LESSON ASSIGNMENT. After completing this lesson, you should be able to: 8-1. Prepare and calculate acid and base solutions.

Transcription

1 LESSON ASSIGNMENT LESSON 8 Concentrated Acids and Bases. TEXT ASSIGNMENT Paragraphs 8-1 through 8-6. LESSON OBJECTIVE After completing this lesson, you should be able to: 8-1. Prepare and calculate acid and base solutions. SUGGESTION After reading the assignment, complete the exercises at the end of this lesson. These exercises will help you to achieve the lesson objective. MD

2 LESSON 8 CONCENTRATED ACIDS AND BASES 8-1. DISCUSSION As discussed in paragraph 2-31, concentrated liquids are manufactured as weight-per-weight solutions. An example of this type of solution is concentrated hydrochloric acid that contains 37 grams of HCl per 100 grams of total solution (w/w). These figures indicate that concentrated HCl has a percent purity of 37 percent or that 37 percent of the weight of 100 grams of solution is HCl. Since it is impractical to weigh concentrated acids and/or bases, one needs another method to determine how much concentrated liquid to use. The method used includes use of the specific gravity of the solution, which is an expression of density (g/ml), and the percent purity or assay value (%A) that indicates the proportion of the solution by weight that is the desired substance. The specific gravity and percent purity values are found on the label of most reagent-grade acids and bases. Using these values, one can determine the actual amount of the substance in a given volume of the concentrated solution EXAMPLE The values listed on the label of a bottle of nitric acid are specific gravity (S.G.) 1.42 and assay 70 percent. What so these values mean? a. These values mean that 1 milliliter of the solution has a mass of 1.42 grams and that 70 percent of this mass is nitric acid (HNO 3 ). b. When preparing a solution using this concentrated liquid, we are concerned only with how much HNO 3 is present. To determine this, multiply the density times the percent assay. The answer is the number of grams of HNO 3 per milliliter of solution g/ml X 0.70 percent HNO grams of HNO 3 per ml of solution c. Therefore, density times %A is an expression that yields the number of grams of desired substance per milliliter of concentrated substance. NOTE: The %A value is equal to the assay value divided by 100. If the assay value is 70 percent then the %A is 0.70 (70/100). The latter value or its equivalent is to be used in all calculations. (g/ml)(%a) = grams of desired substance per unit volume MD

3 8-3. SPECIFIC GRAVITY AND DENSITY a. Specific gravity is a method of measuring density. Density is the amount of matter in a given volume, that is, D = g/ml. Specific gravity is a ratio between the mass of a given volume of any substance and the mass of an equal volume of pure water at 4º C. density of solid or liquid S.G. = density of water at 4º C NOTE: When determining specific gravity, the units divide out. Hence, specific gravity has no unit of report. b. Since one milliliter of pure water at 4º C has a mass of one gram, specific gravity is equal to the numerical value of the mass in grams of one milliliter of any substance. Materials less dense than water have a specific gravity of less than one, whereas materials more dense than water have a specific gravity greater than one. c. Example. The density of 200 ml of HNO 3 solution is 1.42 g/ml. What is the weight of that solution? Solution. Restating the problem, one milliliter of this solution weighs 1.42 grams; how much would 200 ml of the same solution weigh? This problem can be solved by ratio and proportion g 1 ml = x g 200 ml (x g)(1 ml) = (1.42 g)(200 ml) (1.42 g)(200 ml) x g = 1 ml x g = 284 g NOTE: The 284 grams represents the total weight of the solution and not just the weight of the HNO 3 present in the solution. MD

4 8-4. PREPARING SOLUTIONS FROM CONCENTRATED LIQUIDS When preparing a solution in the laboratory, one would normally think of taking a chemical in powder or crystal form (such as NaCl) and calculate the amount to be weighed. When preparing such a solution, one would ask, "How many grams of this substance do I need to make the desired concentration?" Using appropriate factors, you calculate the desired weight required. With concentrated liquids, this cannot be done since it is impractical and too dangerous to try to weigh these liquids. We do know, however, that each concentrated liquid contains a certain mass of solute per milliliter of solution and that this can be determined using the information found on the label of the bottle. Example. A bottle of nitric acid (HNO 3 ) has a S.G. of 1.42 and an assay of 70.0 percent. How many grams of HNO 3 are contained within 1.0 ml of HNO 3? Solution. The given volume times the density of the concentrated liquid times the %A yields the desired results g ml X X = g ml PROBLEM SOLVING You have seen in earlier lessons that the mass of a solute needed to prepare a solution of a certain concentration may be determined by the use of various expressions, dependent upon the desired concentration unit. These various problemsolving expressions all yielded the mass in grams needed to prepare a solution of known concentration and volume. Similar problem solving techniques may be used to determine the amount of concentrated liquid needed to yield a desired amount of solute. The examples that follow will illustrate several methods needed to solve problems involving concentrated liquids. MD

