Monday January 9, 2023 Day 1 Solubility and Models of Solution Formation |
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Textbook Readings 11.3: Hydration of Ions (LibreText) 13.2: Types of Solutions and Solubility |
Course Lectures 1.1 pdf Video* Solutions 1.2 pdf Video* Like Dissolves Like 1.3 pdf Video* Solution Formation Energetics 1.4 pdf Video* Solution Energetics Examples |
Objectives 1. Know and identify the different types of intermolecular forces and their effects on solubility 2. Predict relative solubilities based on molecular polarity and intermolecular forces. "like dissolves like" 3. Understand the "Energetics of solution formation" and the factors that make solution formation endothermic or exothermic" |
Chemistry: Enthalpy of Hydration |
Homework Problems 1.1 Construct Lewis structures for each of the following and identify which are polar. a. H2O b. NH3 c. CH3OCH3 d. CH3CH2OH e. HF 1.2 Construct Lewis structures for each of the following and put them in order from least polar to most polar. a. CH3CH2CH2OH b. CH3CH2OH c. CH3OH d. CH3CH2CH2CH2OH e. CH3CH2CH2CH2CH2CH2OH 1.3 Use what you've learned from question #2 to arrange the molecules from least to most soluble in water, a very polar solvent. 1.4 Use what you've learned from question #2 to arrange the molecules from least to most soluble in liquid hexane (C6H14), a non-polar solvent. |
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1.5 To better understand solution
formation, we break the process up into the three steps illustrated at right. Each step is associated with a ΔH value that tells us how much heat energy is either required or released for that specfic change to occur. The textbook refers to these heat energy terms as ΔH1, ΔH2 and ΔH3. We will instead refer to these values as ΔHsolvent, ΔHsolute and ΔHmixing as you see in the figure at right. Describe, in your own words, the process that each of these terms refers to. |
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1.6 As you can see from the figure above, the ΔHsolution can be determine by adding up the ΔH values for each of the three steps. Calculate a value for ΔHsolution using the values given above. Is the process endothermic or exothermic? 1.7 Draw the graph corresponding to the following values ΔH values using the one above as a model: ΔHsolute = 54 kj/mol ΔHsolvent = 23 kj/mol ΔHmixing = -34 kJ/mol Determine ΔHsolution and compare to the situation above. Explain the differences. 1.8 For ionic solutions we identify ΔHsolute = -ΔHlattice . We also define (ΔHsolvent + ΔHmixing ) = ΔHhydration . Thus, ΔHsolution = - ΔHlattice + ΔHhydration Calculate the ΔHhydration values for LiCl and NaCl given the following information: LiCl: ΔHlattice = -834 kJ/mol ΔHsolution = -37.0 kJ/mol NaCl: ΔHlattice = -769 kJ/mol ΔHsolution = 3.88 kJ/mol Which of the two cations has stronger ion-dipole attractive forces with water? Why? Click and drag the region below for correct answers 1.1 a. polar b. polar c. polar d. polar e. polar (That's right, they're all polar!) 1.2 Least polar ...CH3CH2CH2CH2CH2CH2OH CH3CH2CH2CH2OH CH3CH2CH2OH CH3CH2OH CH3OH ...most polar 1.3 Least soluble ...CH3CH2CH2CH2CH2CH2OH CH3CH2CH2CH2OH CH3CH2CH2OH CH3CH2OH CH3OH ...most soluble in water 1.4 Least soluble ... CH3OH CH3CH2OH CH3CH2CH2OH CH3CH2CH2CH2OH CH3CH2CH2CH2CH2CH2OH ...most soluble in hexane. 1.5 ...looking forward to your explanation! :) 1.6 ΔHsolution = -35 kJ/mol Exothermic 1.7 Graph will be different in that the final enthalpy of the solution is above the initial level. Consequently, this will be an endothermic process with ΔHsolution = +43 kJ/mol 1.8 For LiCl ΔHhydration = -871 kJ/mol For NaCl ΔHhydration = - 765 kJ/mol Lithium cations experience a more owerful interaction with water molecules because the lithium ion, Li+, is small and allows water molecules closer to the center of the positive charge. The closer the water molecules get, the more energy that's released and the more negative the ΔHsolution value. |
Tuesday January 10, 2023 Day 2 Solubility and Henry's Law |
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Textbook Reading 11.3 Solubility (OpenStax) |
