Question

compare the values of the enthalpy of neutralization for the three acids( H3CCOH, Na2S2O3, HCl). explain any similarities and/or differences

Answer 1

enthalpy of neutralization of any strong acid like HCl with a strong base (like LiOH,NaOH,KOH) or vice versa is always the same i.e. 57.1 kJ. This is because strong acids ,strong bases and salt that they form are all completely ionized in dilute aqueous solutions ...thus the reaction between any strong acid and strong base for example in the above case may be written as :
NaOH (aq) + HCl(aq) -----> NaCl (aq) + H2O (l)... ΔH = -57.1 kJ/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 and OH- ions to form one mole of H2O.....since strong acid and strong base completely ionize in aqueous solution number of H(+) and OH(-) produced by 1 gram equivalent of strong acid and strong base is always the same ...hence enthalpy of neutralization between a strong acid and strong base is always constant .

The enthalpy of neutralization for a weak acid like CH3COOH will be less than that of a starong acid.

Na2S2O3 is a salt of strong base and weak acid. When dissolved in water it will have a basic property and completely dissociated. Its enthalpy of neutralization will be similar to strong acid/strong base enthalpy.