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The Effect of Ionic Strength on the Solubility of an Electrolyte

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Title of Experiment 12:
The Effect of Ionic Strength on the solubility of an Electrolyte

Aim:
The purpose of this experiment was to determine the thermodynamics variable of enthalpy, ∆H for the dissolution reaction of boric acid in water. The solubility of boric acid was measured over a range of various temperatures by finding out at what temperature crystallization began for solution of different molarities. A graphical relationship between the natural logarithm of molal concentration and the inverse of the temperature gave a linear plot to determine the enthalpy changes associated with the reaction.

Methodology:

The procedure was carried out as follows in the CHEM2210 Laboratory Manual from page 58 to 59.

Results

| Weight of …show more content…

What is the solubility of boric acid in water at 20°C?

From the graph, when temperature = 20°C,
Solubility = x
When y = 20
20 = 15.0975x ± 15.32 + 15.3231 ± 2.08 x = 0.31 ± 0.583 mol/kg

3. What is the solubility of boric acid in water at 100°C?

From the graph, when temperature = 100°C,
Solubility = x
When y = 100
100 = 15.0975x ± 15.32 + 15.3231 ± 2.08 x = 5.61± 1.66 mol/kg

4. What is the total heat evolved when a saturated solution of boric acid in water is cooled from 100°C to 20°C? Note: The specific heat capacity of the saturated solution is approximately that of water in that temperature range
(4.20 ± 0.02 Jg-1K-1) and the specific heat of the solid is approximately 0.

20°C = 273.15 + 20 = 293.15K
100°C = 273.15 + 100 = 373.15K

ΔT = 373.15 - 293.15 = 80K
Molar enthalpy of solution: ln s = -∆HR + c
Gradient of line = -∆HR
-∆HR= -3662.74406.323
∆H = (3662.74406.323) x 8.314 = 304520.02±33781.67 J/mol = 304.52 ± 33.78 kJ/mol

Total heat evolved when a saturated solution of boric acid in water is cooled from 100°C to 20°C.

Assuming 50g of water is used, ∆H(H2O) = m.c.ΔT = 50 × (4.20 ± 0.02) × -80 = - 16800 ± 80 J

Difference in solubility of boric acid = solubility at 20°C – solubility at 100°C = 0.31 ± 0.583 mol/kg – 5.61± 1.66 mol/kg = –5.30 ± 2.243 mol/kg
1000g –5.30 ± 2.243 mol
50g –0.265 ± 0.112 mol

∆H(H3BO3) = ∆H ×n = 304520.02±33781.67 J/mol = – 80697.81 ± 8952.14 J

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