Question

In a coffee cup calorimeter, 50.0 mL of 1.00 M NaOH and 50.0 mL of 1.00 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Given that the density of the NaCl solution is 1.038 g/mL and he specific of NaCl solution is 3.87 J/g-°C, calculate the ΔHneut/mole for the reaction of HCl with NaOH. Assume that no heat is lost to the surroundings.

Part A. ) Why are the density and specific heat of NaCl solution (instead of those of sodium hydroxide and hydrochloric acid) given in the question?

Which one is correct answer?

Sodium hydroxide and hydrochloric acid react to give sodium chloride and water.

Sodium hydroxide is acidic.

hydrochloric acid is basic

not sure

Part B.) Why is the density of the sodium chloride higher than that of water?

Which one is the correct answer?

The volume of the sodium chloride solution is much larger than that of the water., resulting in higher density for the solution.

The mass of the sodium chloride is added to water but the volume of the solution does not change much, resulting in higher density for the solution.

The mass of the sodium chloride is added to water but the volume of the solution also significantly become larger, resulting in higher density for the solution.

not sure

Part C.) If heat is lost to the surroundings, will the ΔHneut/mole for the reaction of HCl with NaOH be higher than the actual value?

Which one is the correct answer:

higher

lower

no change

not sure

Part D.)

Assume that no heat is lost to the surroundings.

Which one is the correct answer:

-54 kJ/mol

+54 kJ.mol

-52 kJ/mol

+ 52 kJ/mol

Answer 1

Part A.

Sodium hydroxide and hydrochloric acid react to give sodium chloride and water.

Part B

The mass of the sodium chloride is added to water but the volume of the solution does not change much, resulting in higher density for the solution.

part C

lower

part D

q = m*s*DT

m= mass of the mixer = 100*1.038 = 103.8 g

s = specific heat of mixer = 3.87 j/g.c

DT = 31.3 - 24.6 = 6.7

Q = 103.8*3.87*6.7 = 2691.43 joule

no of mole of NaOH reacted = 50*1/1000 = 0.05 mole

DHneu = -q/n = -2.69/0.05 = -53.8 kj/mol

answer: -54 kj/mol