Heat of Precipitation - SPM Chemistry

The heat of precipitation is the energy change when one mole of precipitate is formed from its ion.

Precipitation reaction occurs when two solutions containing cations and anions of insoluble salts added together.
This reaction is known as the double decomposition reaction and is used to prepare any insoluble salt.
The heat of precipitation is the energy change when one mole of precipitate is formed from its ion.
For the heat of precipitation, it can be an exothermic or endothermic reaction.
The example below shows how the heat of precipitation can be calculated from the result of an experiment.

Example:
In an experiment to find the heat of precipitation of magnesium carbonate, 25 cm³ of magnesium nitrate 2.0 mol/dm³ is added into 25 cm³ of sodium carbonate solution, 2.0 mol dm³ in a container. The temperature of the solution decreases by 6°C. What is the heat of precipitation of magnesium carbonate? [Specific heat capacity of the solution = 4.2 Jg^-1°C^-1;; density of the solution =1 g/cm³ ]
Answer:

Step 1: Equation of the reaction

Mg(NO₃)₂ + Na₂CO₃ → MgCO₃+ 2NaNO₃

Step 2:

Number of mole of Mg(NO₃)₂

$\begin{gathered} n = \frac{{MV}}{{1000}} \hfill \\ n = \frac{{(2.0)(25)}}{{1000}} \hfill \\ n = 0.05mol \hfill \\ \end{gathered}$

Number of mole of Na₂CO₃ solution

$\begin{gathered} n = \frac{{MV}}{{1000}} \hfill \\ n = \frac{{(2.0)(25)}}{{1000}} \hfill \\ n = 0.05mol \hfill \\ \end{gathered}$
Number of mole of magnesium carbonate formed = 0.05mol

Step 3

Amount of heat energy been released

= mcθ

= (25+25)(4.2)(6)
= 1260J

Step 4

The heat of precipitation of magnesium carbonate

$\begin{gathered} \Delta H = \frac{{{\text{Heat Change}}}}{{{\text{Number of mole of precipitate form}}}} \hfill \\ = \frac{{1260}}{{0.05}} \hfill \\ = 25200Jmo{l^{ – 1}} = 25.2kJmo{l^{ – 1}} \hfill \\ \end{gathered}$

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