Course Lectures 2.1 pdf Video* Solution Formation Spontaneity 2.2 pdf Video* Solubility 2.3 pdf Video* Gas Solubility and Henry's Law |
Solubility Curves |
Henry's Law |
Objectives 1. Define the terms unsaturated, saturated & super saturated 2. Identify a solution as unsaturated, saturated or supersaturated using solute and solvent information and a solubility curve. 3. Describe how changes in pressure and temperature affect the solubility of solutes in solvents 4. Use solubility curves to predict the maximum amount of solute that can be dissolved in a prescribed amount of solute. 5. Use Henry's law to determine the solubility of a gas using the Henry's law gas constant and known gas pressures. |
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Homework Problems 2.1 Referring to the solubility graph at right, how many grams of KCl will dissolve in 100. grams of water at 40oC? 2.2 70. grams of KCl are added to 100. grams of water at 10oC. How many grams of KCl will dissolve and how many grams will remain undissolved at the bottom of the beaker at this temperature? 2.3 80.0 grams of KCl are added to 200. grams of water at 50oC. The resulting solution is unsaturated. Why? 2.4 Does the presense of solid KCl at the bottom of a KCl solution guarentee the solution is saturated? Why? Why not? |
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2.5 A solution formed by mixing 40.0 grams of KCl in 100. grams of water at 20 oC should be supersaturated according to the graph above. However, simply mixing the KCl and water will not produce a supersaturated solution. What additional steps are required to prepare the supersaturated solution? |
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2.6 Because it's non-polar, oxygen isn't very soluble in water. However, enough oxygen does dissolve to support aquatic life. Use the graph at right to determine the mass of oxygen gas dissolved in 1 liter of water at 25oC and 35oC. 2.7. Will a fish that requires 12 mg/L of dissolved oxygen to survive be able to live in water at 10 oC? |
https://www.ck12.org/book/ck-12-chemistry-basic/section/16.1/ |
2.8 Just like O2, CO2
is a non-polar molecule that shouldn't dissolve very much in liquid water, a polar solvent. However, by increasing the pressure of CO2 at the surface of the liquid more CO2 can be dissolved. This idea is the basis of beverage "carbonation" Why does increasing the pressure of the carbon dioxide (or any other) gas, increase its solubility? 2.9 Henry's law allows us to determine how much gas will dissolve in a specific amount of water. How many grams of carbon dioxide gas is dissolved in a 1.0 L bottle of carbonated water if the manufacturer uses a pressure of 2.4 atm in the bottling process at 25 °C? Useful information: kH of CO2 in water = 3.4 x 10-2 M/atm at 25 °C 2.10 Oxygen is much less soluble in water than carbon dioxide, 0.00412 g/100 mL at 20 °C and 760. mm Hg. Calculate the MOLAR solubility of oxygen gas in water at 20 °C and a pressure of 1150 mm Hg. |
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Click and drag the region below for correct
answer 2.1 Approximately 40 grams of KCl will dissolve in 100. g of water at 40oC 2.2 30 grams of KCl should dissolve at this temperature leaving 40 grams as undissolved ppt found at the bottom of the container. 2.3 80 grams of KCl in 200 g of water is equivalent to 40gKCl/100gH2O. At 50oC, the solubility of KCl is approximately 43 gKCl/100gH2O. Thus, the solution still has room for additional solute and is considered unsaturated. 2.4 In most cases, the presense of a solid solute at the bottom of the container suggests the solution is saturated. However, to be sure, the solution should be thoroughly stirred at constant temperature. 2.5 A supersaturated solution is prepared by dissolving a lot of solute in solvent by increasing the temperature. Then the solution is slowly cooled in hopes that as the saturation temp. is reached, the dissolved solute STAYS dissolved. Sometimes it works and the resulting solution is supersaturated. A slight bump, seed crystal, or dust speck is all that's required to "trigger" the supersaturated solution causing it to release the solute that isn't supposed to be dissolved. 