Chemistry 20 Student Notes PDF

**Chemistry 20**  **Part A: Solutions** NOTES AND PRACTICE Name: \_\_KEY\_\_\_ **[Lesson 1: Properties of Solutions]** **[Solutions ]** - Solutions are \_\_Homogeneous\_\_ mixtures. - A solution consists of at least one solute dissolved in a solvent. - ![](media/...

**Chemistry 20**  **Part A: Solutions** NOTES AND PRACTICE Name: \_\_KEY\_\_\_ **[Lesson 1: Properties of Solutions]** **[Solutions ]** - Solutions are \_\_Homogeneous\_\_ mixtures. - A solution consists of at least one solute dissolved in a solvent. - Solutes and solvents can be solids, liquids or gases. - The substance that is present in the largest quantity (whether by volume, mass of amount) is called the \_\_SOLVENT\_\_\_. - The substances that are dissolved in the solvent are called the \_SOLUTE\_\_\_. Solute in Solvent Example of Solution Source or Use ------------------- ----------------------------- -------------------------------------- Gas in gas **oxygen + acetylene** **welding fuel** Gas in liquid **carbon dioxide + water** **carbonated drinks** Gas in solid **hydrogen + platinum** **coating electrodes in fuel cells** Liquid in liquid **ethylene glycol + water** **antifreeze** Liquid in solid **mercury + silver** **dental fillings** Solid in liquid **sugar + water** **iced tea** Solid in solid **copper + zinc** **brass** **[Discovery Activity: PhET: Investigating Sugar and Salt]** Go to [[https://phet.colorado.edu/en/simulation/sugar-and-salt-solutions]](https://phet.colorado.edu/en/simulation/sugar-and-salt-solutions) and download the simulation.  **Macro View:** The first tab, "Macro" will allow you to shake either salt or sugar into the water container. Click "Show values" to get concentration values to appear in the graph. Then, drag the conductivity meter on the right-hand side of the screen so that the two electrodes are dipped into the solution.  1. What happens to the light bulb when the concentrations of each substance are zero? Select the **SALT** shaker and shake some into the water. 2. Record the concentration: \_\_\_\_\_\_\_\_\_\_ mol/L. What happened to the light bulb?  The light bulb begins to glow. 3. Shake more salt into the water. Record the concentration:  \_\_\_\_\_\_\_\_\_ mol/L. What is the relationship between salt concentration and light-bulb brightness? As salt concentration increases the brightness of the bulb increases. Click "Remove salt," and then switch the shaker from "Salt" to "Sugar."  4. Shake into the water. Record the concentration: \_\_\_\_\_\_\_\_\_ mol/L. What happened to the light bulb? 5. Shake more sugar into the water. Record the concentration:  \_\_\_\_\_\_\_\_\_ mol/L. What is the relationship between sugar concentration and light-bulb brightness? As sugar concentration increases the bulb remains unlit. Sugar will not conduct electricity. 6. Use the following terms to classify each substance. Terms: electrolyte, nonelectrolyte, conductive, nonconductive, molecular, ionic. Salt --  \_\_\_\_Ionic\_\_\_\_\_\_\_electrolyte\_\_\_\_\_\_\_\_\_ conductive\_\_\_\_\_\_\_\_ Sugar - \_\_\_\_Molecular\_\_\_\_\_non-electrolyte\_\_\_\_\_\_\_ non-conductive\_\_\_\_\_\_\_ **Micro View:** Click over to the second tab, "Micro." The shaker present should be "Sodium Chloride." If not, select this substance in the "Solute" box. Shake the sodium chloride into the water. 7. Watch how the substance appears before entering the water and after it enters the water. Describe that change.  The salt comes out as a crystalline structure from the shaker, once it enters the water the Na+ and Cl- ions dissociate and move around the water free from one another. Click "Remove solute" and switch the shaker over to "sucrose." Shake the sucrose into the water. 8. Watch how the substance appears before entering the water and after it enters the water. Describe that change. The sucrose comes out as a crystalline structure from the shaker, once it enters the water it stays together in its molecular form and does not dissociate into ions. 9. Which of the above has gone through "dissociation"? Write the dissociation equation below. 10. Is it more similar to salt or sugar from the previous question? Explain. Similar to NaCl -- both solutes will move around the solvent freely and randomly \- both solutes will separate into ions up contact with the solvent 11. Write the dissociation equation for calcium chloride. Look at the chart to the right with the bars. How do these two items relate?  Click "Reset All." There is an arrow pointing to the word "Solute" in the box. Click once to the right so the two substances are sodium chloride and sodium nitrate. Click on sodium nitrate and shake some into the water. 12. What part is similar to the salt dissolution you noticed previously? -- both solutes will move around the solvent freely and randomly \- both solutes will separate into ions up contact with the solvent 13. What part is similar to the sugar dissolution you noticed previously?  -- sodium nitrate separates into one sodium ion and one nitrate ions that does not break up any further but does break away from the sodium. 14. Write the dissociation equation for sodium nitrate. Look at the chart to the right with the bars. How do these two items relate?  15. Nitrate, NO~3~^-^, is known as a polyatomic ion. Based on your observations and prior knowledge, how would you describe the various bonding that occurs in a salt that contains a polyatomic ion?  