In This Lesson: Molarity (Lesson 2 of 4) Today is Wednesday, January 10 th, 2018 Stuff You Need: Calculator Periodic Table Polyatomic Ion List Worksheet Pre-Class: Earlier in the year I used hydrogen peroxide in a lab but mentioned it was different from normal peroxide. How come? Also: Please find a worksheet and have your solubility curve packets out.
Today s Agenda Concentrations Molarity Percent solutions Dilutions In other words, today we re going to talk about the strength of solutions. Colligative Properties Where is this in my book? P. 525 and following
By the end of this lesson You should be able to calculate the concentration of a solution in several different ways. You should be able to calculate freezing-point depression and boiling-point elevation.
Practice Molarity Practice Chart and Problems First three rows. Use the first column as a sample!
Molarity Practice Chart and Problems Rest of key online Name Calcium iodide solution Sodium chloride solution Potassium carbonate solution Aluminum hydroxide solution Glucose solution Sulfuric acid Formula CaI 2 NaCl K 2 CO 3 Al(OH) 3 C 6 H 12 O 6 H 2 SO 4 Electrolyte Electrolyte Electrolyte Electrolyte Electrolyte or nonelectrolyte Nonelectrolyte Electrolyte Ions in solution Ca 2+, 2 I - Na +, Cl - 2 K +, CO 3 2- Al 3+, 3 OH - N/A 2 H +, SO 4 2-
Let s start with the easy, shall we? Recently we talked about solutions, and today we started by discussing the difference between 3% and 30% H 2 O 2. How are those numbers calculated? http://www.acne.org/images/products/img-442.jpg
Percent Solutions There are two types of % Solutions. Percent by volume. Percent by mass. Each typically measures using g and/or ml. Decide which to use by looking at the denominator s units. Let s start with volume, and here s your practice problem: What is the percent by volume of 5 g of CaCl 2 salt in 250 ml solution?
Percent by Volume What is the percent by volume of 5 g of CaCl 2 salt in 250 ml solution? This is a percent by volume problem, since we re looking at a 250 ml solution. 5 g out of 250 ml is 2%. (5 g / 250 ml) * 100 = 2% If the solute is a solid or gas: mass or volume of solute Percent by volume volume of solvent 100
Percent by Volume If the solute is a liquid, you have to use this formula: volume of solute Percent by volume 100 volume of solution As an example, imagine putting 5 ml of methanol into 100 ml of water. The volume of solution (denominator) is now 105 ml. Fun fact: Liquids (or other stuff) that can be infinitely mixed with others are known as miscible.
Percent By Mass Percent by mass is very similar. Note, however, that the denominator is the total mass of solution (and not solvent). This is because volume is not meaningfully impacted by a solid or gas solute, but mass is. Percent by mass mass of solute mass of solution 100
Okay. Why? Why the differences in calculations? Imagine a glass of 50 ml of water. It s also 50 g. If I shake 10 g of sodium chloride into it, does the water level change? Nope. But does the mass change? Yep. Hence, volume of solvent but mass of solution.
Practice Molarity and Solution Percent worksheet #1, 4, 5 (#1 is challenging)
Molarity At some point this year you may have seen me label acids or other liquids with something like: 6 M HCl The big capital M stands for Molar, as in not your teeth. 6 M HCl is pronounced Six molar hydrochloric acid. Molarity [that s the name of this concept] is another way to measure the concentration or dilution of a solution. The higher the molarity, the stronger the solution.
Calculating Molarity Calculating molarity is fairly simple. Here s the equation: M is the symbol for Molarity M moles of solute liters of solution
Need a Reminder? Molarity Equation Mole Conversion mol g M L mol molar mass
Molarity Practice Problem Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100 ml of solution. What is the molarity of the solution? 0.90 g NaCl = 0.0154 mol NaCl 100 ml = 0.100 L (0.0154 / 0.100) = 0.154 M
Molarity Practice Problem Just for good measure, if you put 0.90 g NaCl in 100 ml of water (a 0.154 M solution), what is the percent mass by volume? NaCl is a solid here, so we can use this equation: Percent by volume mass or volume of solute volume of solvent 100 (0.90 g NaCl / 100 ml water) * 100 = 0.9%
Molarity Practice Problem Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO? 0.7 M = moles of NaClO / 1.5 L Solve for moles of solute: 1.05 mol NaClO
Practice Molarity and Solution Percent worksheet #2, 3
Practice Molarity Practice Chart and Problems Shaded four rows. Use the first column as a sample! Molarity Practice Chart and Problems Bottom two rows. For cation/anion molarity, multiply the coefficient in front of each ion in the Ions in Solution row by the molarity in the Molarity of Solution row. Use the first column as a sample!
