Question

The neutralization of H 3 PO 4 with KOH is exothermic. H 3 PO 4 ( aq ) + 3 KOH ( aq ) ⟶ 3 H 2 O ( l ) + K 3 PO 4 ( aq ) + 173.2 kJ If 55.0 mL of 0.227 M H 3 PO 4 is mixed with 55.0 mL of 0.680 M KOH initially at 22.62 °C, predict the final temperature of the solution, assuming its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C). Assume that the total volume is the sum of the individual volumes.

Answer 1

Explanation:

It is known that,

      No. of moles = Molarity × Volume

So, we will calculate the moles of $H_{3}PO_{4}$ as follows.

         No. of moles = $0.227 \times 0.055 L$

                                = 0.0125 mol

Now, the moles of KOH are as follows.

       No. of moles = $0.680 \times 0.055 L$

                                = 0.0374 mol

And, $3 \times \text{moles of} H_{3}PO_{4}$ = $3 \times 0.0125$

                             = 0.0375 mol

Now, the balanced reaction equation is as follows.

     $H_{3}PO_{4}(aq) + 3KOH(aq) \rightarrow 3H_{2}O(l) + K_{3}PO_{4}(aq) + 173.2 kJ$

This means 1 mole of $H_{3}PO_{4}$ produces 173.2 kJ of heat. And, the amount of heat produced by 0.0125 moles of $H_{3}PO_{4}$ is as follows.

         M = $\frac{0.0125 mol \times 173.2 kJ}{1}$

             = 2.165 kJ

Total volume of the solution = (55.0 + 55.0) ml

                                               = 110 ml

Density of the solution = 1.13 g/ml

Mass of the solution = Volume × Density

                                  = $110 ml \times 1.13 g/ml$

                                  = 124.3 g

Specific heat = 3.78 $J/g^{o}C$

Now, we will calculate the final temperature as follows.

              q = $mC \times \Delta T$

          2165 J = $124.3 \times 3.78 \times (T - 22.62)^{o}C$

        2165 - 469.854 = $T - 22.62^{o}C$

           17.417 = $T - 22.62^{o}C$

               T = $40.04^{o}C$

Thus, we can conclude that final temperature of the solution is $40.04^{o}C$.