To Determine the Heat of Neutralization of a Strong Acid with a Strong Base and a Weak Acid with a Strong Base
psychological study of i am sam movieDepartment of Pure and Applied Chemistry
College of Arts and Sciences
Visayas State University
Visca, Baybay City, Leyte, 6521-A
Rowena M. Perilla
Date Performed: July 6, 2010
Date Submitted: October 1, 2010
Experiment No. 4
Heat of Reaction
Objective
To determine the heat of neutralization of a strong acid with a strong base and a weak acid with a strong base.
Introduction
Calorimeter Constant
All chemical reactions involve changes in energy as a result of bond breaking and bond formation during the chemical change. This energy change is an important parameter when studying chemical reactions and is normally measured in an insulated vessel called a calorimeter. When using a calorimeter, the chemical reagents being studied are mixed directly in the calorimeter with the temperature recorded both before and after the reaction. If the mass of the material is also recorded, then the change in the energy occurring in the calorimeter can be calculated by the relationship:
Δq = m • Δt • cp
where Δq is the change in energy, m is the mass of the solution, Δt is the temperature change of the solution, and cp is the specific heat of the solution. Since most reactions are carried out in dilute water solutions, thespecific heat of the solution is assumed to be the same as that of water (1.00 cal/g°C or 4.185 J/g°C).
The underlying principle utilized in calorimetry is the law of Conservation of Energy. The basic premise of this principle is that “energy can neither be created nor destroyed but may be converted from one form to another.” In other words, the heat lost (or gained) by the chemical reaction is equal to the heat gained (or lost) by the solution in the calorimeter.
In this experiment we are going to simply mix hot and cold water, determine the change in the temperature, and compare the total amount of energy lost with the total amount of energy gained. If the Law of Conservation of Energy is valid, then
Δqhot water = Δqcold water
where Δqhot water is the energy lost by