Nitrate is ionically bonded to sodium since the water will break them apart when in solution, but the nitrogen and oxygen atoms of the polyatomic ion are covalently bonded to each other and therefore will not break up further in water and NO~3~^-^ will remain together in aqueous solutions. **[Properties of Aqueous Solutions]** - An **\_\_ [aqueous] \_\_\_** solution is any solution for which water is the solvent. Water is sometimes called the universal solvent because so many substance, both ionic and molecular, dissolve in water. Most solutions used in chemistry and most liquid solutions you encounter every day are aqueous solutions. - Substances that dissolve to form solutions that conduct electricity are called **\_\_ [ electrolytes].\_\_.** - **Electrolytes** disperse electrically charged particles called **\_\_ [ ions]\_\_**. - **\_ [ Dissociation] \_\_** describes the separation of ions that occurs when an ionic compound dissolves in water. - **Non‐electrolytes** disperse electrically **\_ [ neutral] \_\_** particles and therefore **\_\_ [ do not] \_\_** conduct electricity. - Most **\_ [ molecular] \_\_** compounds are non-electrolytes. - **\_\_ [ Acids] \_\_** are an exception because they are mostly molecular but will ionize in water and can conduct electricity. - Ionization the reaction of **\_\_ [ neutral]\_\_\_** molecular compounds forming **\_ [ charged]\_\_** ions.  **[Dissolving Equations]** - Represents when a substance simply dissolves (usually molecular compounds). - The formula for the solvent (H~2~O~(l)~) does not appear as a reactant in the equation. Although water is necessary for the process, it is not consumed and not a reactant. - The presence of water molecules surrounding the ions is indicated by (aq). - In dissolving the molecule remains together and will NOT be separated. **[Dissociation Equations ]** - Shows what happens when ionic compounds dissolve. - The formula for the solvent (H~2~O~(l)~) does not appear as a reactant in the equation. The presence of water molecules surrounding the ions is indicated by (aq). - In dissociation the molecule splits into its ions. **[Lesson 2: Solubility]** **[Discovery Activity: POGIL -- Saturated and Unsaturated Solutions]** Complete the handout depicting models of solubility. **[Solubility]** - The solubility of a solute is the amount of solute that dissolves in a given quantity of solvent at a given temperature. For example: NaCl~(s)~ has a solubility of \~36g/100mL at 20^o^C and \~39g/100mL at 100^o^C - A **\_\_[saturated.]** [ ] solution is a solution that contains the **\_[maximum]\_\_** amount of dissolved solute at a given temperature in the presence of undissolved solute. For example: 100mL of a saturated solution of NaCl at 20^o^C contains 36g of dissolved sodium chloride. If more sodium chloride is added to the solution, it will not dissolve. - An \_**\_ [ unsaturated] \_\_** \_\_ solution is a solution that is not saturated and, therefore, can dissolve more solute at that particular temperature. For example: a solution that contains 20g of sodium chloride dissolved in100mL of water at 20^o^C is unsaturated. - A **\_\_supersaturated\_\_\_** solution contains more of the dissolved solute than could be dissolved by the solvent under normal circumstances. **[Factors that affect solubility include: ]** 1. Nature of the Solute and Solvent - "Like Dissolves Like" -- refers to the \_\_[polarity] \_\_ of the substances. Water is a polar molecule. Substances that are polar will dissolve in water. These are said to be \_[miscible] \_\_ with water. Substances that are nonpolar will not dissolve in water. These are said to be \_\_[immiscible] \_ with water. 2. Temperature - For solids -- an \_\_[increase] \_\_ in temperature usually results in an [increase] \_ in solubility. - For gases -- the reverse is true. The \_[colder] \_ the solvent, the \_[greater] \_\_ the solubility of gases. - Explain WHY? If the heat given off in the dissolving reaction is less than the heat required to break apart the solid, the net dissolving reaction is endothermic. The addition of more heat facilitates the dissolving reaction by providing energy to break bonds in the solid. This is the most common situation where an increase in temperature produces an increase in solubility for solids. Increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in the gas molecules which break intermolecular bonds and escape from solution. 3. Pressure - For solids -- very little effect - For gases -- gases dissolve better under \_\_[higher] \_\_ pressures [Saturated Solutions and Equilibrium] - A dissociation equation can be written for the dissolving on an ionic compound: CuSO~4(s)~ Cu^2+^~(aq)~ + SO~4~^2-^~(aq)~ - Crystallization is the reverse of dissolve and can be written as: Cu^2+^~(aq)~ + SO~4~^2-^~(aq)~ CuSO~4(s)~ - In a saturated solution, which contains both dissolved and undissolved solute, both of these physical changes are taking place at the same time. - A saturated solution is considered to be in a state of \_\_[equilibrium] \_\_. Dynamic equilibrium occurs when a process and the reverse of the process take place at the same rate in a closed system. - An equilibrium reaction can be represented by combining both the forward and reverse processes into one equation with a double-headed arrow: **CuSO~4(s)~**  **Cu^2+^~(aq)~ + SO~4~^2-^~(aq)~** **[Lessons 1/2: Properties of Solutions and Solubility PRACTICE]** **[Exercise \#1: Electrolytes and Dissociation Equations]** 1. Classify each of the following as an electrolyte or non-electrolyte substances, forming conducting or non-conducting solutions? +-----------------------+-----------------------+-----------------------+ | | **Electrolyte or | **Conducting or | | | Non-Electrolyte?