Question

B. ΔH_solution for neutralization of HCl_(aq) and NaOH_(aq):
Volume of HCl: 0.05 L
Volume of NaOH: 0.05 L
Volume total: 0.10 L
Molarity of HCl: 2.08 M
Molarity of NaOH: 2.08 M
ΔT for reaction B: 10.8°C

1. Calculate the molarity of the resulting sodium chloride solution.

3. Calculate the value (calories) for the heat of reaction.

4. Calculate ΔH for neutralization.

Answer 1

mole of NaOH = 0.05 L * 2.08 mole / L = 0.104 mole

and

mole of HCl = 0.05 L * 2.08 mole / L = 0.104 mole

One mole NaOH react with one mole HCl to form one mole NaCl.

thus

mole of NaCl formed = 0.104 mole.

thus

molarity of NaCl = 0.104 mole / 0.10 L = 1.04 M (answer)

assuming the specific heat of solution = that of water = 1.00 cal / g-oC

and

density of solution = 1.0 g / ml.

0.10 L = 0.10 * 1000 = 100 ml

mass of solution = 100 g

we know,

heat = m * s * ΔT = 100 g * 1.0 cal /g- oC * 10.8 oC = 1080 cal

and

delta H for neutralization = - 1080 cal / 0.104 mole = - 10384.6 cal / mole = - 10.38 Kcal / mole