"Introduction for molar solubility and common ion effect" Essays and Research Papers

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    Molar Mass

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    What is molar mass?Molar mass is the weight of one mole (or 6.02 x 1023 molecules) of any chemical compounds. Molar masses of common chemical compounds that you might find in the chemistry laboratory can range between 18 grams/mole for compounds like water to hundreds of grams per mole for more complex chemical compounds.The lightest possible chemical that one can have under normal conditions is hydrogen gas‚ or H2. There is no limit to how heavy a chemical compound can be - it is not uncommon for macromolecules (large

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    Solubility and Grams

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    Name SOLUBILITY CURVES Answer the following questions based on the solubility curve below. Which salt is least soluble in water .. at 2O° C? 2. How many grams of potassium chloride can be dissolved in 200 g of water at 80° C? IO 3. At 40° C‚ how much potassium _ __nitrate coin be dissoiu$tl ^n 30D.g of water? ------W- ’1 80 70 ...- O --60 0 5© 40 4. Which salt shows the least change 30 In solubility from 0° - 100° C? 20 10 At 30° C‚ 90 g of sodium

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    Factors Affecting Solubility

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    FACTORS AFFECTING SOLUBILITY There are three main factors that control solubility of a solute. (1) Temperature (2) Nature of solute or solvent (3) Pressure EFFECT OF TEMPERATURE Generally in many cases solubility increases with the rise in temperature and decreases with the fall of temperature but it is not necessary in all cases. However we must follow two behaviours: In endothermic process‚ solubility increases with the increase in temperature and vice versa. For

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    Solubility Curves

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    wont dissolve. Solubility of a solute = mass of a solid required to a saturate 100g of water at a particular temperature. Calculating Solubility 2g potassium chlorate dissolves in 20g water at 28oC what is its solubility? 2 x 100/20 = 10.0g potassium chlorate/100g water 4g potassium sulphate dissloves in 30g water at 50oC what is its solubilty? 4 x 100/30 = 13.33g potassium sulphate/100g water 30g sodium chloride dissolves in 75g water at 10oC what is its solubility? 30 x 100/75

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    dependant towards solution concentration. The molar conductivity (Λm) measurements would be more appropriate. Λm could be determined from the conductivity value: [pic] where C is the electrolyte concentration In mol/L. the unit is S cm2mol-1 If the dependant of conductivity towards concentration were to be studied‚ we would observe that the conductivity increases along with solution concentration due to the increase in the number of ions. However the relationship is not linear. So

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    Solubility Lab

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    354-355 Experiment 2 SOLUBILITY 1. Part A. Solubility of Solid Compounds. Use your observations to complete the following table‚ rating each system as soluble‚ insoluble‚ or partially soluble. Organic Compound Benzophenone Water Methyl Alcohol Hexane Malonic acid Biphenyl 2. Considering the polarities of the compound and the solvent and the potential for hydrogen bonding‚ answer the following: a) There should be a difference in your results between the solubilities of biphenyl and benzophenone

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    Solubility of CO2 in water Aim: To decarbonate a bottle of soft drink and find out the amount of CO2 in the drink. * Principle: The reaction between carbon dioxide and water is an example of an equilibrium reaction: Materials: * * 3 soft drink bottles (300ml) * 6g of salt (NaCl) * Triple beam balance scale * Thermometer * Digital scale * Watch glass * Electric hotplate Method: Standing up method 1. An unopened bottle of carbonated drink

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    10-3 mol 3. Molar mass – mass in grams of one mole of a substance. Example 3-5 page 76 4.62 g Na3PO4 Molar Mass Na3PO4 = (22.9898 gNa X 3) + (30.9738 gP) + (15.9994 gO X4) = 163.9408 g per mol Na3PO4 Moles Na3PO4 = 4.62 g X 163.9408 g/ mol = 2.818 X 10-2 mol Na3PO4 Moles Na = 2.818 X 10-2 mol Na3PO4 X 3 mol Na / mol Na3PO4 = 8.45 X 10-2 mol Na Na+ ions = 8.45 X 10-2 mol Na X (6.022 X 1023) = 5.08 X 1022 ions C. Solutions and Their Concentrations 1. Molar Concentration

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    Experiment 1: Study of Solubility Equilibrium Data Treatment and Analysis Section 1: Solubility Product Constant Temperature (˚C) | Volume of NaOH used (mL) |   | |   | Titration 1 | Titration 2 | Average | 28 | 12.7 | 12.8 | 12.75 | 9 | 10.5 | 10.5 | 10.5 | 19 | 11.3 | 11.2 | 11.25 | 40 | 16.2 | 16.2 | 16.2 | 50 | 22.8 | 22.9 | 22.85 | Table 1: The volume of NaOH used in the titration at various temperatures. No. of moles of KHC4H4O6 = 1.45 g ÷ 188.177g/mol = 7.71 x 10-3mol

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    Solubility Lab

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    1. Provide a general discussion of the solubility/miscibility behavior observed in procedure A-D. For part A of the procedure we worked with the solubility of solid compounds in various solvents. The three solid compounds that were worked with during this procedure were benzophenone‚ malonic acid‚ and biphenyl. These three solids were then mixed with water (highly polar)‚ methyl alcohol (intermediately polar)‚ and hexanes (nonpolar). When benzophenone is mixed with water the results turned out to

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