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Solutions II

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Solubility curves Note that a typical solubility curve shows the total mass (in g) of a solute that will completely dissolve in 100 g of water at a given temperature. Note that a typical solubility curve shows the total mass (in g) of a solute that will completely dissolve in 100 g of water at a given temperature. The solid line shows a saturated solution. If the amount of solute dissolved is above that line, the solution is supersaturated. If the amount of solute dissolved is below that line the solution is unsaturated. Look at the graph in your notes.

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Analogy Solving for solubility problems is like changing the serving size of a recipe for kool-aid. The amount of water is 100mL (same as 100g) Use your solubility graph Use your solubility graph The amount of kool-aid powder is a value on the curve. However, depending on the temperature of your water, the amount of powder can change. Not in your notes!

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A. Using solubility curves Whenever you obtain a value from a solubility curve (the actual curve), it is always for a saturated solution and it describes: Whenever you obtain a value from a solubility curve (the actual curve), it is always for a saturated solution and it describes: X g solute 100 g H 2 O If your problem does not have 100 g of water, you will need to use a proportion. If your problem does not have 100 g of water, you will need to use a proportion. If your problem involves two different temperatures, you will need to subtract. If your problem involves two different temperatures, you will need to subtract. Remember, 1 g H 2 O = 1 mL H 2 O Remember, 1 g H 2 O = 1 mL H 2 O

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Example 1: I have a solution at 60 º C with 80 g of sodium nitrate dissolved in 100 g of water. Is it saturated, supersaturated, or unsaturated? Hint: Is the intersecting point above, on, or below the line? BelowTherefore... Unsaturated

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Example 2: How much more sodium nitrate could I add to the previous solution (80 g NaNO 3 in 100g H 2 O at 60 º C) to form a saturated solution? Hint: How many more grams of solute are needed from the intersecting point to the line of saturation? 124g – 80g = 44 grams

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= ? g KCl 300mL H 2 O 300mL H 2 O Ex: How many grams of potassium chloride will dissolve in 300 mL of water at 50.0ºC? x 40g KCl 40g KCl 100g H 2 O 120 g KCl per 300mL water 120 g KCl

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Ex: I have a saturated solution with 9.2 g of solute dissolved in 10 g of water at 25 o C. What is the solute? Hint: The solubility curves only tell us the solubility of solute in 100g of water. = ? g 100 g H 2 O 100 g H 2 O 9.2 g 9.2 g 10 g H 2 O The solute is NaNO 3 92g What is the solute? 25 ° C

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Practice 1. I have a solution with 120 g of sodium nitrate dissolved in 100 mL of water at RT. Is it saturated, unsaturated, or supersaturated? 2. a) How much potassium nitrate is dissolved in 100 g of water at 50 C to make a saturated solution? b) If I have 65 g of potassium nitrate dissolved in 100 g of water at 50 C, how much more could be added and have a saturated solution? 3. What volume of water do I need to make a saturated solution containing 1000. g of potassium chloride at 50 C? Supersaturated 80g 80g – 65g = 15g X = 2500 g water 2.50 x 10 3 g H 2 O 1000 g KCl = X g H 2 O 40 g KCl 100 g H 2 O

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Solutions II In this lesson we will learn how to make and calculate % and molar solutions. Chemists must know how much stuff is dissolved in a solvent so there are simple ratios to express the concentration that everyone has agreed to use. So these are just more definitions but they’re written as equations. Not in your notes!

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Solutions II If a chemist encounters a solution with an unknown concentration of “X,” methods have been devised to measure how much “X” is present. The concentration of “X” is then reported the way you will learn in this lesson. Not in your notes!

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Don’t lose sight of this: Most of the calculations we will do are just ratios (dividing one number by another) turned into percentages. IT’S SIMPLE. DON’T FREAK OUT! IT’S SIMPLE. DON’T FREAK OUT! Not in your notes!

