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.)

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 =

                     =  

                     = 0.00888299198 °C

Hence ,

           Final temperatre = 24.66 + .00888299198

                                        = 24.668883  °C

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

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