Solubility Of Benzoic Acid Essay

Decent Essays
The purpose of this experiment is to find the solubility of benzoic acid and measuring the differential heat of solution.
Solubility of a salt is usually known as the number of grams of the salt that will dissolve in 100 grams of water. Solubility is also defined as the maximum amount of solute dissolves in a solvent to make it a saturated solution at a particular temperature. The temperature is an important part of the solubility data because, like most chemical processes, dissolution of a salt is temperature dependent1. However, the behaviour depends on whether the solubility reaction is an exothermic reaction or an endothermic reaction. The increase in temperature will favour a solubility reaction which is an endothermic process while it
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Solubility in molarity: -
Sample calculation
At 50 °C,
Average volume of NaOH used = (31.00+33.10+34.60)/3 = 32.90 ml = 0.0329 L
Number of moles of NaOH = Concentration × Volume = 0.01 × 0.0329 = 3.29 × 10-4 moles
Number of moles of C6H5COOH = 3.29 × 10-4 moles (stoichiometric ratio is 1:1)
Molarity = (number of moles)/(volume of solution) = (3.29 ×〖10〗^(-4))/(5 × 〖10〗^(-3) ) = 0.0658 mol/L

In grams per 100 g of water: -
Sample calculation
At 50 °C,
Mass of sodium salt of benzoic acid = moles × molar mass = 3.29 × 10-4 × 144.11 = 0.0474 g
Solubility = (mass of salt)/(100 g of water) = 0.0474/100 = 4.74 × 10-4 g per 100 g of water at 50 °C

Temperature / °C Molarity / mol/L Grams per 100 g of water Temperature in Kelvin / K Ln of solubility 1/T / K-1
50 0.0658 4.74 × 10-4 323.2 -7.65 0.00309
45 0.0592 4.27 × 10-4 318.2 -7.76 0.00314
40 0.0488 3.52 × 10-4 313.2 -7.95 0.00319
35 0.0428 3.08 × 10-4 308.2 -8.09 0.00324
30 0.0348 2.51 × 10-4 303.2 -8.29 0.00330

From the graph,
Slope = -3099 K × gram per 100 g of water
Intercept = 1.94 gram per 100 g of water
Calculating the differential heat of solution, using Equation 3 from above: (- ∆H)/R =
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This will increase the volume added, thus decreasing the molarity to too low. This is why, while titrating, the end point should be a very “light” colour, not a dark rich one. The inaccuracy of the standard solution. If the standard solution, that is NaOH is not completely pure, then the molarity would be much small. Because there are other impurities present in the same amount of solution. This too results in low calculated molarity. The end point is not same with equivalence point and the colour change of the indicators are not instant2. All the volume of benzoic acid solution is not transferred. This might happen when we move the pipette between the flasks and may lose a drop of the solution2. Losing the solution. When the solution is swirled vigorously in the conical flask while titration, the solution might splash before the end point has been reached2.
Before taking out the benzoic acid solution from the flask, the temperature in the water jacket was recorded. This is the way, we made sure that the solution was saturated and at equilibrium at the start of the

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