat change is equal to the enthalpy change.

Answer 1

To calculate the heat of mixing, we first need to determine the moles of HCl and NaOH in the solution. Then, we can use the enthalpy of neutralization to find the heat released during the reaction.1. Calculate moles of HCl and NaOH:moles of HCl = volume  L   concentration  M moles of HCl = 0.050 L  1.00 M = 0.050 molesmoles of NaOH = volume  L   concentration  M moles of NaOH = 0.050 L  1.00 M = 0.050 moles2. Determine the limiting reactant:Since both HCl and NaOH have the same number of moles, they will react completely with each other, and there will be no limiting reactant.3. Calculate the enthalpy change:The enthalpy of neutralization for the reaction between a strong acid  HCl  and a strong base  NaOH  is approximately -57.32 kJ/mol. Since both reactants have 0.050 moles, the enthalpy change for the reaction is:H = moles  enthalpy of neutralizationH = 0.050 moles  -57.32 kJ/mol = -2.866 kJ4. Calculate the heat of mixing:Since the heat change is equal to the enthalpy change, the heat of mixing is -2.866 kJ.Note: The heat capacity and density of the solution were not needed for this calculation, as we used the enthalpy of neutralization to find the heat released during the reaction.