The reaction here is:
Ba(OH)2 + 2HNO3 = Ba(NO3)2 + 2H2O
4*10^2 mL of 0.3M Ba(OH)2 solution contains 0.12 moles of Ba(OH)2.
4*10^2 mL of 0.6M HNO3 solution contains 0.24 moles of HNO3, so there is exactly enough nitric acid present to neutralize the Ba(OH)2.
Neutralization of this quantity of Ba(OH)2 releases (note that the enthalpy of neutralization is negative, so energy is released):
0.12 moles * -56.2 kJ/mol = 6744 J of energy as heat
The total volume of solution is the sum of the volumes of the barium hydroxide and nitric acid solutions = 400 mL + 400 mL = 800 mL. With the assumptions given, the heat capacity of this volume of solution is:
cp = 800 mL * 1 gm/mL * 4.184 J/(g*degC) = 3347 J/degC
The increase in temperature if given by:
6744 J/cp = 6744*10^3 J / 3347 J/degC = 2.015 deg C
The final temperature is given by:
18.76 C + 2.015 C = 20.78 C
could someone help me with this question?
A quantity of 400 mL of .6 M HNO3 is mixed with 400 mL of .3 M Ba(OH)2 in a constant pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46 degrees Celsius. What is the final temp of the solution?
the heat of neutralization is -56.2 kJ/mol.
assume the densities and specific heats of the solution are the same as for water. (1.00 g/mL and 4.184 J/gC)
thanks!
1 answer
0 answers