Aim of Experiment: To determine the enthalpy of neutralization of a weak acid and a strong base by using Dewar's Flask.

Theory: A known volume of an weak acid, CH₃COOH solution of exactly known conc. is allowed to react completely with a strong alkali (of known concentration) in dilute solution and the rise in temperature is determined. Knowing the heat capacity of calorimeter with the thermometer immersed, the masses of alkali and acid solution and their specific heats, the amount of heat liberated can be obtained. With the volume and concentration of CH₃COOH solution known, the heat of neutralization can be calculated. If Q is the amount of heat evolved when V cc. of CH₃COOH solution of conc. c (in normailty i.e., gram-equivalent dm^-3) is neutralized completely then,

Heat of neutralization = (1000 Q) / (V.c )

Requirements:  1) Dewar's vacuum flask
                           2) Ring stirrer
                           3) Thermometer
                           4) 0.25 N HCl solution
                           5) 0.25 N NaOH solution

Procedure: Since the heat capacity of Dewar's flask can not be determined in the usual way, it is necessary to determine the water equivalent of the flask along with thermometer and stirrer.

Determination of water equivalent of Dewar's flask by using a reaction of well establish heat effect (the strong acid - strong base neutralization reaction):

1) 250 cc. of 0.25 N NaOH is placed in the flask and fitted with a stirrer, a thermometer and a tap funnel in the lid of the flask and the temperature is measured.
2) Taking 250 cc. of 0.25 N HCl ,the temperature is measured by immersing another thermometer in to it.
3) The acid solution is allowed to run quickly in to the flask and the amount of mix up is recorded . The mixture is stirred quickly and the temperature reading every minute for about 8 minutes after mixing is recorded.
4) After this an appropriate temperature correction is applied (by plotting temperature against time) and the correct highest temperature at the moment of mixing is found.

Calculation:

Let  t₁ , t₂ and t₃ are the temperature of alkali, acid and highest temperature reading at the mixing. The water equivalent "C" of flask together with thermometer and stirrer will be calculated by the following equation:

                   ( m S + C' ) [ t₃ - ( t₁ + t₂) / 2 ] = (1 / 16) (13.67) (10³) cal.................(1)

                    Where  m = Total volume of acid and alkali mixed

                                S = The heat capacity of solutions and it's value is taken to be equal to  1 cal/g of the solution

                                Q =     The amount of heat evolved during the reaction ,we have

                                Q = 250 .S.d.( t₃ - t₁) + 250 .S.d ( t₃ - t₂) + C' ⁽ t₃ - t₁)

Taking heat capacity per gram and density equal to unity, we get

                                Q = 250 (t₃ - t₁) + 250 (t₃ - t₂) +  C ^{' (} t₃ - t₁) .......................(2)

Observation:

The temperature of NaOH (t₁)    :
The temperature of HCl     (t₂ )    :

Table for determination of temperature reading during the reaction of time intervals: 

These values of temperature reading are plotted against the temperature intervals and from these the higher temperature reading  t₃ is measured and is found to be ..........

Now from equation (1) mentioned earlier ,we can calculate the first water equivalent of the flask ,together with thermometer and stirred by putting the value of  t₁ , t₂ , t₃ and S  as:

( m S + C' ) [ t₃ - ( t₁ + t₂) / 2 ] = (1 / 16) (13.67) (10³) cal 

or  .....................................................................................(getting the value of C' )

Then putting the value of C' in equation (2),we can calculate the amount of heat evolved in the reaction as:

Q = 250 (t₃ - t₁) + 250 (t₃ - t₂) +  C^{' (} t₃ - t₁) 

or ..................................................................

Therefore, the heat of neutralization of CH₃COOH (weak acid) is:

                        (1000.Q) / (V.c ) = .................