1. Calculate the heat absorbed by the solution:
* Specific heat capacity of water: Assume the solution has the same specific heat capacity as water (4.184 J/g°C).
* Mass of the solution: We'll assume the density of the solution is close to that of water (1 g/mL). Since we have 100 cm³ (which is equal to 100 mL), the mass is approximately 100 g.
* Temperature change: ΔT = 26.8°C
* Heat absorbed (q) = mass × specific heat capacity × ΔT
* q = 100 g × 4.184 J/g°C × 26.8°C
* q = 11200 J ≈ 11.2 kJ
2. Calculate the moles of NaOH:
* Molarity of NaOH: 1.0 M
* Volume of NaOH: 100 cm³ = 0.1 L
* Moles of NaOH = Molarity × Volume
* Moles of NaOH = 1.0 M × 0.1 L = 0.1 mol
3. Calculate the standard heat of neutralization (ΔH):
* ΔH = -q / moles of NaOH (The negative sign is because the reaction releases heat, meaning it's exothermic)
* ΔH = -11.2 kJ / 0.1 mol
* ΔH = -112 kJ/mol
Therefore, the standard heat of neutralization for the reaction of NaOH and HCl is approximately -112 kJ/mol.
Important Notes:
* This calculation assumes that the reaction goes to completion, meaning all of the NaOH and HCl react to form water and salt.
* In reality, the heat of neutralization may vary slightly depending on the specific concentrations of the reactants and the temperature of the surroundings.
* This calculation is an approximation. For more precise results, you would need to consider the heat capacity of the calorimeter used to perform the experiment, and account for any heat losses to the surroundings.