** | Non-Conducting?** | +=======================+=======================+=======================+ | a. **KI** | **Electrolyte** | **Conducting** | +-----------------------+-----------------------+-----------------------+ | b. **ICl** | **Non-electrolyte** | **Non-conducting** | +-----------------------+-----------------------+-----------------------+ | c. **CaBr~2~** | **Electrolyte** | **Conducting** | +-----------------------+-----------------------+-----------------------+ | d. **CH~4~** | **Non-electrolyte** | **Non-conducting** | +-----------------------+-----------------------+-----------------------+ | e. **LiOH** | **Electrolyte** | **Conducting** | +-----------------------+-----------------------+-----------------------+ | f. **CH~3~OCH~3~** | **Non-electrolyte** | **Non-conducting** | +-----------------------+-----------------------+-----------------------+ | g. **N~2~O** | **Non-electrolyte** | **Non-conducting** | +-----------------------+-----------------------+-----------------------+ 1. Write dissociation equations for the following substances when mixed with water. Show **physical states** of the products and balance the equations. h. hydrochloric acid i. sodium hydroxide  j. ammonium acetate k. potassium dichromate  l. nitric acid **[Exercise \#2 - Saturation Application Questions]** 1. A flask contains a saturated solution of NaCl in water. You carefully pour off 100mL of the solution, taking care not to let any crystals of salt fall into the new container. Is the salt solution in the new container saturated? Explain. Saturated as the number of dissolved particles is at its max for that temperature. 2. A student half-filled a 100mL beaker with water and added a few grams of NaCl crystals. Seeing the crystals settle immediately to the bottom of the beaker, the student said the solution was saturated because some undissolved solid was present. Was the student correct? Why? 3. Refer to the solubility graph to answer the following. a. Identify the substance in the graph above that is most soluble at 60°C. NaNO~3~ b. Identify the substance in the graph above that is least soluble at 60°C. SO~2~ ***Look at the curve for NH~4~Cl for questions c-f:*** c. If 12 g of solute is added to 100 g of water at 20°C, what type of solution will result? Saturated, unsaturated or supersaturated? Will dissolved easily. Unsaturated d. At 50°C, what is the maximum amount of this solute that can be dissolved in 100 g of water? 52 g e. If 50 g of solute is added to 100 g of water at 20°C, what type of solution will result? Saturated, unsaturated or supersaturated? f. Describe what you would observe in question 'e'. g. Identify and state the difference between the solubility curves for ammonia and sodium nitrate. *Note*: ammonia is a gas and sodium nitrate is a solid at room temperature. h. Suggest a reason why solubility decreases with increasing temperature for gaseous solutes but increases for solid solutes.  4. Using the diagram as a reference, describe the meaning of "supersaturated". Is it a condition that can last long? Why or why not? You may refer to the following link ([[http://preparatorychemistry.com/Bishop\_supersaturated.htm]](http://preparatorychemistry.com/Bishop_supersaturated.htm)) **[Lesson 3: Energy Changes]** - Dissolving is a \_ [ physical]\_\_\_ change. The molecules or ions of a solid solute are held together by bonds. When dissolving occurs, these bonds break and the ions or molecules of the solute become attracted to the solvent particles. When a solute dissolves in a solvent, such as water, the three processes of dissolving occur simultaneously: 1. The forces between the particles in the solute break. This process always requires energy and is an - In an ionic solid, the forces holding the ions together are \_ [ ionic]\_ bonds. In a molecular solid, the forces holding the moles together are \_ [London] \_ dispersion and \_\_ [ dipole-dipole]\_ forces. 2. Some of the intermolecular forces between the particles in the solvent (often water) also break. Because this process involves breaking bonds, it is an \_ [ endothermic] \_ process. - For example, in an aqueous solution, some of the \_ [ hydrogen]\_\_ bonds between water molecules break. 3. The attraction between the particles of solute and the particles of solvent result in the formation of new bonds. This step always \_\_ [ releases]\_\_ energy. Processes that release energy are called \_ [ exothermic] \_\_ processes. 4. Using the law of conservation of energy, the overall energy change observed when a solute dissolves in a solvent is the sum of the energy changes in the three steps. **Energy required to break bonds in solute** ------- ------------------------------------------------------------------- **+** **Energy required to break bonds in solvent** **-** **Energy released from bonds forming between solute and solvent** **=** **Total energy change**  - If the amount of energy required to break bonds is [more]\_\_ than the amount of energy required to make bonds, the overall change is \_\_ [ endothermic]\_\_. - If the amount of energy required to break bonds is \_ [ less] \_ than the amount of energy required to make bonds, then overall change is \_ [ exothermic]\_\_\_. **[\*\*\*EXTENSION EXAMPLE\*\*\* NOT A PART OF CHEM 20:]** The chart to the left shows the lattice energy (energy required to break bonds in salt) and the hydration energy (energy released when ions dissolved) for various salts. The energy required to break the intermolecular bonds in water is 21 kJ/mol. Determine if the following salts undergo and endothermic or exothermic process when then dissolve. a. LiF = (1032 + 21) -- 1005 = +48 kJ/mol ENDO b. SrCl~2~ = (2110 + 21) -- 2161 = -30 kJ/mol EXO c. KI = (632 + 21) -- 617 = +36 kJ/mol ENDO d. NaOH = (737 + 21) -- 799 = -41 kJ/mol EXO e. Which samples above will feel cold? Which ones will feel warm? COLD = LiF and KI Warm = SrCl~2~ and NaOH **[Lesson 4: Expressing Concentration of Solutions -- % Concentration and ppm]** Most solutions are similar in that they are colourless and aqueous, so there is no way of knowing, by looking at them, how much of the solute is present in the solution.   