A quantity of 8.00 x 10 2 mL of 0.600 M HNO 3 is mixed with 8.00 x 10 2 mL of 0.

Question

A quantity of 8.00 x 102 mL of 0.600 M HNO3 is mixed with 8.00 x 102 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initital temperature of both solutions is the same at 18.46 degrees C. The heat of neutralization when 1.00 mol of HNO3 reacts with 0.500 mol Ba(OH)2 is -56.2 kJ/mol. Assume that the densities and specific heats of the solution are the same as for water (1.00 g/mL and 4.184 J.g* degrees C, respectively). What is the final temperature of the solution?

_____ degrees C

Explanation / Answer

Given:

Given volume of HNO3 = 8.00 X 102 mL = 0.8 L

Molarity of HNO3 = 0.600 M

Volume of Ba(OH)2 = 0.8 L

Molarity of Ba(OH)2 = 0.300 M

We calculate the moles of each

Mole of HNO3 = Volume of HNO3 * molarity of HNO3 = 0.8 L * 0.600 M = 0.48 mol

Mol Ba(OH)2 = 0.8 L * 0.300 M = 0.24 mol

Reaction :

Ba(OH)2 +2 HNO3 -->Ba(NO3)2 + 2H2O

Mol ratio of Ba(OH)2 : HNO3 is 1 : 2

Lets calculate limiting reactant.

1 mol Ba(OH)2 needs 2 mol HNO3

Mol of HNO3 required to react completely with Ba(OH)2

= 0.24 mol Ba(OH)2 * 2 mol HCl / 1 mol Ba(OH)2

= 0.48 mol HCl

Mole of HCl = 0.48 mol

So both are present correct proportion

Lets calculate the mass of the solution

Mass of the solution = volume of solution in mL * density

=( 800 mL + 800 mL ) * 1 g /mL

= 1600 g

Now q = m c delta T

m = mass in g , c = specific heat , delta T = change in T = Tf- Ti

q is heat absorbed / evolved.

Lets calculate q

q =- ( heat of neutralization * mol HCl )

= 56.2 kJ / mol * 0.48 mol HCl           

= 26.9760 kJ

= 26976.0 J

q = 26976.0 J

Lets plug all the values in above equation

26976 J = 1600 g * 4.184 J / deg C * ( Tf- 18.46 deg C )

4.029 deg C = Tf- 18.46 deg C

Tf =22.89 deg C

Final temperature of the solution = 22.89 deg C

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