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A. Vocabulary What is concentration? 1. The word concentration is used to describe how much solute is dissolved in a solvent to make a solution. solvent solute into

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A. Vocabulary What is concentration? 2. Qualitatively, we use the word “concentrated” to imply a lot of dissolved solute. 3. A solution is “dilute” if it contains a relatively small amount of dissolved solute. 4. If a solution is diluted, it means the original concentration value has been lowered by the addition of solvent to a portion of the original solution. 5. A concentrated solution that is used to make more diluted solutions is often called a stock solution.

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Pictorial Representation

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B. Formulas for calculating and reporting Concentration as a Percentage There are several ways to express concentration. They are RATIOS of either: mass to mass, mass to volume, or volume to volume. *KEY POINT: Recognize the denominators. The represent the WHOLE SOLUTION, NOT just the solvent (this is a common student error!). Percent ratios % m/m = Mass (in grams) of SOLUTE Mass (in grams) of SOLUTION x 100 % m/v = Mass (in grams) of SOLUTE Volume (in milliliters) of SOLUTION x 100 % v/v = Volume (in milliliters) of SOLUTE Volume (in milliliters) of SOLUTION x 100 The denominator is always SOLUTION. Always specify which ratio you are calculating Volumetric flask or graduated cylinder is used for m/v or M solutions. Graduated cylinder is used for v/v solutions Fix this in your notes!!

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So, when we talk about the mass or volume of the SOLUTION, it’s either: 1. the sum of the mass values for the solute and solvent (% m/m) 2. the sum of the volume values for the liquid solute and the solvent (% v/v) 3. the TOTAL volume that both the solute and solvent together occupy (% m/v)

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Example 1: A solution contains 7 g of NaCl in 165 g total. X m/m = 7 g LiCl 165 g solution x 100 = 4.24% NaCl by mass (m/m) % m/m = Mass (in grams) of SOLUTE Mass (in grams) of SOLUTION x 100

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Example 2: A saline solution contains 3.5 g of NaCl in 62.5 mL of solution. X m/v = 3.5 g NaCl 62.5 mL solution x 100 = 5.60% NaCl by mass (m/v) % m/v = Mass (in grams) of SOLUTE Volume (in milliliters) of SOLUTION x 100

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Example 3: How many milliliters of a 5.00% m/v solution will Dr. Poche make using 23.4 g of LiCl? 5.00% m/v = 23.4 g LiCl X mL solution x 100 X mL = 23.4 g LiCl 5.00 = 468 mL solution % m/v = Mass (in grams) of SOLUTE Volume (in milliliters) of SOLUTION x 100 ( X mL solution) 5.00% m/v = x 100

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Example 4: What volume of a 2.8% (m/v) glucose solution would you need to deliver to a patient who needs 750 mg of glucose? 2.8% m/v = 750 mg glucose X mL solution x 100 X mL = 750 mg 2.8 = 26786 2.68 x 10 4 mL solution % m/v = Mass (in grams) of SOLUTE Volume (in milliliters) of SOLUTION x 100 ( X mL solution) 2.8% m/v = x 100

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Example 4: 20 mL of alcohol is diluted with water to a total volume of 65 mL. What is the percent alcohol by volume? X v/v = 20 mL 65 mL solution x 100 = 30.8% alcohol by volume (v/v) % v/v = Volume (in milliliters) of SOLUTE Volume (in milliliters) of SOLUTION x 100

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Example 5: What volume of ethanol is needed to produce 120 mL of a 22.3% (v/v) ethanol solution? 22.3% v/v = X mL 120 mL solution x 100 = 26.8 mL ethanol by volume (v/v) % v/v = Volume (in milliliters) of SOLUTE Volume (in milliliters) of SOLUTION x 100 X mL = 22.3% v/v x 120 mL 100

