Question

Determine the enthalpy of neutralization in Joules/mmol for a solution resulting from 19 mL of 1.4 M NaOH solution and 19 mL of a HCl with the same molarity. If separately, each had a temperature of 27.3 degrees Celsius, and upon addition, the highest temperature reached by the solution was graphically determined to be 38 degrees Celsius. Round to the nearest whole number.

Answer 1

Step-by-step explanation:

Total volume of the solution is as follows.

19 ml + 19 ml = 38 ml

As, density is the mass divided by volume.

Mathematically, Density =
`(mass)/(volume)`

Now, calculate the mass of solution as follows.

mass of solution = Density × volume

= 1.00 g/ml × 38 ml

= 38 g

Specific heat of the solution is 4.184 J/g°C.

Relation between heat energy, mass, specific heat and temperature change temperature as follows.

Q =
`m * C * \Delta T`

=
`38 g * 4.184 J/g^(o)C * (38 - 27.3)^(o)C`

= 1701.21 J

Now, milimoles = molarity × volume

=
`1.4 M * 19 ml`

= 26.6 mmol

Enthalpy of neutralization is as follows.

`(1701.21 J)/(26.6 mmol)`

= 63.95 J/mmol

or, = 64 J/mmol

Thus, we can conclude that the enthalpy of neutralization in Joules/mmol for given solution is 64 J/mmol.