Question

Sodium hydroxide (NaOH) has a lattice energy of -887 kJ/mol and a heat of hydration of -932 kJ/mol.

How much solution could be heated to boiling by the heat evolved by the dissolution of 24.5 g of NaOH? (For the solution, assume a heat capacity of 4.0 J/g⋅∘C, an initial temperature of 25.0 ∘C, a boiling point of 100.0 ∘C, and a density of 1.05 g/mL.)

Express your answer using two significant figures.

Answer 1

Solution-

We can write the lattice energy as-
Na+ + OH- ----> NaOH.. .dHrxn = -887 kJ/mole

And the heat of hydration can be given as
Na+ + OH- + mH2O ----> infinitely dilution solution of Na+ + OH-....dHhyd = -932 kJ/mole


We can find the find the heat of dissolution as follow

NaOH ---> Na+ + OH- ---------- dHrxn = +887 kJ/mole
Na+ + OH- + mH2O ---> dilution-----dHrxn = -932 kJ/mole (heat of hydration)
________________________________________________.
NaOH ----> diluted Na+ + OH- solution... dHrxn = -45 kJ/mole

from a heat balance
heat evolved by dissolution of NaOH = heat gained by solution


(m x dHdissolution) = (m Cp dT) solution
24.5g x (1 mole / 40.00g) x (45 kJ/mole) = X mL x (1.05g / mL) x (0.0040 kJ/gC) x (100.0°C - 25.0°C)
X = (27.56 kJ / (0.315 kJ/mL) = 87.49 mL