Homework Answers
Moles of acid =1mol × .08 L/ 1 L
= .08 mol
Moles of base ( as it is dibasic acid its moles will be double then molarity)
=2× 1 mol × .08 L/ 1L
= .16 mol
Moles of base involved in reaction= .08 mol
Total number of moles of acid+ base=
0.08+0.08= 0.16 mol
Volume of solution= 80+80= 160 ml
Mass of solution= 160 g
q= mc( T2-T1)
q= 160g× 4.184J/g°C×(35°C- 20°C)
q=10041.6 J
Enthalpy of neutralisation= q/n
∆H= 10041.6/ moles of acid+ base
= 62760 J /mol
= 62.760 kJ/ mol
This value is little high as heat capacity of calorimeter is included in it.
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