Concentration is a ratio comparing the quantity of \_[solute] \_\_\_ to the quantity of \_\_\_[solution] \_\_. - *Dilute* -- a relatively \_[small] \_\_ quantity of solute per volume of solution - *Concentrated* -- a relatively \_[large] \_\_ quantity of solute per volume of solution There are many different ways to express concentration: - percent concentration - parts per million / billion - amount concentration / molar concentration / molarity **[Percent Concentration:]** 1. **%W/W** \ [ ]{.math.display}\ **[Example \#1:]** Calcium chloride, CaCl~2(s)~ can be used instead of road salt to melt the ice on roads during the winter. To determine how much calcium chloride had to been used on a nearby road, a student took a sample of slush to analyze. The sample had a mass of 23.47g. When the solvent was evaporated, the residue had a mass of 4.58g. What was the percent by mass of calcium chloride in the slush? 2. **%V/V** \ [ ]{.math.display}\ **[Example \#2:]** A photographic soap bath is made by mixing 140mL of pure acetic acid with 360mL of water. What is the percentage by volume concentration of acetic acid?  3. **%W/V** \ [ ]{.math.display}\ **[Example \#3:]** A topical solution of hydrogen peroxide is typically 3.0%W/V. What mass of hydrogen peroxide must be mixed with water to produce a 150mL bottle of the ointment? **[Concentration in Parts per Million:]** This unit gives more reasonable numbers to very dilute concentration. (One ppm is only a drop of solute in a full bathtub.)  **[Example \#4:]** "Hard" water contains a relatively large quantity of dissolved minerals, one of which is calcium carbonate. Hard water in Alberta may contain up to 300 ppm calcium carbonate. Older models of toilets use up to 20L per flush. How much calcium carbonate could be crystallized from the water in one flush of an older model of toilet? **[Lesson 4: % Concentration and ppm PRACTICE]** 1. Glucose is a sugar that is found abundantly in nature. What is the percent by mass of a solution made by dissolving 163 g of glucose in 755 g of water? Do you need to know the formula of glucose? **\[17.8%\]**       2. Determine the volume percent of toluene in a solution made by mixing 40.0 mL toluene with 75.0 mL of benzene. **\[34.8%\]**       3. Describe the process you would use in order to prepare 5.00 kg of an aqueous solution that is 8.00% NaCl by mass.                       **   \[400g NaCl in 4.60 kg of water\]**         4. What is the mass percent of solute when 4.12 g is dissolved in 100.0 g of water?  **\[3.96%\]**       5. What is the volume percent of 10.00 g of acetone (d=0.789 g/mL) in 1.55 L of solution? **\[0.818%\]** ** **       6. If I make a solution by adding 83 grams of sodium hydroxide to 750 mL of water. What is the percent by mass of sodium hydroxide in this solution? **\[10%\]** 7. If I make a solution by adding water to 35 mL of methanol (CH~3~OH) until the final volume of the solution is 275 mL. What is the percent by volume of methanol in this solution? **\[13%\]**     **Extra practice (%, ppm, ppb)**: *In all of these assume that every ml of water has a mass of 1 gram.* 1. Calculate the concentration of salt in a solution of water in percent if 45 grams is dissolved in 1200 ml of water. **\[3.6%\]** 2. Calculate the mass of solvent in a 6.0 ppm solution of a drug if the mass of the solute is 0.050 milligrams. (In this case the amount of the solute is very small and treat the solution as though the mass of solvent=mass of the solution.) **\[8.3 g\]** 3. What is the concentration in ppm of selenium if 1.3 milligrams is found in 2500 kg of soil? 4. What is the concentration in ppb of PCB\'s in a chemical spill, if there is 0.060 mg in 4,600 Kg of soil? 5. Calculate the number of moles of H~2~O~2~ in 450ml of a 35% H~2~O~2~ solution.       **\[4.6 moles\]** 6. Calculate the mass of solute PCB\'s in a 65 Kg person, if the concentration is 4 ppm?             **\[0.26 g\]** 7. What mass of solution would be needed to deliver 3.0 mg of a drug if the concentration of the drug in the solution was 3.5%?           **\[0.086 g\]** 8. What is the concentration, in ppm, if 0.808 g of CaCl~2~ is dissolved in 250.0 ml of water?**\[3.22x10^3^ ppm\]** 9. 25 grams of a chemical is dissolved in 75 grams of water. What is the % of solute in this solution? **\[25%\]** 10. Suppose 17 grams of sucrose is dissolved in 183 grams of water. What is the concentration of sucrose in ppm? **\[85000 ppm\]** 11. 35 grams of ethanol is dissolved in 115 grams of water.  What is the concentration of ethanol in parts per billion (ppb)?   **\[2.3x10^8^ ppb\]** 12. A certain pesticide has a toxic solubility of 5.0 grams/Kg of body weight.  What is this solubility in ppm? Assume 1Kg of body weight is the solution.                   **\[5000 ppm\]** **[Lesson 5: Expressing Concentration of Solutions -- Molar Concentration and Molarity]** **[Discovery Activity: PhET Concentratin and Molarity]** ***[Part A: Exploring Concentration:]*** Familiarize yourself with the operation of the various functions of the simulation. You can always press the RESET button to get back to the original situation.  