Question

suppose that you repeated PART B of the experiment using 1M HNO3(aq). How would you expect the △Hneut to compare with that for the neutralization of HCL? explain.

△H = Hfinal - Hinitial = qp = △E + P △V

PART B - Enthalpy of Neutralization of a Stong Acid

The heat of neutralization of a strong acid, hydrochloric acid, will be determined. The strong base is sodium hydroxide of a concentration slightly less than that of the acid.

Procedure

1.In a clean dry graduated cylinder, measure out exactly 50 mL of the NaOH solution and transfer completely to the clean dry calorimeter. Record this temperature and the concentration of the NaOH solution.

2.Meanwhile, rinse out the graduated cylinder with tap water, distilled water and a few mL of the HCl solution. Measure out 50 mL of the HCl solution. Allow it to stand until its temperature is constant and equal to room temperature. Record this temperature and the concentration of the acid solution.

3.Add the acid as quickly as possible to the calorimeter, counting the time from the addition of the first drop of acid. Note the time and temperature after the addition of the last drop, and at 15 sec intervals from then on for 4 minutes. Stir the contents of the calorimeter constantly.

4.Test the contents of the calorimeter with litmus paper. The solution should be slightly acidic(pH<7.0), indicating that the known amount of base was completely consumed in the neutralization

Answer 1

Enthalpy of neutralization of an acid by a base is defined as heatc when one gram equivalent of acid is neutralized by a base,the reaction being carried out in dilute aqueous solution.

Enthalpy of neutralization of base by an acid is defined in asimilar manner.

enthalpy of neutralization of HCl with NaOH is 57.1 kj

enthalpy of neutralization of any strong acid (like HCl,HNO3,H2SO4)with a strong base (like LiOH,NaOH,KOH) or vice versa is always thesame i.e. 57.1 kj.

This is because strong acids ,strong bases andsalt that they form are all completely ionized in dilute aqueous solutions.

NaOH (aq) + HCl(aq) -----> NaCl (aq) + H2O (l) ; H= -57.1kj/mole

They will dissociate as:

Na(+) (aq) + OH(-) (aq) + H(+) (aq) + Cl(-) (aq) ---> Na(+) (aq)+ Cl(-) (aq) + H2O (l)
common ions will cancel out..
H(+) (aq) + OH(-) (aq) ----> H2O (l)

thus neutralization is simply a reaction between H(+) ions given by acids and OH(-) ions given by base to form one mole of H2O.

Since strong acid and strong base completely ionize inaqueous solution number of H(+) and OH(-) produced by 1 gramequivalent of strong acid and strong base is always the same...hence enthalpy of neutralization between a strong acid andstrong base is always constant.

So using HNO3 instead of HCl won't have any effect on neutralization enthalpy.