Question

The neutralization of 50.0 mL of 1.00 M hydrochloric acid with 50.0 mL of 1.00 M sodium hydroxide causes a 6.7°C increase in temperature. Predict how the following changes to the experimental protocol would affect the value of the change in temperature.

The temperature change depends on the moles of water produced (moles of reaction) and the mass of the solution. Assume that the density of each solution is 1.0 g/mL.
a. using 100.0 mL of 1.00 M hydrochloric acid and 100.0 mL of 1.00 M sodium hydroxide
b. using 50.0 mL of 2.00 M hydrochloric acid and 50.0 mL of 2.00 M sodium hydroxide
c. using 25.0 mL of 1.00 M hydrochloric acid and 75.0 mL of 1.00 M sodium hydroxide
d. using 25.0 mL of 1.00 M sulfuric acid and 75.0 mL of 1.00 M sodium hydroxide (Be careful. Think about the value of n!)

Answer 1

HCl +NaOH = NaCl +H2O

Here, 50.0 mL of 1.00 M hydrochloric acid = 0.05 mol and 50.0 mL of 1.00 M sodium hydroxide = 0.05 mol.

Thus reaction will

0.05 HCl + 0.05 NaOH = 0.05 NaCl + 0.05 H2O

First the total volume of the solution is 50+50 =100 mL. Now calculate its mass by density as follows:

Density = 1.0 g/mL

Mass = volume ,100 ml *density = 100 g

The temperature change is 6.7°C. Because the temperature increases, the reaction is an exothermic.

Now calculate the total heat of this reaction =

Q = -(specific heat of solution * mass of solution *temperature change )

Q= -4.18 J /g °C *100 g *6.7 °C = -2800.6 J or delta H

The temperature change depends on the moles of water produced (moles of reaction). here 0.05 mole H2O produced.

Thus, the enthalpy change per mole of H2O is

delta H = --2800.6 J /0.05 = 56012 J / mol or 56.0123 kJ /mol.

a. using 100.0 mL of 1.00 M hydrochloric acid and 100.0 mL of 1.00 M sodium hydroxide

HCl +NaOH = NaCl +H2O

Here, 100.0 mL of 1.00 M hydrochloric acid = 0.10 mol and 100.0 mL of 1.00 M sodium hydroxide = 0.10 mol.

Thus reaction will

0.1 HCl + 0.1 NaOH = 0.1 NaCl + 0.1 H2O

The enthalpy change per mole of H2O is 56.0123 kJ /mol.

here 0.1 mol H2O produce then the enthalpy change for this reaction is

56.0123 kJ /mol *0.1 mol = 5.6 kJ

b. using 50.0 mL of 2.00 M hydrochloric acid and 50.0 mL of 2.00 M sodium hydroxide

HCl +NaOH = NaCl +H2O

Here, 50.0 mL of 2.00 M hydrochloric acid = 0.10 mol and 50.0 mL of 2.00 M sodium hydroxide = 0.10 mol.

Thus reaction will

0.1 HCl + 0.1 NaOH = 0.1 NaCl + 0.1 H2O

The enthalpy change per mole of H2O is 56.0123 kJ /mol.

here 0.1 mol H2O produce then the enthalpy change for this reaction is

56.0123 kJ /mol *0.1 mol = 5.6 kJ

c. 25.0 mL of 1.00 M hydrochloric acid and 75.0 mL of 1.00 M sodium hydroxide

HCl +NaOH = NaCl +H2O

Here, 25.0 mL of 1.00 M hydrochloric acid = 0.025 mol and 75.0 mL of 1.00 M sodium hydroxide = 0.075 mol.

here 0.025 mol HCl is limiting agent.

Thus reaction will

0.025 HCl + 0.025 NaOH = 0.025 NaCl + 0.025H2O

The enthalpy change per mole of H2O is 56.0123 kJ /mol.

here 0.025 mol H2O produce then the enthalpy change for this reaction is

56.0123 kJ /mol *0.025 mol = 1.4 kJ

d. 25.0 mL of 1.00 M sulfuric acid and 75.0 mL of 1.00 M sodium hydroxide

H2SO4 +2 NaOH = Na2SO4 + 2H2O

Here, 25.0 mL of 1.00 M H2SO4= 0.025 mol and 75.0 mL of 1.00 M sodium hydroxide = 0.075 mol.

here 0.025 mol H2SO4 is limiting agent.

Thus reaction will

0.025 H2SO4 + 0.050 NaOH = 0.025 Na2SO4 + 0.050 H2O

The enthalpy change per mole of H2O is 56.0123 kJ /mol.

here 0.050 mol H2O produce then the enthalpy change for this reaction is

56.0123 kJ /mol *0.050 mol = 2.8 kJ