Question

Calculate the heat of mixing when 100 mL of 0.2 M HCl is mixed with 100 mL of 0.1 M NaOH, assuming that the density of the resulting solution is 1 g/mL and the specific heat capacity is 4.18 J/g℃.

Answer 1

To calculate the heat of mixing, we first need to determine the amount of heat released during the neutralization reaction between HCl and NaOH. The balanced chemical equation for this reaction is:HCl + NaOH  NaCl + H2OThe limiting reactant in this reaction is NaOH, as it has a lower concentration  0.1 M  compared to HCl  0.2 M . First, let's find the moles of NaOH and HCl:Moles of NaOH = Volume  L   Concentration  M Moles of NaOH = 0.1 L  0.1 M = 0.01 molesMoles of HCl = Volume  L   Concentration  M Moles of HCl = 0.1 L  0.2 M = 0.02 molesSince NaOH is the limiting reactant, the moles of HCl that will react with NaOH are equal to the moles of NaOH:Moles of HCl that react = 0.01 molesNow, we can calculate the heat released during the reaction. The heat of neutralization for a strong acid and strong base is approximately -57.3 kJ/mol. Heat released = Moles of limiting reactant  Heat of neutralizationHeat released = 0.01 moles  -57.3 kJ/mol = -0.573 kJNow, we need to calculate the temperature change of the solution. The total volume of the solution is 200 mL, and since the density is 1 g/mL, the mass of the solution is 200 g. We can use the formula: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. We can rearrange the formula to solve for T:T = q /  mc T = -0.573 kJ /  200 g  4.18 J/g T = -573 J /  836 J/ T = -0.685 Since the heat of mixing is the heat released during the reaction, the heat of mixing is -0.573 kJ.