Question

	Acetic Acid	NaOH	Ti°C	Tf°C
Trial 1	5mL	5mL	23.5	28.6
Trial 2	5mL	5mL	23.8	29.3

Write a balanced equation chemical equation for the neutralization reaction.

Using the density of the salt solution, calculate the mass of the salt solution for each trial.

Calculate the heat (qrxn) for each trial

Calculate the average(qrxn).

From the balanced equation, calculate the moles of water formed in the reaction.

Calculate Enthalpy of Neutralization (△H neutralization) in kJ/mol.

Calculate % error by comparing the calculated value to the actual value.

-density of sodium acetate solution= 1.53g/mL

-Enthalpy of Neutralization for acetic acid with NaOH is 55.2kJ

Answer 1

a) The balanced chemical equation for the reaction between acetic acid (CH₃COOH) and NaOH is

CH₃COOH (aq) + NaOH (aq) -------> CH₃COONa (aq) + H₂O (l)

b) Total volume of the solution for each trial = (5 + 5) mL = 10 mL.

Density of sodium acetate (CH₃COONa) is 1.53 g/mL.

Mass of CH₃COONa = (volume of salt solution)*(density of CH₃COONa) = (10 mL)*(1.53 g/mL) = 15.3 g.

c) Assume the specific heat of the salt solution to be the same as the specific heat of water, i.e, C = 4.18 J/g.⁰C (you haven’t provided a different value).

Trail 1: q_rxn = (mass of CH₃COONa)*(specific heat of CH₃COONa)*(change in temperature) = (15.3 g)*(4.18 J/g.⁰C)*(28.6 – 23.5)⁰C = 326.1654 J (ans).

Trial 2: q_rxn = (mass of CH₃COONa)*(specific heat of CH₃COONa)*(change in temperature) = (15.3 g)*(4.18 J/g.⁰C)*(29.3 – 23.8)⁰C = 351.7470 J (ans).

d) Average q_rxn = ½*[(q_rxn, Trial 1) + (q_rxn, Trial 2)] = ½*(326.1654 + 351.7470) J = 338.9562 J (ans).

e) As per the balanced stoichiometric equation in part (a) above,

1 mole CH₃COONa = 1 mole H₂O

We have formed 15.3 g CH₃COONa.

Molar mass of CH₃COONa = (2*12.01 + 3*1.008 + 2*15.9994 + 1*22.9897) g/mol = 82.0325 g/mol.

Moles of CH₃COONa = moles of H₂O = (15.3 g)/(82.0325 g/mol) = 0.1865 mole.

f) Enthalpy of neutralization = (Average q_rxn)/(moles of H₂O) = (338.9562 J)/(0.1865 mole) = 1817.4595 J/mol = (1817.4595 J/mol)*(1 kJ/1000 J) = 1.8174595 kJ/mol ≈ 1.82 kJ/mol (ans).