Move the concentration probe into the liquid. Then shake some drink mix into the water. 1. What happens to the concentration as more solute is added? **The concentration increases as more solute is added.** 2. What units of the concentration on the meter measuring "concentration"? **mol/L** 3. What does it mean and how is it different than the unit g/L? 4. What is the relationship between concentration and color? How do you know? **The higher the concentration, the darker the colour. It is visible in the simulation.** Reset and choose (from the drop-down menu) a chemical "solid" solute to shake in. Be sure the concentration probe is in the solution. Add water using the tap above. Then try sliding the "Evaporation" tab. Repeat using the same chemical "solution" from the eye dropper. 5. What do you find the relationship to be between volume of water and concentration? Explain. 6. Now try the above again, but this time instead of opening the tap above, open the tab below on the right. What happens to the concentration? Why do you think so? ***[Part B: Saturation:]*** Press Reset then and select CuSO~4~ from the drop-down menu and move the concentration probe into the solution. Create a saturated solution. 7. How do you know when a saturated solution is created? Try shaking more. Does the concentration change? **When the solution reaches its saturation point  (1.380 mol/L), adding more solute will not change the concentration.** 8. What is the concentration value? \_\_**1.380 mol/L** \_\_\_ 9. Now reset and use the "solution" form of copper(II) sulphate. Make a saturated solution. Explain what you needed to do. 10. Do you think this concentration will be the same value if you made a saturated solution of potassium permanganate? Why or why not? Try it. **We think that the saturated concentration of potassium permanganate will be different than the saturated concentration of CuSO4. This is because different substances have different saturation points.** **[*Part C: Exploring Molarity:*]** Familiarize yourself with the operation of the various functions of the simulation. You can always press the RESET button to get back to the original situation. Be sure the box for "Solution Values" is checked on. In Chemistry, we often want to do numerical calculations. You should see that the Solute Amount (moles) = 0.500 mol, the Solution Volume (Liters) = 0.500 L, and the Solution Concentration (Molarity) = 1.000 M. Play with the sliders and find other value combinations for moles and Liters that will result in a Molarity of 1.000 M Solute Amount (moles) Solution Volume (Liters) Solution Concentration (Molarity) ----------------------- -------------------------- ----------------------------------- **0.494 mol** **0.494 L** 1.000 M **0.635 mol** **0.635 L** 1.000 M **0.904 mol** **0.904 L** 1.000 M **0.200 mol** **0.200 L** 1.000 M 1. Did you have to move the sliders in the same or opposite directions to achieve your goal each time? **We have to move the sliders in the same direction to achieve our goal.** The concentration or strength of a solution is called its **Molarity**. We can write the formula: \ [\$\$\\mathrm{Solution\\ Concentration\\ (Molarity)}\\mathrm{\\ }\\mathrm{=}\\mathrm{\\ }\\frac{\\mathrm{Solute\\ Amount\\ (moles)}}{\\mathrm{Solution\\ Volume\\ (Liters)}}\\mathbf{\\text{\\ \\ \\ \\ \\ \\ \\ }}\\mathrm{\\text{or}}\\mathbf{\\ \\ \\ \\ \\ \\ \\ \\ c =}\\frac{\\mathbf{n}}{\\mathbf{V}}\$\$]{.math.display}\ 2. Look at the formula for concentration. Explain why the action in the question above makes sense. Using the equation and your calculator, determine the missing values in the table. Then check with the simulation. +-----------------+-----------------+-----------------+-----------------+ | moles | Liters | Molarity | Show | | | | | Calculations | +=================+=================+=================+=================+ | 0.887 mol | **0.558 L** | 1.590 M | **V= n/C** | | | | | | | | | | **V= (0.887mol) | | | | | / (1.590 | | | | | mol/L)** | +-----------------+-----------------+-----------------+-----------------+ | 0.299 mol  | **0.830 L** | 0.360 M | **V= n/C** | | | | | | | | | | **V= (0.299 | | | | | mol) / (0.360 | | | | | mol/ L)** | +-----------------+-----------------+-----------------+-----------------+ | 0.750 mol | 0.295 L | **2.542 M** | **C= n/V** | | | | | | | | | | **C= (0.750 | | | | | mol) / (0.295 | | | | | L)** | +-----------------+-----------------+-----------------+-----------------+ | 0.908 mol | **0.585 L** | 1.552 M | **V= n/C** | | | | | | | | | | **V= (0.908 | | | | | mol) / (1.552 | | | | | mol/ L)** | +-----------------+-----------------+-----------------+-----------------+ | 0.205 mol | 0.880L | **0.233 M** | **C= n/V** | | | | | | | | | | **C= (0.205 | | | | | mol) / (0.880 | | | | | L)** | +-----------------+-----------------+-----------------+-----------------+ | 0.205 mol | **0.200 L** | 1.025 M | **V= n/C** | | | | | | | | | | **V= (0.205 | | | | | mol) / (1.025 | | | | | mol/ L)** | +-----------------+-----------------+-----------------+-----------------+ [Calculating Grams]: A problem, like the one below, first requires you to identify how many moles of NiCl₂ using a molarity formula, and then to convert that number of moles to grams using the molar mass of NiCl₂. If you wanted to make 0.800 Liters of a 0.531 M solution of Nickel(II) chloride, NiCl₂, how many grams of NiCl₂ would you need? Calculate using formulas and your calculator. Then use the simulation to check your answer. **V= 0.800 L                          C=n/V** **C= 0.531 mol/L                   n=CV** **n=?                                      