2.6 Here's the graph. At 25oC, S = 8.1 mg/L ... at 35 oC, S = 7.0 mg/L This demonstrates that the solubility of a gas actually decreases as temperature increases. 2.7 The O2 solubility at this temperature is approximately 10.8 mg/L (See graph). This level is below the required 12 mg/L for the fish and will be too low for survival. 2.8 The amount of CO2 dissolved in the liquid solvent depends on the number of collisions gas phase CO2 experiences with the liquid's surface. Higher gas pressures mean more frequent CO2/liquid-surface collisions and an increased likelyhood of a CO2 molecule dissolving in the solvent. 2.9 3.59 grams of CO2 2.10 1. 95 x 10-3 mol/L. |
Wednesday January 11, 2023 Molarity Manipulations and Dilution |
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Textbook Readings 6.3 Molarity |
Course Lectures 3.4 pdf Video* Solutions: Molarity and Titrations 3.1 pdf Video* Concentrations and Conversions |
Molarity (Chemistry Solution) |
Dilution (Chemistry Solution) |
Objectives 1. Calculate the molarity of a solution given moles or grams of solute and total solution volume. Molarity = molessolute/Literssolution 2. Determine moles, volume or Molarity given any two parameters. 3. Determine the new concentration after dilution (addition of pure solvent) 4. Determine the volume of solvent required to dilute a solution to obtain a new concentration 5. Determine the concentration of a solution made by mixing two different solutions of different concentrataions. 6. Determine the individual ion concentrations of species present in solutions. |
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Homework Problems 3.1 What is the molar concentration of the solution formed when 2.54 grams of KNO3 are dissolved in 325 mL of water? 3.2 145 mL of distilled water is added to 500. mL of 4.50 M HCl. What is the new concentration of the solution? 3.3 How many mL of distilled water must be added to 335 mL of 0.500 M NaOH to lower the solution's concentration to 0.100 M? 3.4 What are the individual ion concentrations present in 4.50 M K3(PO4)2? 3.5 Why must you never use the dilution equation, M1V1 = M2V2 in a titration calculation? Click and drag the region below for correct answers 3.1 0.0773 M 3.2 3.49 M 3.3 1340 mL additional water 3.4 [ K+] = 13.5 M [PO4-3] = 9.00 M 3.5 The dilution equation assumes a 1:1 mole ratio. However, in many titration calculations, the mole ratio is NOT 1:1. |
Thursday January 12, 2023 Day 4 Concentration: Molality ... Mole Fraction ... Mass Percent |
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Textbook Readings 11.4 Colligative Properties |
Course Lectures 3.1 pdf Video* Concentrations and Conversions |
i. Molality |
ii. Mole Fraction |
Objectives 1. Use the Molarity definition to determine molessolute, volumesolution or Molarity of the solution. Molarity = molessolute/Literssolution 2. Use the Molality definition to determine molessolute, volumesolution or Molality of the solution. Molality = molessolute/kgsolvent |
iii. Mass Percent |
3. Use the Mass Percent definition to determine masssolute, masssolvent or Mass Percent of the solute and solvent Mass % = masssolute / masssolution x 102 4. Use the mole fraction (X) definition to determine molessolute, moles solvent or mole fraction of the solute and solvent. Xsolute = molessolute/molesTotal Xsolvent = molessolvent/molesTotal 5. Use the ppm or ppb definitions to determine masssolute, masssolvent or ppm/ppb of solute. ppm = masssolute / masssolution x 106 ppb = masssolute / masssolution x 109 Homework Problems 4.1 ppm is approximately the same as mgsolute/LiterH2O. Why is it safe to assume the mass of the solution is the same as the mass of the water?? 4.2 What are the NaCl mass percent, ppm and ppb levels in a solution that is prepared by dissolving 0.30 mg of NaCl in 1.00 Liter of water? (assume Dsolution = DH2O) 4.3 How are molarity and molality concentration units the same? How are they different? 4.4 Consider 350. mL of a 0.100 M Ca(NO3)2 solution a. How many moles of Ca(NO3)2 are there in this solution? b. How many grams of Ca(NO3)2 are there in this solution? c. What is the molar concentration of the solution after the addition of 500. mL of distilled water? 