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Practice 1. If 10 mL of pure acetone is diluted with water to a total solution volume of 200 mL, what is the percent by volume of acetone in the solution? 10mL x 100 = 200mL 2. A bottle of hydrogen peroxide is labeled 3.0% (v/v). How many mL of H 2 O 2 are in a 400.0 mL bottle of this solution? 3% v/v = X x 100 400mL 3. Calculate the grams of solute required to make 250 g of 0.10% MgSO 4 (m/m). 0.10% m/m = X x 100 250g 4.A solution contains 2.7 g CuSO 4 in 75 mL of solution. What is the percent (m/v) of the solution? 2.7 g x 100 = 75 mL 5.00% acetone (v/v) X = 12.0mL H 2 O 2 X = 0.250g MgSO 4 3.60% (m/v)

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Practice 5.What volume of a 2.8% (m/v) glucose solution would you need to deliver to a patient who needs 750 mg of glucose? 2.8% (m/v) = 750 mg glucose x 100 X = 2.68 x 10 4 mL X mL glucose solution

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III. MOLARITY (RATIO OF MOLES OF SOLUTE TO LITERS OF SOLUTION) *If you are given grams, convert it to moles using the MOLAR MASS of the substance. Molarity- unit abbreviation is M Molarity = Moles of SOLUTE Liters of SOLUTION

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What is the molarity of a solution containing 12.3g of Na 2 CO 3 in 675 mL of total volume? 12.3 g Na 2 CO 3 11 mol Na 2 CO 3 = 0.172 M Na 2 CO 3.675 L 105.988 g Na 2 CO 3 1 Example 1:.675 L Molarity- unit abbreviation is M Molarity = Moles of SOLUTE Liters of SOLUTION Molar Mass

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A saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? 0.90 g NaCl11 mol NaCl = 0.154 M NaCl.100 L58.443 g NaCl1 Example 2:.100 L Molarity- unit abbreviation is M Molarity = Moles of SOLUTE Liters of SOLUTION Molar Mass

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How many grams of solute are present in 562 mL of 0.24M Na 2 SO 4 ? 0.135 mol Na 2 SO 4 1 mol Na 2 SO 4 = 19.2 g Na 2 SO 4 142.042 g Na 2 SO 4 1 Example 3:.562 L Molarity- unit abbreviation is M Molarity = Moles of SOLUTE Liters of SOLUTION Molar Mass.24 M = Moles.562 L Moles = 0.135 Na 2 SO 4

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Practice 1. A solution has a volume of 2.0 L and contains 36.0 g of glucose. If the molar mass of glucose is 180.156 g/mol, what is the molarity of the solution? 36.0 g glucose|1 mole glucose| 1 = 1 180.156 g/mol 2.0L 2. How many moles of ammonium nitrate are in 335 mL of 0.425 M NH 4 NO 3 ?.425M NH 4 NO 3 = X moles.335 L 3. How many grams of solute are in 250 mL of 2.0 M CaCl 2 solution? 2.0 M CaCl 2 = X moles X = 0.5 moles CaCl 2.250 L.5 moles CaCl 2 | 110.986 g CaCl 2 | = 1 mole CaCl 2 4. Describe how you would prepare 250 mL of a 0.2 M NaOH solution. Need 2.00g NaOH in 250mL of solution.2 M NaOH = X g NaOH.250 L 0.100M glucose X = 0.142 moles NH 4 NO 3 55.5 g CaCl 2 X = 0.0500g NaOH in 250mL of solution

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IV. DILUTIONS FROM A STOCK SOLUTION USING MOLARITY *If you are given grams, convert it to moles using the MOLAR MASS of the substance. Dilutions (molarity) (M 1 > M 2 and V 1 < V 2 ) M1V1M1V1 Moles of Stock used Moles of diluted solution M2V2M2V2 = =

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Example 1: What volume of 2.5 M stock solution is needed to make 75.0 mL of 0.0234 M NaCl? M1V1M1V1 M2V2M2V2 = V 1 = 0.0234M x 75.0mL V 1 = 0.0234M x 75.0mL V 1 = 0.0234M x 75.0mL 2.5 M 2.5 M V 1 = 0.702 mL 2.5 M 2.5 M x