n= (0.531 mol/L) ( 0.800L)         =              0.4248 mol** **0.4248 mol x [129.59 g] =    55.0 g NiCl₂** **                        1 mol** **Test your understanding: You may use the simulations to check your answers.**  **1 =   [B]    2 =    [C]     3 =    [A]    4 =  [B]      5 =   [A]     6 =   [E]    7 = [E]     8=  [E]** **[Amount Concentration / Molar Concentration / Molarity]**  - The most commonly used unit of concentration in chemistry is amount concentration. Text Description automatically generated Shorthand notation for molar concentration = square brackets. Ie. \[NaCl\] represents the concentration of sodium chloride in mol/L **[Example \#1:]** **A laboratory technician made a solution using 0.255 mol of sodium hydroxide in 1.50 L of solution. Calculate the amount (molar) concentration of sodium hydroxide.** **[Example \#2:]** A sample of ammonia solution has a concentration of 11.5 mol/L. What chemical amount of ammonia is present in a 3.5 L bottle? **[Example \#3:]** A student dissolves 5.00 g of solid sodium carbonate to make 125 mL of a solution. What is the concentration of this solution?  **[Molar Concentration of Ions in Solution]** In solution, ionic compounds separate by dissociation into individual ions. Writing a dissociation equation will help you relate the concentration of ions to the concentration of solute. **[Example \#4:]** Find the molar concentration of aluminum and sulfate ions in a 0.30 mol/L solution of aluminum sulfate. **[Example \#5:]** Determine the molar concentration of barium and hydroxide ions in a solution made by dissolving 7.34g of barium hydroxide to make a volume of 150 mL.  **[Example \#6:]** What mass of magnesium bromide must be dissolved to make 1.20L of solution with a bromide ion concentration of 0.40 mol/L? **[Lesson 5: Molar Concentration and Molarity PRACTICE]** **[Exercise \#4 -- Concentration of Solutions Problem Set]** 1. Calculate the molar concentration for the following: a. 2.3 moles of sodium chloride in 0.45 liters of solution. **\[5.1 M\]** b. c. d. e. f.   2. How many moles of beryllium chloride are needed to make 125 mL of a 0.050 M solution? **\[6.3x10^-3^ mol\]** 3. What is the volume (in mL) of a 0.500 M solution if it contains 25.0 grams of calcium hydroxide? 4. How many grams of ammonia are present in 5.0 L of a 0.050 M solution? **\[4.3 g NH3\]**  **For the next questions, write the dissociation equations first, and then find the concentration of ions asked.** 5. What is the concentration of each ion in a 10.5 M sodium sulfite solution? **\[21.0 M Na^+^, 10.5 M SO~3~^2-^\]** 6. What is the concentration of the sulphate ion formed in solution when 94.78 g of iron (III) sulfate is dissolved into 550.0 mL of water?  **\[1.293 M SO~4~^2-^\]** 7. What is the concentration of each ion in a 5.55 M zinc phosphate solution? **\[16.7M Zn^+2^, 11.1M PO~4~^3-^\]** 8. What is the concentration of each ion in a 1.22 M zinc acetate solution? **\[1.22M Zn^+2^, 2.44M CH~3~COO^-^\]** **\ ** 9. What is the concentration of the sodium ion in a 0.20 M sodium phosphate solution? 10. What is the concentration of the Al^3+^ formed when 16.5 g of aluminum sulfate is dissolved into 600.0 mL of water?  **\[0.161 M Al^+3^\]** **[Lesson 6: Net Ionic Equations]** Solubility of a salt depends upon the type of ions in the salt. Some salts are soluble in water and others are not. When two soluble salts are mixed together in water they may form a third insoluble salt. Net ionic equations are a way of showing the reactions that take place between two substances dissolved in water.  An **\_\_aqueous \_\_\_** solution is a solution with water as the solvent.  A compound is said to be **\_soluble \_** if it readily dissolves in water and does not *precipitate* if left undisturbed for an extended period of time.  A **\_\_\_\_ spectator ion \_** is an ion that is present during a reaction but does not take part in the reaction.  A \_\_\_\_\_net ionic equation \_\_\_ *is an* equation that only shows the ions that undergo changes during a chemical reaction. (Spectator ions are omitted from net ionic equations.)  **[Writing Net Ionic Equations:]** 1. Write a complete balanced chemical equation. 2. Dissociate all high-solubility ionic compounds, and ionize all strong acids to show the complete ionic equation. 3. Cancel spectator ions that appear on both the reactant and product sides. - When cancelling spectator ions, they must be identical in every way: chemical amount, form, and state of matter 4. Write the net ionic equation, reducing coefficients if necessary. When a soluble salt is placed in water, it separates into its ions. For example, sodium chloride is soluble. **[Demonstration \#1:]** **Sodium nitrate reacts with potassium acetate in an aqueous solution. ** In *double displacement (replacement)* reactions, two ionic compounds react and switch ions.  According to this "pencil and paper" reaction, potassium nitrate and sodium acetate are produced. However, if this reaction is actually carried out in an aqueous solution, nothing appears to happen. If we investigate this system using the concept of a net ionic reaction we can see why it appears that nothing happens. First, we write all of the compounds in the equation, showing the ions that are formed when the reaction is carried out in water.  Next, cross out any ions that are present on both the reactant side and product side of the reaction.  