Molarity Practice Chart and Problems Rest of key online Name Calcium iodide solution Sodium chloride solution Potassium carbonate solution Aluminum hydroxide solution Glucose solution Sulfuric acid Ions in Solution Molarity of Solution Ca 2+, 2 I - Na +, Cl - 2 K +, CO 3 2- Al 3+, 3 OH - N/A 2 H +, SO 4 2-0.045 M 1 M 0.50 M 0.75 M 1.5 M 0.51 M Molarity of Cation 1 x.045 M = 0.045 M 1 M 2 x 0.5 M 0.75 M Na+ = 1 M K + Al 3+ - 2 x 0.51 M = 1.02 M H + Molarity of Anion 2 x.045 M = 0.090 M 1 M 0.5 M Cl- CO 2-3 3 x 0.75 M = 2.25 M OH - - 0.51 M SO 4 2-
Computer Practice Okay, now it s time to do a little reinforcement using the computers. Head to Quia and open the quiz called: Salts and Solubility From there, it s a pretty self-explanatory quiz that will have you visiting another page at the same time (PhET, an awesome website). You will be looking at various salts dissolving into water and making molarity calculations based on them. Username: [firstnamelastname]832 Password: Gleicher[house/apt. #]
Dilutions of Molar Solutions Sometimes chemists need to create dilutions using known molarities and volumes. For example, during our Baking Soda/HCl lab, I used 6 M HCl. However, the school receives hydrochloric acid in big jugs of 12 M HCl (nasty stuff). I needed to dilute it down to 6 Molar, and here s how I did
Molar Dilutions Because we re not changing how much solute is around during a dilution (we re just adding water, the solvent), we can use the following formula: M 1 V 1 = M 2 V 2 In other words, like with gas laws, when molarity goes up, volume typically goes down. This is Boyle s Law, except with molarity instead of pressure.
Molar Dilution Practice Problem How many milliliters of aqueous 2.00 M MgSO 4 solution must be diluted with water to prepare 100.0 ml of aqueous 0.500 M MgSO 4? M 1 V 1 = M 2 V 2 2.00 M * V 1 = 0.5 M * 100 ml V 1 = 25 ml
Molar Dilution Practice Problem You put 2 moles of HCl into 312 ml of water. If you wanted to make a 1 M dilution, to what volume would the original solution need to be diluted? 2 mol HCl / 0.312 L = 6.41 M HCl solution 6.41 M * 312 ml = 1 M * V 2 V 2 = 2000 ml
Practice Molarity Practice Chart and Problems Reverse side (all)
Finally our last topic of the lesson. Let s introduce it with a photo: http://www.bceo.org/images/saltbrinesystem_e.jpg
So why use salt on a road? Salt on a road lowers the freezing point of water, thus reducing icing. All those dissolved salt ions interfere with the water s ability to make an orderly crystalline arrangement and thus freeze. You can also use this phenomenon to make ice cream. Put your ingredients in a plastic bag and put the bag in a bucket of ice water. The ingredients (milk, sugar) have lots of dissolved solutes. Add lots of salt to the ice water. The ice water temperature will drop as the ice absorbs heat from the water, but the water can t freeze. In turn, the ice water absorbs heat from the bag of ingredients, which needs to drop significantly below freezing to freeze. Eventually you get that nice softish ice cream texture.
Colligative Properties In fact, this is one of three main colligative properties we re going to learn: 1. Freezing-Point Depression 2. Boiling-Point Elevation 3. Vapor-Pressure Lowering A colligative property is one that depends only on how much solute you have, not what kind of solute it is. Like osmosis. You remember biology, right?
1. Freezing-Point Depression Every mole of dissolved solute particles lowers the freezing point of 1000 g of water by 1.86 C.
2. Boiling-Point Elevation Similarly, for every mole of solute in 1000 g of water, the boiling point increases by 0.512 C. This is NOT why some recipes call for salting water before boiling. Think of how much salt you d have to add. It s just for taste. Our last colligative property is closely related.
