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Solutions Part II DHS Chemistry Chapter 15 Slide 2 I. Concentrations of Solutions The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. Amount of solute vs. amount of water Slide 3 Slide 4 Dilute vs Concentrated Little solute a lot of solute Slide 5 Very Concentrated Slide 6 Less Concentrated Slide 7 Concentrated solutions A concentrated solution is one that contains a high concentration of solute. Slide 8 Dilute solutions A dilute solution contains a small concentration of solute. Slide 9 Pictorial Representation Slide 10 Slide 11 There are several ways to express concentration. These include: percent solutions (by volume or mass), molarity, or molality. Slide 12 A. Percent Solutions Slide 13 Percent Solutions % solute = amount of solute _ 100 TOTAL amount of solvent solution 3 types: (%m/m) same units (%v/v) same units (%m/v) needs to be g/mL Slide 14 How much vinegar is just acetic acid? 5% of vinegar is acetic acid Slide 15 Percent by Volume % by volume (% (v/v)) = volume of solute 100 Volume of solution units must be the same Tip: watch out for the wording. You may need to add the volume of the solute and solvent to get the volume of the total solution (solute + solvent) Slide 16 Percent by Mass % by mass (% (m/m)) = mass of solute 100 mass of solution *units must be the same Tip: watch out for the wording. You may need to add the mass of the solute and solvent to get the volume of the total solution (solute + solvent) Slide 17 Percent Mass by Volume % mass by volume (% (m/v) = mass of solute (g) 100 volume of solution (mL) *units must g/mL Slide 18 Ex 1: 20 mL of alcohol is diluted with water to a total volume of 65 mL. What is the percentage of alcohol, by volume? %(v/v) = Volume of solute 100 Volume of solution %(v/v)= 20 mL alcohol 10065 mL H 2 O ? % = 30.8% Alcohol by volume Slide 19 65 mL alcohol + water 30.8% of this solution is alcohol. The rest is water. 20 mL alcohol Slide 20 Is white gold really gold? White Gold, 18 Karat Au 75% - Pt or Pd 25% Solute = platinum or palladium Solvent = gold Slide 21 If I had an 18k white gold ring that weighed 20g, how much of the ring isnt pure gold? White Gold, 18 Karat Au 75% - Pt or Pd 25% 25% = ?g 100 20g 5 grams of my ring isnt pure gold. Slide 22 Ex 2 : A solution containing 7 g of NaCl in 165 g of solution. What is the percent of NaCl by mass? 7 g NaCl 100 165 g solution ? % NaCl (m/m) = 4.24% Solution = Solute + solvent Solution = 7 g + 158 g 158 g of solvent (water) Slide 23 Childrens Dose vs Adult Dose Diphenhydramine hydrochloride (active ingredient in allergy medicine like Benadryl) Slide 24 How do you feed a child medicine when one tablet is too strong? Slide 25 Liquid dose for children has been diluted to 12.5 mg for every 5 mL of medicine Slide 26 What percent by mass of diphenhydramine hydrochloride is in the solution? Liquid dose for children has been diluted to 12.5mg for every 5mL of medicine.0125 g =.250% (m/v) 5 mL 100 Slide 27 EX 3: A saline solution containing 3.5 g of NaCl in 62.5 mL of solution. What is the percent of NaCl, by mass. 3.5 g NaCl 10062.5 mL solution ? % = 5.60% Slide 28 EX 3: A saline solution containing 3.5 g of NaCl in 62.5 mL of solution. What is the percent of NaCl, by mass. 3.5 g NaCl 10062.5 mL solution ? % = 5.60% Slide 29 Ex 4: What volume of ethanol is needed to produce 120 mL of a 22.3% (v/v) ethanol solution? %(v/v)= ethanol by volume (v/v) ? mL ethanol 100120 mL solution 22.3 % = 26.8 mL Slide 30 Ex 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? glucose by volume (m/v) 0.750 g glucose 100 ? mL glucose solution 2.8 % = 26.8 mL Slide 31 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? 