Question

the enthalpy change when a strong acid is neutralized by base is -56.1 kJ/mol.
If 135 ml of .450 M HI at 23.15 C is mixed with 145 of .500 M NaOH at 23.15 C What will the max temperature reached by the resulting solution?
tamana: Assume the specific heat is 4.18 J/gC and dencity is the same as H2O

Answer 1

HI + NaOH ==> NaI + H2O
mmoles HI = 135 mL x 0.450 = 60.75
mmoles NaOH = 145 x 0.500 = 72.5
Obviously, the NaOH is is excess, therefore there will be formed 60.75 mmoles of the salt (said another way, the neutralization reaction occurred with 0.06075 moles of the acid/base.
heat produced = 56,100 J/mol x 0.06075 = 3408 J
Then plug that into
3408 = mass water x specific heat water x (Tfinal-Tinitial) and solve for Tfinal.
If you want to do a little closer job, you can note that you will produce an extra 0.06075 moles H2O which is about 1.09 grams and that can be added to the volume from the acid and base. Check my work.

Answer 2

-10.4