4. A student collected the following data while determining the molar heat of ne

Question

4. A student collected the following data while determining the molar heat of neutralization of a mono- protic acid, HA, with NaOH. Calculate the molar heat of neutralization of this acid with NaOH. As- sume that the specific heats of the solutions are the same as water a. temperature of HA and NaOH before mixing b. concentration of the acid, HA c. concentration of the NaOH d. volume of the acid, HA, used 23.7 °C 1.06 M 1.12 M 50.0 mL e. f. g. volume of the base, NaOH, used temperature of calorimeter and mixture at time - 0 heat capacity of calorimeter 50.0 mL 29.7 oc 6.91/°C

Explanation / Answer

Calorimetric data

mass of solution = 100 g

Cp for water = 4.184 J/oC.g

dT = 29.7 - 23.7 = 6 oC

Cp for calorimeter = 6.9 J/oC

moles of limiting reactant (acid) = 1.06 M x 0.050 L = 0.053 moles

So,

molar heat of neutralization = [100 x 4.184 x 6 + (6.9/6)]J/1000 x 0.053 moles

                                            = 47.39 kJ/mol

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