Question

lab data
concentration of hydrochloric acid (m) 2.11
concentration of sodium hydroxide (m) 2.05
calorimeter constant (j/°c) 13.4
volume of hydrochloric acid (ml) 93.0
temperature of hydrochloric acid (°c) 24.0
volume of sodium hydroxide (ml) 90.5
temperature of sodium hydroxide (°c) 24.0
final temperature after mixing (°c) 37.8
mass of mixed solution (g) 183.5
moles of h+ (mol) 0.196
moles of oh- (mol) 0.186
limiting reactant oh-
δhneutralization (kj/mol)
how to calculate the enthalpy of neutralization

Explanation:

Step1: Calculate heat absorbed by solution

The formula for heat absorbed by the solution is $q_{solution} = mc\Delta T$. Assuming the specific heat capacity $c$ of the solution is $4.184\ J/g^\circ C$ (same as water), mass $m = 183.5\ g$, $\Delta T = 37.8 - 24.0 = 13.8^\circ C$.
So $q_{solution} = 183.5\ g\times4.184\ J/g^\circ C\times13.8^\circ C$.
$q_{solution}=183.5\times4.184\times13.8\approx183.5\times57.74\approx10595\ J = 10.595\ kJ$.

Step2: Calculate heat absorbed by calorimeter

The formula for heat absorbed by calorimeter is $q_{calorimeter} = C\Delta T$, where $C = 13.4\ J/^\circ C$, $\Delta T = 13.8^\circ C$.
So $q_{calorimeter}=13.4\ J/^\circ C\times13.8^\circ C = 184.92\ J = 0.18492\ kJ$.

Step3: Total heat released by reaction

The total heat released by the reaction $q_{reaction}=-(q_{solution}+q_{calorimeter})$.
$q_{solution}+q_{calorimeter}=10.595 + 0.18492\approx10.7799\ kJ$.
So $q_{reaction}=- 10.7799\ kJ$.

Step4: Moles of limiting reactant

The limiting reactant is $OH^-$ with moles $n = 0.186\ mol$.

Step5: Calculate enthalpy of neutralization

The formula for $\Delta H_{neutralization}=\frac{q_{reaction}}{n}$.
$\Delta H_{neutralization}=\frac{- 10.7799\ kJ}{0.186\ mol}\approx - 57.96\ kJ/mol$.

Answer:

$\approx - 58.0\ kJ/mol$ (rounded to three significant figures)