https://lh6.googleusercontent.com/Kam7nBwOL3I5uQa9wYTN01yoUhhCvKnk7HWNxABJHIAscF63kwHnZ-Atyd4MpyymFd3ApQNb-sU4kwlgS8\_AsizPa9ZBMyHDU\_0BP9W\_RzQ9ynRAmPZ7bCtYRMqJcqzWshS-2As\_ The ions we cross out, which are the same on both sides, are called *spectator ions* (they are just "standing around watching", hence the term spectator). Both of the compounds on the lefthand side of the reaction (reactants) and the right hand side of the reaction (products) are soluble. Therefore, no solid forms and ***[no reaction occurs]***. The ions are all simply floating around together in the solution.  **[Demonstration \#2:]** **Strontium nitrate solution reacts with potassium sulfate solution. ** Predict products: Sr(NO~3~)~2(aq)~ + K~2~SO~4(aq)~ → 2 KNO~3(aq)~ + SrSO~4(s)~ Write soluble compounds as ions and insoluble compounds in combined form: Cross out spectator ions:  Final net ionic equation: Sr^2+^~(aq)~ + SO~4~^2−^~(aq)~ → SrSO~4(s)~  In this example, writing a net ionic equation is useful because it indicates those chemical species that participate in the chemical reaction and form an insoluble product.  ***Key Questions:*** Graphical user interface, text, application Description automatically generated 5. In Demonstration \#1, solutions of sodium nitrate and potassium acetate are mixed together forming soluble products. If the solution were to be evaporated to dryness, name all of the compounds that might be found left over in the container.   **[Lesson 6: Net Ionic Equations PRACTICE]** 1. Write the **NET ionic equation** for each of the following reactions. Make sure your net ionic equation is properly balanced. Text, letter Description automatically generated  2. Write the **formula equation, complete ionic, and net ionic equations** for the following reactions.  a. A zinc strip is dipped into an aqueous solution of silver nitrate. Text Description automatically generated b.  c. c.  d. Text Description automatically generated e.  **[Lesson 7: Preparation of Solutions and Dilution]** Standard Solution -- \_\_ [ a solution with a precisely known concentration] \_\_ - Used in chemical analysis and to precisely control chemical reactions - Precision equipment is required to measure mass of solute (electronic balances) and volume of solution (volumetric flask) There are two ways to make a standard solution: 1. Dissolve a measured amount of pure solute in a certain volume of solvent 2. Dilute a standard solution Dilution -- \_\_ [ Decreasing] \_\_ the concentration of a solution - Usually accomplished by \_\_ [ adding more solvent] \_\_ to a stock solution. - The number of moles of solute [does not change] \_ when a solution is diluted - Dilution is especially important in manipulating the concentration of solutions in chemistry for better control of reactions. - Concentrated solution reactions can be too \_\_ [ violent] \_\_ to be safe and/or too \_ [ fast] \_ to observe. Where: c~1~ = initial concentration V~1~ = initial volume c~2~ = final concentration V~2~ = final volume 1. **[General Procedure for dissolving a measured amount of solute:]** **Step 1:** - Using an electronic balance, measure the mass of solute into a small beaker. - Place the beaker on the balance. Press "tare" or "zero" to re-zero the balance. - Add the solute using a scoopula until you reach the desired mass. - If a weigh boat is used instead of a beaker, the weigh boat must be rinsed into the beaker to reduce the amount of solute lost during the transfer. **Step 2:** - Add distilled water to the solute in the beaker. Stir with a stirring rod to dissolve. - Add approximately 1⁄2 of the total volume of water desired for the final solution (ie: if you are making 100 mL of solution, add about 50 mL -- this volume does not have to be precise) - At this point, all of the solute must be dissolved. Add more water if necessary or heat the solution to dissolve all of the solute (do not add more than the total volume of the solution and leave some room for the water from rinsing.) **Step 3:** - Transfer your solution to a volumetric flask of appropriate size. - A **volumetric flask** is a pear-shaped glass container with a flat bottom and a long neck. They are used to make up standard solutions because they are carefully calibrated to the precise volume listed. - Ensure that the volumetric flask is clean before using. You may rinse the flask with distilled water. The flask does not need to be dry. - Use a funnel to transfer the solution from the beaker to the volumetric flask. **Step 4:** - Rinse the beaker and stirring rod into the funnel with distilled water. - Rinse all glassware that came into contact with the solution into the funnel. This will reduce the amount of solute lost during the transfer. - Thoroughly rinse the funnel into the flask. - Be careful not to add too much water -- the level MUST remain below the etched line on the neck of the volumetric flask. **Step 5:** - Add distilled water until the bottom of the meniscus reaches the etched line on the volumetric flask. Add the rest of the water slowly. The narrow neck of the volumetric flask will fill up quickly. - When the flask is almost full, add the water drop by drop until the bottom of the meniscus rests at the etched line. The flask must be on a level surface and the line must be at eye level. - Stopper the flask, keep your thumb on the stopper and invert the flask several times to mix the solution. - Depending on the size of the volumetric flask used, the volume will be precise to either one or two digits after the decimal point. Check with your teacher as to the precision of your laboratory glassware. 