Question

Calculate the heat of mixing when 100 mL of 0.1 M HCl are added to 100 mL of 0.1 M NaOH at 25°C, given that the specific heat capacity of the resulting solution is 4.18 J g^−1 K^−1 and the density is 1.00 g mL^−1.

Answer 1

To calculate the heat of mixing, we first need to determine the heat released during the neutralization reaction between HCl and NaOH. The balanced chemical equation for this reaction is:HCl aq  + NaOH aq   NaCl aq  + H2O l Since both solutions have the same concentration  0.1 M  and volume  100 mL , they will react completely with each other, forming an equal amount of NaCl and water.The heat released during the reaction can be calculated using the enthalpy change of the reaction, which is -55.9 kJ/mol for the neutralization of HCl and NaOH. We need to determine the number of moles of HCl and NaOH reacting to find the total heat released.Moles of HCl =  0.1 mol/L  *  100 mL  *  1 L/1000 mL  = 0.01 molMoles of NaOH =  0.1 mol/L  *  100 mL  *  1 L/1000 mL  = 0.01 molSince both moles are equal, the reaction goes to completion. Now, we can calculate the heat released:Heat released =  0.01 mol  *  -55.9 kJ/mol  = -0.559 kJNow, we need to determine the heat absorbed by the resulting solution. The total volume of the solution is 200 mL, and since the density is 1.00 g/mL, the mass of the solution is:Mass of solution =  200 mL  *  1.00 g/mL  = 200 gAssuming the temperature change is negligible, we can calculate the heat absorbed by the solution using the specific heat capacity:q = mcTWhere q is the heat absorbed, m is the mass of the solution, c is the specific heat capacity, and T is the temperature change. Since the temperature change is negligible, T is approximately 0, and the heat absorbed by the solution is also approximately 0.Therefore, the heat of mixing is approximately equal to the heat released during the neutralization reaction:Heat of mixing  -0.559 kJ