In doing the neutralization (using the same calorimeter as in problem 5) of HF w

Question

In doing the neutralization (using the same calorimeter as in problem 5) of HF with NaOH; 50.0mL of 2.08M NaOH at 22.2oC are mixed with 50.0ml of 2.08 M HF at 22.2oC. After the two solutions are mixed, allow to react, and come to equilibrium temperature; the final temperature was found to be 33.2oC. The resulting solution weighed 100.791 g. In order to obtain the specific heat of the resulting NaF solution. 50mL ( 51.264 g) of distilled water at 64.3oC were added to 52.177 g of the NaF solution from the resulting solution at 22.2oC in the same calorimeter. After equilibrium had been reached, the final temperature was found to be 39.7oC. Calculate (a) the specific heat (cal/g-oC) of the NaF solution, (b) qn (heat of neutralization for the amount used in this experiment) in calories, and (c) delta H(n) (Enthalpy of neutralization of HF with NaOH) in the units of (1) cal/mol, (2) kcal/mol, and (3) kJ/mol.

Explanation / Answer

a)

Given,

Mass of NaF solution = 52.177 g

Initial Temp.of NaF solution = 22.2 degree C

Mass of distilled water = 51.264 g

Initial Temp. of distilled water = 64.3 degree C

Final Temp. of Mixture = 39.7 degree C

We know that,

Specific Heat capacity of water = 1 cal / g K

Let specific Heat of NaF solution = c cal / g K

Assuming no heat is lost to the surrounding

Heat lost by water = Heat gained by NaF solution

=> 51.264 x 1 x (64.3 - 39.7) = 52.177 x c x (39.7 - 22.2)

=> c = 1.38 cal / g K = Specific heat of NaF solution.

b) heat of neutralization for the amount used in this experiment

Weight of solution = 100.791 g

specific heat = 1.38 cal / g K

Initial T = 22.2 degree C

Final T = 33.2 degree C

Heat of neutralization = 100.791 x 1.38 x (33.2 - 22.2) = 1530 cal

c)

Moles of NaOH used = 0.05 x 2.08 = 0.104 moles = Moles of HF used

Enthalpy of Neutralization is defined per mole of the reaction

=> Enthalpy of Neutralization = 1530 / 0.104 = 14711.5 cal / mol

= 14.71 kcal / mol

= 14.71 x 4.18 = 61.5 kJ / mol

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