Question

1. A 50.0 mL solution of 0.805 M acetic acid, a monoprotic acid, is completely neutralized by the addition of 23.55 mL of sodium hydroxide. What is the concentration of the initial sodium hydroxide solution? What is the final concentration of the sodium ions in mol/L?
2. In the reaction in question 1, the initial temperature of each solution is 22.1°C. The temperature after mixing is 29.4°C. How much heat is liberated by this neutralization reaction? What is the heat of neutralization expressed in terms of kJ/mol for this reaction?

Answer 1

M1V1 = M2 V2

50.0 x 0.805 = M2 x 23.55

M2 = 1.71 M

initial concentration of sodium hydroxide = 1.71 M

final concentration of sodium ions = 1.71 mol /L

total volume = 50 + 23.55 = 73.55 mL

density = 1 g / mL

mass = density x volume

           = ( 1 g / mL ) x 73.55 mL

           = 73.55 g

dT = 29.4 - 22.1 = 7.3 oC

Q = m Cp dT

Q = 73.55 x 4.184 x 7.3

Q = 2246.45 J

Q = 2.25 kJ

heat is liberated by this neutralization reaction   = 2.25 kJ

moles of acid = 50 x 0.805 / 1000 = 0.04025

heat of neutralization = - Q / n

                                   = -2.25 kJ / 0.04025

                                = -55.9 kJ / mole

heat of neutralization = -55.9 kJ / mol