Question

A a. After having completed your experiment and calculated your Enthalpy of Neutralization, e usu-you are told concentration of the NaOH solution you used was 5% less than indicated on its label and asked to recalculate your enthalpy of neutralization. Will the recalculated cment value be larger, smaller, or the same as you originally calculated? b. Now consider the HCl you used in your experiment, if the concentration was 5% higher than that stated on its label, what will be the effect on the enthalpy of neutralization originally calculated. 5. A 50.0 mL sample of a 1.00 M solution of MgSO, is mixed with 50.0mL of 2.00 M NaOH in a calorimeter. The temperature of both solutions was 23.2 °C before mixing and 29.3 °C after mixing. The heat capacity of the calorimeter is 15 J/°C. From these data, calculate AH for the process: MgSOMa) +2NaOHMg(OH)20 + Na,SO a) 4(ag) (ag) Assume that the specific heat and density of the solution after mixing are the same as those of pure water and that the volumes are additive. (4 pts) on this

Answer 1

Answer (a) Calculation of enthalpy of neutralisation

For maximum enthalpy of neutralisation , moles of acid (HCl) should be equal to the moles of base (NaOH) . If amount of NaOH is 5% less than the original amount , than less no. of mole will neutralised and hence recalculated value of enthalpy of neutralisation will be smaller.

(b) : No, higher concentration of only HCl will not effect on the value of enthalpy of neutralisation. For increase in enthalpy of neutralisation concentration of both acid and base should increase.

5): in this reaction no of equivalents of both the reactants are equal. So, no any limiting reagent.

initial temperature (T1) = 23.2 °C

Final temperature (T2) = 29.3 °C

Heat capacity = 15 J/°C

Enthalpy = heat capacity * temperature difference = 15 * (29.3 - 23.2 ) = 15 * 6.1 = 91.5 J