Calorimeter question!?

Question

a quantity of 400 ml of .6 molar HNO3 is mixed with 400 ml of .3 molar Ba(OH)2 in a constant pressure calorimeter. The initial temperatures of both solutions is the same at 18.46 C. What is the final temp of the solution.

The problem required going back to a previous problem. In the problem, the q rxn = -2.81 kj, and q soln = +2.81 kj

and the heat of neutralization was -56.2 kj/mol.

that answer is supposed to be 22.49 C, but i keep getting around 19.2?

help please if you know how!!

Answer 1

When combining an acid and a base you get water and the heat of reaction is actually the heat of neutralization, aka the heat released as HOH is formed from the ions.

H+ + OH- -->  HOH  DH = -55.84 kJ

With 400 mL of 0.6M and 400 mL of 0.3M you will be combining 0.24 mol H+ and 0.24 mol OH- to make 0.24 mol H2O, with

DH = 0.24 mol x -55.84 kJ/mol = -13.4 kJ.

q = mcDT = 800 g x 4.18J/gC x (Tf-Ti)

DT = q / mc = 13,400 J / 800 g / 4.18 J/gC = 4.0C

If the initial temp is 18.46C then 4 more is 22.49 C.

You're values of the heat of reaction  (2.81 kJ) are too low.  And since I didn't know the density of the solutions, I just assumed that 400 mL of 0.6M HNO3 is 400 grams, and did the same for the NaOH.

The 4.0 C will only be an approximation because of the assumptions about density and specific heat.