A quantity of 5.70 x 10^2 mL of .600 M HNO3 is mixed w/ 5.70 x 10^2 mL of .300 M
ID: 857604 • Letter: A
Question
A quantity of 5.70 x 10^2 mL of .600 M HNO3 is mixed w/ 5.70 x 10^2 mL of .300 M Ba(OH)2 in a constant pressure calorimeter of negligblie heat capacity. The inital temperature of both solutions is the same at 18.46 C. The heat of neutrailzation when 1.00 mol of HNO3 reacts with.500 mol Ba(OH)2 is -56.2 kJ.mol. Assume that the densities and specific heats of the solutionsa are the same as for water (1.00 g/mL and 4.184 J/g x C, respectively) What is the final temperature of the solution?
Please show work, I am so lost!
Explanation / Answer
The reaction here is:
Ba(OH)2 + 2HNO3 = Ba(NO3)2 + 2H2O
5.7*10^2 mL of 0.3M Ba(OH)2 solution contains 0.17 moles of Ba(OH)2.
5.7*10^2 mL of 0.6M HNO3 solution contains 0.34 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):
0.17 moles * -56.2 kJ/mol = 9554 J of energy as heat
The total volume of solution is the sum of the volumes of the barium hydroxide and nitric acid solutions = 570 mL + 570 mL = 1140 mL. With the assumptions given, the heat capacity of this volume of solution is:
cp = 1140 mL * 1 gm/mL * 4.184 J/(g*degC) = 4765.2 J/degC
The increase in temperature if given by:
9554 J/cp = 9554 J / 4765.2 J/degC = 2.004 deg C
The final temperature is given by:
18.46 C + 2.004 C = 20.46 C
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Note added in response to comment.
I checked my arithmetic, and it's ok. The only think I can think of is that they want you to take into account the volume of H2O that is produced by the reaction. This is not really consistent, because if you do that, you should also take into account the heat capacity of the other reaction product, Ba(OH)2.
Nevertheless, let's just take into account the water. Each mole of Ba(OH)2 neutralized produces 2 moles of H2O. We have 0.17 moles of Ba(OH)2 in this case, so 0.34 moles of water are produced.
0.34 moles of water has a mass of 0.34 mol * 18 gm/mol = 6.12 gm and assuming a density of 1 gm/ml for water, this increases the volume of the final solution to 1146.12 ml. (= 1146.12 gm give the assumption about the density of the solutions.
The heat capacity of the final solution is then:
1146.12 gm * 4.184 J/(g*degC) = 4790.78 J/degC
The increase in temperature is given by:
9554 J / 4790.78 J/degC = 1.99 deg C
and the final temperature would be:
18.46 C+ 1.99 C = 20.45 C