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Practice 1.How many mL of a stock solution of 4.00 M KI would you need to prepare 250.0 mL of 0.760 M KI? 4.00M * V 1 = 0.760 M * 250.0 mL V 1 = 0.760 M * 250.0 mL 4.00M 2.What volume must you dilute to make 50.0 mL of 0.20 M KNO 3 from 4.0 M KNO 3 ? 0.20M * 50.0 mL = 4.0 M * V 2 V 2 = 0.20M * 50.0 mL 4.0M 3.What is the molarity of a solution formed when you add 200 mL of water to 50 mL of 5.0 M HCl? M 1 * 200mL = 5.0M * 50mL M 1 = 5.0M * 50mL 200mL V 1 = 47.5 mL of 4.00M KI V 2 = 2.5 mL of 0.20 M KNO 3 M 2 = 1.25M of HCl

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A molar solution has a certain number of moles of solute per one liter of solution. 1. These solutions, where the VOLUME of the whole solution is required, are best done in a piece of glassware called a 1 L volumetric flask. 2. It only has one volume mark but that’s all you need. It is marked on the neck at the volume the flask is calibrated to hold. By the way, that volume mark is only good for a given temperature, 20 o C. If you use the flask at any other temperature, your solution concentration will be off a little.By the way, that volume mark is only good for a given temperature, 20 o C. If you use the flask at any other temperature, your solution concentration will be off a little. V. Making a Solution

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3. If given moles, convert to grams. 4. Measure the appropriate mass and add it to the volumetric flask. 5. Next, add enough solvent to reach the mark (meniscus!) 6. You now have a TOTAL volume of a 1 L for the solution. V. Making a Solution

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A. % m/m Though you probably would not want to do this, making a %m/m solution could be done in an empty peanut butter jar! The point is that an expensive piece of glassware is not required although an expensive lab balance might be. V. Making a Solution

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A. % m/m i. The key is to weigh the solute in a container and record the mass. ii. Then, weigh the solvent in the same container (if possible) and record the mass. The sum of the two masses is the mass of the solution. iii. If you were careful, you also recorded the mass of the EMPTY container and lid, just in case you need these numbers later to correct a mistake. V. Making a Solution

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B. % m/v or Molar or Dilutions i. These solutions, where the VOLUME of the whole solution is required, are best done in a piece of glassware made exactly for this job. ii. The glassware is called the volumetric flask. iii. It only has one volume mark but that’s all you need. It is marked on the neck at the volume the flask is calibrated to hold. By the way, that volume mark is only good for a given temperature, 20 o C. If you use the flask at any other temperature, your solution concentration will be off a little.By the way, that volume mark is only good for a given temperature, 20 o C. If you use the flask at any other temperature, your solution concentration will be off a little. V. Making a Solution

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Image for NOTES

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B. % m/v or Molar or Dilutions iv. The solute is weighed into the volumetric flask. This accounts for the volume it occupies in the final solution. v. Next, enough solvent is added to dissolve the solute and the volume is brought up to the volume mark (meniscus!) with additional solvent. vi. You now have a TOTAL volume for the solution. vii. The volume of everything is accounted for in the volumetric flask. V. Making a Solution

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B. % m/v or Molar or Dilutions For DILUTIONS: i. A pipette or graduated cylinder is used to accurately remove a given volume of stock solution. ii. This solution is placed into a volumetric flask and solvent is added to the volume mark. iii. The dilution reduces the grams of solute per unit volume, but the total amount of solute in the solution does not change. V. Making a Solution Before dilution After dilution 1 2 3

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C. % v/v 1. These solutions should be made using a graduated cylinder. This glassware is marked off with accurate volume increments. 2. Since you need to measure the volume of the solute and the final solution, this can be done by adding the liquid solute to a graduated cylinder to the desired volume and then adding the solvent until the final volume mark is reached. V. Making a Solution

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C. % v/v 3. You might also consider adding the solute into a volumetric flask via a buret. Then the solvent could be added to the volumetric flask up to the volume mark. V. Making a Solution

