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.

Explanation:

It is known that,

      No. of moles = Molarity × Volume

So, we will calculate the moles of as follows.

         No. of moles =

                                = 0.0125 mol

Now, the moles of KOH are as follows.

       No. of moles =

                                = 0.0374 mol

And, =

                             = 0.0375 mol

Now, the balanced reaction equation is as follows.

This means 1 mole of produces 173.2 kJ of heat. And, the amount of heat produced by 0.0125 moles of is as follows.

         M =

             = 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

                                  =

                                  = 124.3 g

Specific heat = 3.78

Now, we will calculate the final temperature as follows.

              q =

          2165 J =

        2165 - 469.854 =

           17.417 =

               T =

Thus, we can conclude that final temperature of the solution is .