2. **[General Procedure for dilution:]** **Step 1:** - Rinse the pipette with the standard solution (see technique in Step 2). Discard. - A **pipette** is a thin glass tube that is used to measure a precise volume of solution. There are two types of pipettes: graduated and volumetric. - Pipette A is a graduated pipette. It is calibrated to measure a number of volumes, like a graduated cylinder. Pipette B is a volumetric pipette. It is calibrated to measure one volume only, like a volumetric flask.  **Step 2:** *For a manual pipette:* - Squeeze some air from the pipette bulb before lightly placing it over the pipette to draw up the solution (draw up more solution than desired without overfilling the pipette and flooding the bulb). - Remove the bulb and quickly place your index finger over the top of the pipette. - By slightly rolling your index finger to the side, release the solution from the pipette until the bottom of the meniscus is on the line etched into the glass. (be sure to look at the line at eye level) *For a 3-valve pipette:* - Insert the top of the pipette in the bottom of the pipette filler. - Release air from the pipette filler by squeezing valve "A" on the top of the pipette filler while simultaneously squeezing the bulb. - Insert the tip of the pipette into the liquid to be dispensed. - Siphon liquid into the pipette to the desired level by squeezing valve "S" on the bottom of the pipette filler. Be careful not to draw liquid into the pipette filler. - Empty the pipette by squeezing valve "E" on the side‐tube. **Step 3:** - Release the standard solution into a clean volumetric flask. - **DO NOT** force the last drop from the pipette using the bulb. The pipette should be vertical and should touch the side of the flask as shown below. - After the solution has visibly stopped flowing, keep the pipette touching the side of the flask for three seconds to ensure the full volume has had time to exit the pipette. - A small volume of standard solution will remain in the pipette. Pipettes are calibrated in the factory based on how much solution they *deliver*, not on how much solution they *contain*. **Step 4:** - Add distilled water until the bottom of the meniscus reaches the etched line on the volumetric flask. Stopper the flask, keep your thumb on the stopper, and invert several times to mix the solution. **[Example \#1:]** A jug contains a solution with a concentration of 3.46 mol/L. A student wishes to make a 100 mL solution with a concentration of 0.100 mol/L. Show calculations and briefly describe how to make this dilution. \-- Take 2.89 mL of the original solution from the jug and place in a 100 mL volumetric flask. \-- Add enough distilled water to reach a volume of 100 mL \-- Now you have a 100 mL solution with a concentration of 0.100 mol/L **[Example \#2:]** A chemistry student took 13.2 mL of solution from a stock bottle and added enough water to this to create a 1.15 mol/L solution with a volume of 250 mL. Calculate the initial concentration. **[Lesson 7: Preparation of Solutions and Dilution PRACTICE]** **[Preparation of Solutions:]** 1. I have two solutions. In the first solution, 1.0 g of sodium chloride is dissolved to make 1.0 liters of solution. In the second one, 1.0 g of calcium chloride is dissolved to make 1.0 L of solution. Is the molarity of each solution the same?  Explain your answer.   2. I have two solutions. In the first solution, 1.0 moles of sodium chloride is dissolved to make 1.0 liters of solution. In the second one, 1.0 moles of sodium chloride is added to 1.0 liters of water. Is the molarity of each solution the same?  Explain your answer. 3. For the next questions, explain the procedure you would take to make the following solutions in the lab. Include the equipment you would use and the amounts you would need to measure out. a. 450 mL of a 0.250 M NaOH solution. **\[4.50 g NaOH\]** b. c. **Dissolve 437 g HCl, dilute to 2 L** d. 4. Explain why this experimental procedure is incorrect: To make 1.00 L of a 1.00 M NaCl solution, I will dissolve 58.5 grams of sodium chloride in 1.00 L of water.      **[Dilutions]**: 1. If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? **C~1~V~1~ = C~2~V~2~** **(0.15 M)(0.125 L) = x (0.150 L)** **x = 0.125 M** 2. If I add water to 100.0 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? **\[0.10 M NaOH\]** **C~1~V~1~ = C~2~V~2~** **(0.15 M)(0.1000 L) = x (0.150 L)** **x = 0.10 M** 3. How much 0.0500 M HCl solution can be made by diluting 250 mL of 10 M HCl? **\[50.0 L\]** **C~1~V~1~ = C~2~V~2~** **(10 M)(0.250 L) = (0.05 M) x** **x = 50.0 L** 4. I have 345 mL of a 1.5 M NaCl solution.  If I boil the water until the volume of the solution is 250 mL, what will the molarity of the solution be? **\[2.07 M NaCl\]** **C~1~V~1~ = C~2~V~2~** **(1.5 M)(0.345 L) = x (0.250 L)** **x = 2.07 M** 5. How much water would I need to add to 500. mL of a 2.4 M KCl solution to make a 1.0 M solution? **\[700 mL\]** **C~1~V~1~ = C~2~V~2~** **(2.4 M)(0.500 L) = (1.0 M) x** **x = 1.2 L** **1200 mL will be the final volume of the solution. However, since there's already 500 mL of solution present, you only need to add 700 mL of water to get 1200 mL as your final volume. The answer: 700 mL.** 6. Can you prepare 6.0 M NaI solution from 9.0 M NaBr solution by dilution? Explain. 7. To what volume will you have to dilute 30.0 mL of a 12 M HCl solution to make a 0.35 M HCl solution? **\[1.03 L\]** **C~1~V~1~ = C~2~V~2~** **(12 M)(0.0300 L) = (0.35 M) X** **x = 1.03 L**