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

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 = $\frac{mass}{volume}$

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 = $m \times C \times \Delta T$

                   = $38 g \times 4.184 J/g^{o}C \times (38 - 27.3)^{o}C$

                   = 1701.21 J

Now,   milimoles = molarity × volume

                            = $1.4 M \times 19 ml$

                             = 26.6 mmol

Enthalpy of neutralization is as follows.

                      $\frac{1701.21 J}{26.6 mmol}$

                          = 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.