Question

If 50.00 mL of 1.05 M sodium hydroxide is added to 25.00 mL of 1.88 M hydrochloric acid, with both solutions originally at 24.66°C, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02 g/mL and a specific heat of 3.98 Jg⁻¹°C⁻¹.)

Answer 1

Answer:

The final temperature of solution is 24.668883°C

Explanation:

The enthalphy change for the reaction of 1 mol of strong acid and strong base is -57.62 J/mol.

moles of NaOH = 0.0525 mol

moles of HCl = 0.047 mol

Hence ,

          Heat released in this reaction is .047 × 57.62 joules

                                                             = 2.70814 joules

 Final mass of solution is density × final volume.

                                        = 1.02 × 75

                                        =76.5 grams

And ,

       MS = 76.6 × 3.98

              = 304.868

Since,

         Q = M × s × ΔT

Therefore,

                ΔT = $\frac{Q}{MS}$

                     =  $\frac{2.70814}{304.868}$

                     = 0.00888299198 °C

Hence ,

           Final temperatre = 24.66 + .00888299198

                                        = 24.668883  °C