Question

A student mixes 50.0ml of 0.10M HCl and 150.0ml of 0.10M NaOH solutions. The calorimeter is assumed to be an ideal calorimeter. The molar heat of neutralization of this reaction is -56.5kj/mol. Determine the change in temperature in this experiment.

Answer 1

we are adding less of HCl.

So, HCl is limiting reagent.

mol of HCl reacting = M(HCl)*V(HCl)

= 0.10 M * 50.0 mL

= 0.10 M * 0.050 L

= 5.0*10^-3 mol

delta H neutralisation = -56.5 KJ/mol

Q = mol of acid/base reacted * delta H neutralisation

= (5.0*10^-3 mol)*(-56.5 KJ/mol)

= -0.2825 KJ

= -282.5 J

This Q will be absorbed by solution and increase the temperature.

For solution,

Q = 282.5 J

volume of solution = 50.0 mL + 150.0 mL

= 200.0 mL

since density of solution is 1g/mL,

mass of solution = 200.0 g

specific heat capacity of water,

C = 4.184 J/g.oC

use:

Q = m*C*delta T

282.5 J = 200.0 g * 4.184 J/g.oC * delta T

delta T = 0.338 oC

Answer: 0.338 oC