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. Calculate the grams of solute required to make 250 g of 0.10% MgSO 4 (m/m). 4. A solution contains 2.7 g CuSO 4 in 75 mL of solution. What is the percent (m/v) of the solution? Slide 32 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? %(v/v)= acetone by volume (v/v) 10 mL acetone 100 200 mL solution ? % = 5.00 % Slide 33 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? %(v/v)= H 2 O 2 by volume (v/v) ? mL H 2 O 2 100 400. mL solution 3.00 % = 12.0 mL Slide 34 3. Calculate the grams of solute required to make 250 g of 0.10% MgSO 4 (m/m). %(m/m)= MgSO 4 by volume (v/v) ? g MgSO 4 100 250 g solution 0.10 % = 0.250 g Slide 35 4. A solution contains 2.7 g CuSO 4 in 75 mL of solution. What is the percent (m/v) of the solution? %(m/v)= CuSO 4 by volume (m/v) 2.7 g CuSO 4 100 75 mL solution ? % = 3.60 % Slide 36 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? 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. Calculate the grams of solute required to make 250 g of 0.10% MgSO 4 (m/m). 4. A solution contains 2.7 g CuSO 4 in 75 mL of solution. What is the percent (m/v) of the solution? 5.00% acetone (v/v) 12.0mL H 2 O 2 0.250g MgSO 4 3.60% (m/v) Slide 37 B. Molarity Slide 38 Molarity Molarity (M) is the number of moles of a solute dissolved per liter of solution. Slide 39 Molarity Molarity is also known as molar concentration and is read as __#__ molar (Ex. a 2M HCl solution is read as two molar HCl Note that the volume involved is the total volume of solution, not just the solvent. Slide 40 Molarity Molarity (M) = moles of solute M =mol Liters of solution 1 L *if given grams, convert if to moles using the molar mass of the substance Slide 41 Why are grams important? You can not directly measure moles, you must calculate the mass in grams first Grams Moles Molar mass ___g = 1 mole Slide 42 How to Prepare a Solution To make 1.00 liter of a 1.00 molar (1.0 M) solution: 1) add 1.0 mol of solute to a volumetric flask 2) add about flask of distilled water. Swirl the flask till the solute is dissolved. 3) slowly add water until the final volume reads 1.00 L Slide 43 Slide 44 Molarity EX 1. What is the molarity of a solution that contains 8 moles of CaCl 2 in 50 mL of solution? M = 1 mol L 8 mol 0.05 L 160M CaCl 2 Slide 45 Molarity EX 2. How many grams of NaCl are needed to make 500mL of a 0.2 M solution? M = 1 ? mol L.2 M 0.5 L.1 mol 0.1 mol NaCl 1 mol NaCl 58.443 g NaCl=5.84 mol NaCl Slide 46 Using Molarity Ex 3: A saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? 0.90g NaCl x 1 mol NaCl 58.443 g NaCl =0.0154 mol NaCl 100 mL x 1 L 1000 mL = 0.100 L NaCl Step 1: Calculate # moles Step 2: mL L Slide 47 Ex 3 continued Step 3: Calculate Molarity M = 1 0.0154 mol L ? M 0.1 L 0.154 M Slide 48 Ex 2: How many grams of solute are present in 562 mL of 0.24 M Na 2 SO 4 ? M = mol mol = M L L mol = 0.24M Na 2 SO 4 x.562L = 0.135mol Slide 49 Convert from Moles to Grams 0.135mol Na2SO4 | 142g Na 2 SO 4 = | 1 mol Na 2 SO 4 = 19.2g Na 2 SO 4 Slide 50 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 g/mol, what is the molarity of the solution? 