Question

Suppose you perform a calorimeter experiment to determine the molar heat of neutralization of an unknown acid, H A HA, with sodium hydroxide, N a O H NaOH. You mix 37.2 mL of 0.50 M H A HA with 56.8 mL of 0.75 M N a O H NaOH and calculate the heat of reaction as -1.6 kJ. What is the molar heat of neutralization (in kJ/mol) for the unknown acid

Answer 1

Answer:

-86.02 kJ/ mole

Explanation:

The moles of the acid used = Molarity × Volume (L) =

= 0.50 (0.0372 L)

= 0.0186 moles

The heat released = -1.6 kJ

∴ 0.0186 moles neutralization of HA heat is: -1.6 kJ

The molar heat of neutralization due to one mole of the unknown acid = -1.6/0.0186

= -86.02 kJ/ mole