3. Vapor-Pressure Lowering Recall that vapor pressure is the pressure of a gas above a liquid in a container. Dissolved solutes will lower this pressure, since lots of solutes will cause solvent molecules to bind to them, making it harder for evaporation to occur. Don t worry about numbers here.
So In order to calculate these changes, there are two more formulas we should know how to do. The first of those is molality. Molarity is mol/l solution Molality is mol/kg solvent These are equal when water is the solvent, but different when it s not. Molality is particularly useful for calculations of colligative properties, especially for non-h 2 O stuff. Molality, oddly, uses the symbol m (like meters) or b.
Colligative Property Equation ΔT = K * i * m ΔT = change in temperature K = constant K b for boiling-point elevation K f for freezing-point depression i = number of dissociated ions m = molality K is different based on the solvent and is usually in units of C/m. For water: K b = 0.512 C/m K f = 1.86 C/m Solvent K b ( C/m) Water 0.512 Ethanol 1.19 Benzene 2.53 Cyclohexane 2.79 Acetic Acid 3.07 Phenol 3.56 Nitrobenzene 5.24 Camphor 5.95 Solvent K f ( C/m) Water 1.86 Acetic Acid 3.90 Benzene 5.12 Nitrobenzene 7.00 Phenol 7.40 Cyclohexane 20.2 Camphor 37.7
Example What is the freezing-point depression of a glucose solution containing 20.0 g of C 6 H 12 O 6 in 50.0 g H 2 O? (K f = 1.86 C/m) 20 g C 6 H 12 O 6 = 0.111 mol C 6 H 12 O 6 0.111 mol C 6 H 12 O 6 / 0.05 kg H 2 O = 2.22 m ΔT f = K f * i * m ΔT f = 1.86 * 1 * 2.22 (since it s molecular) ΔT f = 4.13 C (realistically it s -4.13 C)
Example Deux 500 kg of CaCl 2 are spread across Rt. 611 in winter. If we assume that it comes in contact with 1000 kg of H 2 O, what will the freezing point of that water become? We have 500 kg of CaCl 2 (solute). Let s make it moles. 500 kg CaCl 2 = 500,000 g CaCl 2 = 4505.32 mol CaCl 2 4505.32 mol in 1000 kg H 2 O is 4.505 m. ΔT = K f * i * m ΔT = 1.86 C/m * 3 * 4.505 m [because CaCl 2 splits into 3] ΔT = 25.14 C So the new freezing point is -25.14 C.
Example Three One last freezing-point depression question: Suppose you re the person in charge of Doylestown s road department, and therefore are in charge of the de-icing process that happens usually every winter. You have two salt choices to spread on the roads: NaCl or CaCl 2. Assuming they re equal in cost, which do you choose? CaCl 2 every one mole of calcium chloride dissociates into three moles of ions, whereas only two are formed from sodium chloride. In reality, either one is used.
Practice Colligative Properties Worksheet (except #3) Note: Some of these are challenging. Skip #3 for now. About #3: 25.0 mass percent solution means that 25% of the total solution mass is, in this case, ethylene glycol. You ll have to make up a kg value for water, but be careful with it.
Closure TEST QUESTION Determine the molar mass of a molecular compound that will raise acetic acid s boiling point to 121 C when 25.8 grams of solute are added to 135.0 g acetic acid. K b = 2.9 C/m, acetic acid s boiling point is 118.0 C. ΔT b = K b * i * m ΔT b = 3 C 3 C = 2.9 C/m * 1 * m m = 1.03
Closure TEST QUESTION Okay, so the molality of the compound is 1.03. We know that there were 135.0 g acetic acid. m = mol / kg 1.03 = mol / 0.135 kg mol = 0.13905 g / molar mass = mol. We know 25.8 g = 0.13905 mol. 0.13905 = 25.8 g / molar mass Molar mass = 185.54 g/mol
Another way to think of this Colligative property problems almost always come down to the two main equations: T K b i m m mol kg solute solution From only reading the problem, I can fill in this much: 121 118 2.9 C/ m 1 m m Start with the left equation and you re off! mol solute 0.135 kg
Closure One last little thing. Once you have that solution, now freezing at a lower temperature than normal, you have yourself a eutectic mixture. You don t need that word for a test, but it s fun nonetheless.
Closure Deux An enormously wide-ranging review video in under five minutes: TED: Cynthia Chubbuck The Chemistry of Cars