4.5 Your goal is to prepare a 0.500 m KCl solution in 350. mL of water. a. How many kg of solvent will there be? b. How many moles of KCl must you add to the water? c. How many grams of KCl must you add to the water? d. What is the mass of the solution once you've made it? 4.6 Use your results from question 5 to determine the mole fraction of KCl (XKCl) and mole fraction of H2O (XH2O). What do the two mole fraction values add up to? 4.7 Describe the actual steps taken to prepare 500. mL of a 1.50 M CuSO4 solution? 4.8 Describe the actual steps taken to prepare approximately 500. mL of a 1.50 m CuSO4 solution? Answers: Click and drag in the space below. 4.1 For dilute solutions, such as those appropriately described by ppm or ppb concentration units, the mass of dissolved solute is so low that it is reasonable to assume the mass of the entire solution is approximately equal to the mass of the solvent. 4.2 3.00 x 10-5 %, 0.300 ppm, 300. ppb 4.3 Molarity and molality both utilize "moles" of solute in their determinations. However, molarity is calculated using the solution's volume in liters whereas molality is determined using the solvent's mass in kg. 4.4. a. 0.0350 moles b. 5.743 grams c. 0.04117 M 4.5 a. 0.350 kg H2O (...using DH2O = 1.00 g/mL @ 25oC) b. 0.175 moles of KCl c. 13.04 d. 363 grams 4.6 XKCl = 0.00892 XH2O = 99.1 Mole fractions add up to 1.00 4.8 In a large beaker, weigh out 500. grams of distilled water. Add 119.71 g (0.750) moles of CuSO4 and stir until all solid is dissolved. |
Friday January 13, 2023 Day 5 Concentration Conversions |
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Textbook Readings 11.4 Colligative Properties |
Course Lectures 3.1 pdf Video* Concentrations and Conversions |
Molarity-Molality-Mass percent |
Mole Fraction & Solution Concentration Practice Problems |
Objectives 1. Convert between concentration systems using... a. solution density b. the 100 g assumption c. the 1.00 Liter assumption |
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Homework Problems 5.1 Concentrated hydrochloric acid is received as an 11.65 M solution with a density of 1.180 g/mL. a. Assume V = 1.00 L and determine the mass of the solution b. Assume V = 1.00 L and determine the moles & grams of HCl present in the solution. c. Use your results from a & b to determine the mass & moles of H2O present in the solution d. Use your results above to calculate the molality and mole fraction for HCl. e. Why was it both acceptable and important to assume V= 1.00 L in the above calculations? 5.2 An aqueous NaCl solution is 2.55 %NaCl by mass. a. Assuming there are 100 grams of solution, determine the mass of NaCl and H2O present in the solution b. Use your results from a & b above to determine the moles of NaCl and H2O present in the 100 gram solution sample. c. Use your results from "b" to determine the molality of the solution , XNaCl & XH2O d. This is a very dilute solution. Make an appropriate assumption and determine the molarity of the solution. 5.3 Concentrated acetic acid solution is 99.5% by mass CH3COOH and has a density of 1.05 g/mL What is the molarity and XCH3COOH of the solution? (Assume water is the solvent) 5.4 Consider an aqueous solution of Ca(NO3)2 where the mole fraction, XCa(NO3)2 = 0.25 What is the molality of the solution? Answers: Click and drag in the space below. 5.1 a. 1180. gramssolution b. 11.65 molesHCl = 424.759 gramsHCl c. 755.241 gramsH2O = 41.9222 molesH2O d. 15.425 m XHCl = 0.2175 e. The solution's concentration is independent of the volume we consider. Assuming 1L as the solution's volume, we can easily calculate the solution's mass. Additionally, the assumption let's us know there are 11.65 moles of solute present which is necessary for later calculations. 5.2 a. 2.55 grams NaCl 97.45 grams H2O b. molNaCl = 0.043634 molH2O = 5.409297 c. molality = 0.4477 m XNaCl = 8.002 x 10-3 & XH2O = 0.9919 d. Assume solution has same density as water (i.e. the solution is very dilute) Molarity = 0.4363 M 5.3 [CH3COOH] = 17.399M (Note the brackets are used to designate Molar Concentration) XCH3COOH = 0.9835 5.4 18.50 m |