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SOLUTIONS II STUDY GUIDE Percent ratios % m/m = Mass (in grams) of SOLUTE Mass (in grams) of SOLUTION x 100 % m/v = Mass (in grams) of SOLUTE Volume (in milliliters) of SOLUTION x 100 % v/v = Volume (in milliliters) of SOLUTE Volume (in milliliters) of SOLUTION x 100 Molarity- unit abbreviation is M Molarity = Moles of SOLUTE Liters of SOLUTION Dilutions (molarity) M 1 V 1 = M 2 V 2 Moles of Stock used = Moles of diluted solution (M 1 > M 2 and V 1 < V 2 ) The denominator is always SOLUTION. Always specify which ratio you are calculating Volumetric flask or graduated cylinder is used for m/v or M solutions. Graduated cylinder is used for v/v solutions In dilutions, moles from stock = mole in diluted solution

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Solutions Study Guide con’t PROBLEM SOLVING: ALL UNITS MUST BE SHOWN Examples: How many milliliters of a 5.00 % m/v solution will I make using 23.4 g of LiCl ? 5.00 % m/v = 23.4 g LiCl x mL solution x 100 x mL = 23.4 g LiCl.0500 = 468 mL solution What is the molarity of a solution containing 12.3 g of Na 2 CO 3 in 675 mL of total volume? 12.3 g Na 2 CO 3 1 1 mole Na 2 CO 3 106 g Na 2 CO 3 1 0.675 L = 0.172 M Na 2 CO 3 Molar mass liters What volume of 2.5 M stock solution is needed to make 75.0 mL of 0.0234 M NaCl ? M 1 V 1 = M 2 V 2 2.50M x V 1 = 0.0234M x 75.0 mL V 1 = 0.702 mL stock solution

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SOLUTIONS II STUDY GUIDE Percent ratios % m/m = Mass (in grams) of SOLUTE Mass (in grams) of SOLUTION x 100 % m/v = Mass (in grams) of SOLUTE Volume (in milliliters) of SOLUTION x 100 % v/v = Volume (in milliliters) of SOLUTE Volume (in milliliters) of SOLUTION x 100 Molarity- unit abbreviation is M Molarity = Moles of SOLUTE Liters of SOLUTION Dilutions (molarity) M 1 V 1 = M 2 V 2 Moles of Stock used = Moles of diluted solution (M 1 > M 2 and V 1 < V 2 ) The denominator is always SOLUTION. Always specify which ratio you are calculating Volumetric flask or graduated cylinder is used for m/v or M solutions. Graduated cylinder is used for v/v solutions In dilutions, moles from stock = mole in diluted solution

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Solution Concentration

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What is different between the glasses of Kool-aid?

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Solution concentration can be described generally Dilute - reduced in strength, weak, watered down. Concentrated – stronger, pure. Has less water.

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What’s the problem with just using dilute and concentrated as descriptions of the solution concentration?

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Is solution B dilute or concentrated? The terms dilute and concentrated are relative. Scientists need a more precise way of referring to the concentration of a solution. ConcentratedDilute Solution A Solution B Solution C

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Solution concentration can be described specifically Do you remember the “mole” from Stoichiometry? What is a mole? How might you use it to describe the concentration of a solution?

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Concentration can be represented by… Molarity Molality Percent by Mass

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Molarity The ratio of the moles of solute to the volume of solvent in liters. Symbol: M You must be careful to label the molarity with a capital M Molarity (M) = Moles of solute Volume in Liters of Solvent

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How to read: 6M NaCl Read: “6 molar solution of NaCl” Can be abbreviated 6M solution Be careful to label the molarity with a capital M

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How to make a 6M NaCl solution (a) Add 6 moles NaCl to the volumetric flask. How would you measure that? 6 moles NaCl 58.443 g NaCl 1 mole NaCl = 351 g NaCl

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How to make a solution using molarity (6M NaCl) (b) Add dH 2 O to dissolve and mix the NaCl (c) Fill the flask with dH 2 O until you reach the 1000mL line.

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