- a) Write the net ionic equation for the reaction between hydrochloric acid and aqueous sodium hydroxide, then
calculate the heat of neutralization (AH) for the net ionic equation using the standard enthalpies of formation ( found in the Appendix of the textbook. ΔΗ
Molecular equation: HCl(aq) + NaOH(aq) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(1)
Net ionic equation: H+(aq) + OH-(aq) → H2O(1) ΔΗ
b) Do the same for the reaction between nitric acid and aqueous sodium hydroxide.
Molecular equation: HNO (aq) + NaOH(aq) HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(I)
Net ionic equation: H+(aq) + OH-(aq) → H2O(1) ΔΗ=
c) Calculate the percent error for your experimental AH, for each reaction using the following formula. Show method.
% error = jexperimental value – true value
E
% error for HCI + NaOH:
% error for HNO, + NaOH:
The Correct Answer and Explanation is :
a) Hydrochloric Acid and Sodium Hydroxide
Molecular Equation:
[
\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}
]
Net Ionic Equation:
[
\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O(l)}
]
Heat of Neutralization (ΔH):
The heat of neutralization for strong acid-strong base reactions (such as HCl and NaOH) is typically about (-57.3 \, \text{kJ/mol}) because both the acid and base fully dissociate in aqueous solution. To calculate the heat of neutralization, you need the standard enthalpy of formation (ΔHf) for each reactant and product, which are typically given in textbooks.
For this reaction:
- (\Delta H_f) for ( \text{HCl(aq)} ) and ( \text{NaOH(aq)} ) are provided in your textbook.
- The final state, ( \text{NaCl(aq)} ) and ( \text{H}_2\text{O(l)} ), have their own standard enthalpies.
The total heat of neutralization can be calculated using the equation:
[
\Delta H = \sum \Delta H_f (\text{products}) – \sum \Delta H_f (\text{reactants})
]
The reaction releases heat due to the exothermic nature of acid-base neutralization.
b) Nitric Acid and Sodium Hydroxide
Molecular Equation:
[
\text{HNO}_3\text{(aq)} + \text{NaOH(aq)} \rightarrow \text{NaNO}_3\text{(aq)} + \text{H}_2\text{O(l)}
]
Net Ionic Equation:
[
\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O(l)}
]
Heat of Neutralization (ΔH):
The heat of neutralization for nitric acid (HNO₃) and sodium hydroxide (NaOH) is also approximately (-57.3 \, \text{kJ/mol}), similar to the hydrochloric acid and sodium hydroxide reaction. This is because both are strong acid and strong base, which dissociate completely in aqueous solution.
c) Percent Error Calculation
To calculate percent error for the heat of neutralization (( \Delta H )):
[
\text{Percent error} = \frac{| \text{experimental value} – \text{true value} |}{\text{true value}} \times 100\%
]
Example Percent Error Calculation:
Let’s assume the experimental value for the heat of neutralization is ( \Delta H = -55.0 \, \text{kJ/mol} ) for both reactions, and the true value from literature is ( \Delta H = -57.3 \, \text{kJ/mol} ).
For HCl + NaOH:
[
\text{Percent error} = \frac{| -55.0 – (-57.3) |}{-57.3} \times 100\% = \frac{2.3}{57.3} \times 100\% \approx 4.0\%
]
For HNO₃ + NaOH:
[
\text{Percent error} = \frac{| -55.0 – (-57.3) |}{-57.3} \times 100\% = \frac{2.3}{57.3} \times 100\% \approx 4.0\%
]
The percent error for both reactions is approximately 4.0%. This error could be due to experimental inaccuracies, such as heat loss to the environment or measurement errors in temperature.
Now, I will generate an image that illustrates the equations and heat calculations for better understanding.
Here is the diagram showing the molecular and net ionic equations for the reactions between hydrochloric acid and sodium hydroxide, as well as between nitric acid and sodium hydroxide. It also includes a section on calculating the heat of neutralization (ΔH) and how to determine percent error. This should help visualize the concepts more clearly. Let me know if you’d like further clarification or adjustments!
