2 Heat of Precipitation

Topics: Sodium chloride, Solubility, Chlorine Pages: 21 (2120 words) Published: July 7, 2015
Heat of precipitation

- precipitate is unsoluble salt
- precipitate must be prepared through double bond
decomposition or precipitation method

Do you still remember what is meant by double bond decomposition? [please refer to salts notes]

General equation double bond decomposition/precipitation;

Ionic equation for precipitation reaction.

Solubility in water
Li+, Na+, K+, NH4+
All salt dissolve in water
Nitrate, NO3-
All nitrate salt dissolve in water
All chloride salt dissolve dissolve in water except;
PbCl2 - lead(II) chloride (dissolve in hot water)
AgCl - argentums/silver chloride
HgCl - hydroargentum chloride, mercury chloride
All sulphate salt dissolve in water except;
PbSO4 , BaSO4 , CaSO4
All carbonate salt not dissolve in water except;
Li2CO3 Na2CO3 , K2CO3 , (NH4)2CO3
All oxide not dissolve in water except;
Na2O , K2O , CaO
All hydroxide not dissolve in water except;
NaOH, KOH, Ca(OH)2 , Ba(OH)2

Formula to determine the heat change;

Heat released/absorbed, H = mcӨ [unit = J or kJ]

mass of solution
1cm3 = 1 g
specific heat capacity of solution
4.2 J g-1 oC-1
temperature change

To determine precipitation heat of silver chloride, AgCl

In this experiment you must have the following data;

Data tabulation
Initial temperature of sodium chloride NaCl /oC
x oC
Initial temperature of silver nitrate, AgNO­3 /oC
y oC
Average initial temperature for both solution

Highest temperature for the solution
z oC
Temperature change
z – (x + y) oC = Ө oC

Chemical equation for the reaction;

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

Ionic equation for the reaction;

Ag+ (aq) + Cl- (aq)  AgCl (s)

Calculation of heat of precipitation for AgCl;
1. Calculate the number of mole of precipitate formed

No. of mol NaCl = = = 0.0125 mol

No. of mol AgNO3 = = = 0.0125 mol

No. of mol AgCl= 0.0125 mol

2. Calculate the heat released/given out
[From the experiment]
Total volume of the mixture = 25 cm3 AgNO3 + 25 cm3 NaCl
= 50 cm3

Mass of solution= 50 g

Temperature change= Ө oC

Heat given out, H=
= 50 × 4.2 × Ө J

= kJ

Therefore, heat given out during the experiment is kJ

3. Calculate the heat of precipitation

0.0125 mol of AgCl produces kJ

1 mol of AgCl produces = kJ mol-1

= × kJ mol-1

∆H = kJ mol-1

The heat of precipitation of silver chloride, AgCl;

∆H = – kJ mol-1

Example 1: Precipitation for lead(II) sulphate

Chemical equation;
Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 ∆H = -50 kJmol-1 Ionic equation;
Pb2+ + SO42- → PbSO4 ∆H = -50 kJmol-1

50 kJ heat released when 1 mol of lead(II) ions react with 1 mol of sulphate ions to form 1 mol precipitate of lead(II) sulphate.

The heat of precipitation for PbSO4 = – 50 kJmol-1

Example 2: Precipitation for silver chloride

Chemical equation ;
AgNO3 + KCl → AgCl + KNO3
∆H = – 65.5 kJmol-1 Ionic equation;
Ag+ + Cl- → AgCl ∆H = – 65.5 kJmol-1

65.5 kJ heat released when 1 mol of silver ions react with 1 mol of chloride ions to form 1 mol precipitate of silver chloride.

The heat of...
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