0.100M glucose 2. How many moles of ammonium nitrate are in 335 mL of 0.425 M NH4NO3? 0.142mol NH 4 NO 3 3. How many grams of solute are in 250 mL of 2.0 M CaCl 2 solution? 55.5gCaCl 2 4. Describe how you would prepare 250 mL of a 0.2 M NaOH solution. Need 2.00g NaOH in 250mL of solution Slide 51 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 g/mol, what is the molarity of the solution? Molarity = mol L Glucose = C 6 H 12 O 6 Molar mass = 6(12.01g) + 12(1.008g) + 6(15.999g) = Calculate moles: 36.0g C 6 H 12 O 6 | 1 mol C 6 H 12 O 6 = mol C 6 H 12 O 6 | XXX g C 6 H 12 O 6 Calculate Molarity:XXXmol C 6 H 12 O 6 = 0.100M glucose 2.0 L Slide 52 C. Dilutions Slide 53 You can make a less concentrated solution by diluting it with solvent. The dilution reduces the grams of solute per unit volume, but the total amount of solute in solution does not change. Slide 54 Diluted Solutions Before After Dilutions Slide 55 Moles of solute before dilution = Moles of solute after dilution Moles of solute = Molarity x volume Dilutions: M 1 V 1 = M 2 V 2 Slide 56 Ex: How many mL of a stock solution of 2.00 M MgSO 4 would you need to prepare 100.0 mL of 0.400 M MgSO 4 ? (2M)(V 1 ) = ( 0.400M)(100mL) m 1 v 1 m 2 v 2 V 1 = 20mL of stock solution Stock soln Slide 57 Ex. 2: Describe how to prepare 100 mL of 0.400M MgSO 4 from 2M MgSO 4. (see previous example) Add 20mL of 2M stock solution in a container and add solvent up to the 100mL mark Slide 58 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? 47.5mL of 4.00MKI 2. What volume must you dilute to make 50.0 mL of 0.20 M KNO 3 from 4.0 M KNO 3 ? 2.5mL of 4M KNO 3 3. What is the molarity of a solution formed when you add 200 mL of water to 50 mL of 5.0 M HCl? 1.00M Slide 59 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? (4M) (V1) = (0.760M) (250mL) m1 v1 m2 v2 V1 = 47.5mL of 4M stock solution Slide 60 2. What volume must you dilute to make 50.0 mL of 0.20 M KNO 3 from 4.0 M KNO 3 ? (4M) (V1) = (0.20M) (50mL) m1 v1 m2 v2 V1 = 2.5mL of 4M solution Slide 61 3. What is the molarity of a solution formed when you add 200 mL of water to 50 mL of 5.0 M HCl? (5.0M) (50mL) = (M2) (250mL) m1 v1 m2 v2 M2 = 1M of solution Slide 62 EX. A chemist starts with 50 mL of a 0.40M NaCl solution and dilutes it to 1000 mL. What is the concentration of the dilute solution? (0.4 M)(50 mL) = ( ?M)(1000 mL) M 1 V 1 M 2 V 2 M 2 = 0.0200 M is the concentration of the diluted solution Stock soln Slide 63 Practice 1)What volume of a 3.00M KI stock solution would you use to make 0.300 L of a 1.25 M KI solution?0.125 L 2)How many milliliters of a 5.0M H 2 SO 4 stock solution would you need to prepare 100.0 mL of a 0.25M H 2 SO 4 ? 5.00 mL 3)If you dilute 20.0 mL of a 3.0M solution to make 100.0 mL of solution, what is the molarity of the dilute solution? 0.600M Slide 64 Practice 1 (3.00M) (V 1 ) = (1.25M) (.300) m1 v1 m2 v2 V 1 =.125 L of 3M stock solution Practice 2 (5M) (V 1 mL) = (0.25M) (100 mL) m1 v1 m2 v2 V 1 = 5.00mL of 5M solution Slide 65 Practice 3 (3.0M) (20mL) = (M 2 ) (100mL) m 1 v 1 m 2 v 2 M 1 = 0.600M is the new concentration