4.2.1 - Laboratory Activity : Heat of Precipitation of Silver Chloride

Laboratory Activity 4.2.1: Heat of Precipitation of Silver Chloride

Aim: The define the heat of precipitation of silver chloride.
Problem statement: What is the value of heat of precipitation for silver chloride?
Hypothesis: What is the value of heat of precipitation for silver chloride?
Variable:

	»	Fixed variable : Volume and concentration of silver nitrate, chloride salt solutions, polystyrene cup.
	»	Manipulated variable : Chloride salt solution.
	»	Responding variable : Heat of precipitation.

Material: » Sodium chloride 0.5mol dm-3 » Potassium chloride 0.5mol dm-3 » Silver nitrate 0.5mol dm-3

Apparatus: » Polystyrene cup » Thermometer » Measuring cylinder

Procedure:

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1. 25cm3 of silver nitrate 0.5mol dm-3 is measured with measuring cylinder and poured into a polystyrene cup. 2. Its temperature T1 is measured. 3. 25cm3 of sodium chloride is measured with a clean measuring cylinder and poured into the cup containing silver nitrate solution. 4. The mixture is stirred, and the highest temperature is recorded. 5. The experiment is repeated by replacing sodium chloride with potassium chloride at the same concentration.

Observation:

	►	A white precipitate was formed
	►	Temperature of the mixture increases.

Data:

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Solution NaCl KCl Initial temperature of silver nitrate (°C) T1 T5 Initial temperature of chloride salt (°C) T2 T6 Average initial temperature (°C) $\frac{{\text{T}}_1+{\text{T}}_2}{2}\mathrm{=}{\text{T}}_3$ $\frac{{\text{T}}_5+{\text{T}}_6}{2}\mathrm{=}{\text{T}}_7$ Final temperature (°C) T4 T8 Increased in temperature (°C) T4 – T3 = θ1 T8 – T7 = θ2

Calculation:

	►	Number of moles of AgNO3 = $\frac{\text{MV}}{1000}$ = $\frac{0.5\mathrm{\times }25}{1000}\text{mol}$ = 0.0125mol Number of moles of NaCl = $\frac{0.5\mathrm{\times }25}{1000}\text{mol}$ = 0.0125mol Total volume of the reacting solution = (25 + 25) cm3 Total mass of the reacting solution = 50g Heat change = mcΔθ = 50 x 4.2 x Δθ = y joules
		$\begin{array}{c}{\text{AgNO}}_3(\text{s})\\ \text{1 mole}\end{array}+\begin{array}{c}\text{NaCl(aq)}\\ \text{1 mole}\end{array}\to \begin{array}{c}\text{AgCl(s)}\\ \text{1 mole}\end{array}+{\text{NaNO}}_3(\text{aq})$
		From the equation 1 mole of AgNO3 will react with 1 mole of NaCl to form 1 mole of AgCl.
		Therefore 1 mole of AgCl will liberate $(\frac{1}{0.0125})\text{y joules}$ = 80y joules = $\frac{\mathrm{80y}}{1000}\text{kJ}$
		Heat of precipitation of silver chloride, ΔH = $\frac{\mathrm{80y}}{1000}{\text{kJ mol}}^{\mathrm{-}1}$
	►	The calculation is then repeated with potassium chloride solution.

Analysis:

	►	The white precipitate is silver chloride.
	►	The other components in the reacting mixture are all in ionic form. ○ Ag+(aq) + Cl−(aq) → AgCl(s)
	►	The silver ions can be from any soluble silver salts.
	►	The chloride ions can be from any soluble chloride salts.
	►	The formation of AgCl with any silver and chloride salts will the same heat of precipitation, theoretically equal to 65.5kJ mol-1
	►	The energy level diagram for the precipitation of silver chloride, AgCl, is shown below.

Discussion:

	►	Precautionary steps: ○ The initial temperature of silver nitrate, sodium chloride and potassium chloride solution taken only after a few minutes to ensure the solution has reached a uniform temperature. ○ Solution of sodium chloride and potassium chloride solution should be poured into the silver nitrate solution as quickly and carefully as possible to minimize heat loss to the environment and avoid spilling of solution. ○ Mixture of the solution in cup polystyrene should always be stirred to ensure a uniform temperature. ○ The thermometer reading should be observed at all times so that the highest temperature reached by the mixture solution can be recorded.
	►	The experiment value will always be less than the theoretical value, due to the heat lost to the surroundings and heat absorbed by the polystyrene cup.

Conclusion:

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The heat of precipitation of silver chloride is constant, 65.5kJ mol-1.

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