Question

A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capacity of 325 J/°C. If the initial temperatures of both solutions are the same at 18.24°C, what is the final temperature of the mixed solution? The heat of neutralization is −56.2 kJ/mol. Assume the density and specific heat of the solutions are the same as those for water.

Answer 1

Explanation:

The given data is as follows.

         $V_{1}$ = 85.0 ml,        $M_{1}$ = 0.9 M

         $V_{2}$ = 85.0 ml,        $M_{1}$ = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = $0.9 M \times 0.085 L$        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              $56.2 kJ/mol \times 0.076 mol$

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = $mC \Delta T$

                     4271 J = $170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C$    

                     0.0773 = $T_{f} - 18.24$

                    $T_{f}$ = $18.317^{o}C$  

Thus, we can conclude that final temperature of the mixed solution is $18.317^{o}C$.