Experiment (2) Determination of the Heat of Neutralization of Strong Acid with a Strong Base.

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Presentation transcript:

Experiment (2) Determination of the Heat of Neutralization of Strong Acid with a Strong Base

Theory Heat of Neutralization The amount of heat evolved from the reaction of strong acid with strong base is equal 13,700 calories when one mole of water is formed. NaOH + HCl NaCl + H2O

Procedure Weight dry calorimeter (W1). Measure 100 ml of HCl by using measuring cylinder and transfer it into the calorimeter. Measure 100 ml of NaOH by using measuring cylinder. Measure the temperature of HCl and NaOH, the temperature must be the same (Ti). Quickly pour the base solution into the calorimeter, cover the calorimeter and stir. Read the temperature until the temperature become constant (Tf). Reweight the calorimeter (W2).

Results Ws = W2 – W1 ΔT = Tf – Ti W1 = Weight of the system (Dry Calorimeter). W2 = Weight of the system (Calorimeter) + acid base solution. Ti = Initial temperature. TF = Final temperature. Ws = W2 – W1 ΔT = Tf – Ti

Calculations qsystem = qneu. + qcalorimeter + qs The system is isolated => qsystem = 0 qneu. = - ( qcalorimeter + qs ) qs = SNaCl × Ws × ΔTs = 0.931 × (W2 – W1) × (Tf – Ti) qcalorimeter = Ccalorimeter × ΔTcalorimeter

Calculations 𝑀= 𝑛×1000 𝑉 (𝑚𝑙) n= No. of moles 𝑛= 𝑀×𝑉(𝑚𝑙) 1000 qneu.  0.2 mole X  1 mole 𝑋= 𝑞 𝑛𝑒𝑢. 0.2 = Calorie/mol