Lab Report 4 - Determination of solubility of salt in water/ ethanol and percentage error in PDF

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Determination of solubility of salt in water/ ethanol and percentage error in solubility...

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Solubility of salt in water/ethanol Introduction: Solubility of the salt is the ability of salt to dissolve in the solution. It depends on the molarity of the solution. The purpose of the experiment was to calculate the solubility of salt and find the percentage error in solubility between the theoretical value of solubility of salt and experimental value of solubility of salt. It was hypothesized that there will be certain difference in experimental and theoretical value of solubility.

Experimental: To find the solubility of salt in water/ethanol, Sodium Chloride, di-water, ethanol, filter paper, 250ml beaker, 50 ml beaker, 400 ml beaker, glass stirring rod, flat ended spatula, funnel, lab stand, and ring clamp were used. Sodium Chloride was used as salt in the experiment and 37grams of sodium chloride was weighted. The salt was placed in 250ml beaker and di-water was added to 100ml mark on beaker. The solution was swirled to dissolve salt with water. A piece of filter paper was weighted. 20 ml of sample solution was poured into 50 ml beaker and 3ml of absolute ethanol was added to the solution and mixed with stirring rod to make the solution. The funnel was placed on ring clamp of lab stand, filter paper was placed on above funnel and rinsed it with di- water. 400ml beaker was placed under funnel. Now the sample solution was allowed to filtrate and passed through filter paper. When the filtering is finished, the solid left on filter paper was rinsed with 5ml of absolute ethanol. Then filter paper was removed with flat ended spatula and placed under fume hood for approximately 10 minutes to dry. The weight of dry filter paper was recorded and subtracted with the initial mass to obtain the mass of dissolved solid. The same process was repeated using 4ml of absolute ethanol and then 5ml of

absolute ethanol. Hence, the solubility of dissolved salt was calculated and then percentage error was calculated.

Result: The solubility of dissolved salt in each trial was calculated and the percentage error in the solubility of salt was calculated. At first, the mass of NaCl that needs to be dissolved in water was found and then only the solution was prepared. The solubility of dissolved salt in each trial was measured and percentage error was calculated. Solubility of NaCl = 360g/L =0.360g/mL The salt was placed in 250 ml beaker and di-water was added up to 100ml. So, the solubility of the salt in 100 ml of water was calculated and the mass of NaCl dissolved in 100 mL of water was given by, The solubility of NaCl salt in 100ml of water at 20℃ is = 0.360g/mL The mass of NaCl in 100 mL of water = 0.360g/mL * 100 mL = 36.0 g In order to make over saturated solution of NaCl, 1g of extra NaCl was added to the solution. So, the total mass of NaCl salt added in order to make the solution was 37g. Total mass of NaCl added in 100 ml of water= 36g+1g = 37g NaCl Next, the theoretical value of NaCl in 1/5 of 100 mL solution was calculated as, The theoretical value of mass of NaCl in each 20 mL of solution= 20 mL * 0.37 g/mL

= 7.4 g Therefore, the theoretical value of mass of NaCl in each 20 mL portion was found to be 7.4g. After filtration, the mass of residue along with filter paper was weighted. The mass of residue on filter paper was calculated by subtracting mass of residue along with filter paper from mass of filter paper in first trail. The mass of salt dissolved in the solution was calculated by subtracting mass of solid from mass of NaCl in 20 ml of solution. For first trail, Mass of filter paper =0.6g Mass of filter paper with solid = 1.7g Mass of solid NaCl= mass of filter paper with solid – mass of filter paper = 1.7g – 0.6g = 1.1 g Mass of salt dissolved in solution = mass of NaCl in solution – mass of solid = 7.4g – 1.1g = 6.3g Therefore, the solubility of dissolved salt in solution = mass of salt dissolved/ volume of solution (S1) = 6.3g / 20mL = 0.315 g/mL For second trial,

Mass of filter paper =0.68g Mass of filter paper with solid = 2.2g Mass of solid NaCl= mass of filter paper with solid – mass of filter paper = 2.2 g – 0.68g = 1.52 g Mass of salt dissolved in solution = mass of NaCl in solution – mass of solid = 7.4g – 1.52g = 5.88 g Therefore, the solubility of dissolved salt in solution = mass of salt dissolved/ volume of solution (S2) = 5.88 g / 20mL = 0.294 g/mL

For third trial, Mass of filter paper =0.66g Mass of filter paper with solid = 3.21g Mass of solid NaCl= mass of filter paper with solid – mass of filter paper = 3.21 g – 0.66g = 2.55 g

Mass of salt dissolved in solution = mass of NaCl in solution – mass of solid = 7.4g – 2.55g = 4.85 g Therefore, the solubility of dissolved salt in solution = mass of salt dissolved/ volume of solution (S3) = 4.85 g / 20mL = 0.2425 g/mL The average solubility of salt from first, second, and third trial is calculated by dividing solubility of salt from all three trial by total number of trials. Average solubility of salt = (S1+ S2+ S3)/3 = (0.315+ 0.294 + 0.2425)/3 = 0.284 g/mL The percentage error in the solubility of salt was calculated by using experimental value and theoretical value of solubility of NaCl.

Percentage error=

=

theoretical value−experimental value × 100 % theoretical value 0.360 – 0.284 ×100 % 0.360

= 21.1% Hence, the percentage error in solubility of NaCl was found to be 21.1%.

Conclusion: The solubility of NaCl dissolved in the solution was calculated by using mass of NaCl dissolved in the solution and volume of solution. The percentage error in the experimental value of solubility of NaCl and theoretical value solubility of NaCl was also calculated. The experimental helps to know procedure in finding solubility of different substances in solvent and helps to find the percentage error. The major cause of inaccuracy in the value of solubility of salt was due to complete dryness of solid NaCl. The percentage error of the experiment could be reduced by drying the solid completely in fume hood. For this experiment, the mass of NaCl required to dissolve in 100 mL of water was calculated from theoretical value of solubility of NaCl in water. Then, salt was dissolved in 100 mL of water and 20ml of dissolved solution was prepared to filter the solution. Each 20 ml of solution was passed through filter paper with ethanol and after filtration, undissolved residue was dried in hood. The undissolved dry mass of solid gives mass of dissolved salt and solubility of dissolved mass was calculated in each trial. Hence, the average solubility of salt was measured, and percentage error was calculated. Hence, the percentage error was found to be 21.1% and average solubility of salt was found to be 0.284 g/mL. The experiment could be used in future to find the solubility of other substances in different solvent and nature of solubility. The experiment gives idea on minimizing error in the value of experiment....