5 a. Example 1. How much concentrated HNO 3 with a S.G. of 1.42 and an assay of 70% would you need to prepare 1.0 liter of a 1.0 mol/l HNO 3 solution? Solution. As before, with problems that involve molarity, calculating the gram molecular weight of the compound involved is a good starting point. Calculate the GMW of the compound. HNO 3 H 1.0 X 1 = 1.0 N 14.0 X 1 = 14.0 O 16.0 X 3 = g/mol Determine the amount of HNO 3 needed to prepare the solution. 1.0 mol 63.0 g 1.0 L X X = 63 g HNO 3 L mol Calculate the grams of HNO 3 per milliliter of concentrated liquid g 70 X = g HNO 3 /ml concentrated liquid ml 100 Use the grams per milliliter of solute as an appropriate factor times the number of grams of solute needed to determine the required volume of concentrated liquid. ml 63 g X = 63 ml g MD

6 b. Example 2. How much H 2 SO 4 with a S.G. of 1.84 and an assay of 95% is required to make 500 ml of a 4.00 Eq/L H 2 SO 4 solution? Solution. As before, with problems that involve normality, calculating the gram equivalent weight of the compound involved is a good starting point. Calculate the GEW of the compound. H 2 SO 4 H 1.0 X 2 = 2.0 S 32.1 X 1 = 32.1 O 16.0 X 4 = g/mol 98.1 g mol X = 49.0 g/eq mol 2 Eq Determine the amount of solute needed in the desired solution. 1 L 4.00 Eq 49.0 g 500 ml X X X = 98.0 g H 2 SO ml L Eq Calculate the grams of H 2 SO 4 per milliliter of concentrated liquid g 95.0 X = 1.75 g H 2 SO 4 /ml concentrated liquid ml 100 Use the grams of solute per milliliter of concentrated liquid as an appropriate factor times the number of grams of solute needed to determine the required volume of concentrated liquid. ml 98.1 g X = 56.1 ml 1.75 g MD

7 c. Example 3. How much concentrated HCl with an S.G. of 1.19 and an assay of 38.0% is required to make 250 ml of a 20.0 g/dl HCL solution? Solution. Read the problem carefully and determine the desired quantity. Determine the milliliters of concentrated HCl. Calculate the amount in grams of solute needed to prepare the solution. 1 dl 20.0 g 250 ml X X = 50.0 g of HCl needed 100 ml dl Determine the grams of HCl per milliliter of concentrated liquid g 38.0 X = g HCl/mL of concentrated liquid ml 100 Use the grams of solute per milliliter of concentrated liquid as an appropriate factor times the number of grams of solute needed to determine the required volume of concentrated liquid. ml 50.0 g X = 111 ml g 8-6. DETERMINING THE CONCENTRATION OF CONCENTRATED LIQUIDS At times, it may be necessary to determine the concentration of a commercial liquid such as the mol/l, Eq/L, or g/dl concentration. To do this consider the definition of each unit of concentration, then multiply the density (g/ml), given on the bottle times the percent assay (%A) times an appropriate conversion factor to yield the desired unit of concentration. Concentrated commercial liquids are based on g/ml of solution. a. Conversion from g/ml to g/dl. Percent concentration is based on grams of solute per 100 ml of solution. For example: What is the g/dl concentration of a HNO 3 solution that has a specific gravity of 1.42 and an assay of 70.0%. MD

8 Solution. Express the desired unit of concentration (g/dl) using the information given as appropriate factors. Calculate the grams of HNO 3 per milliliter of concentrated liquid g 70.0 X = g HNO 3 / ml concentrated liquid ml 100 Calculate the g/dl concentration by multiplying the grams of solute per milliliter of concentrated liquid times an appropriate unit volume conversion factor g 100 ml X = 99.4 g/dl ml 1 dl b. Conversion from g/ml to mol/l. Molarity is based on moles of solute per liter of solution. For example: What is the mol/l concentration of a HNO 3 solution with a S.G. of 1.42 and an assay of 70.0%. Solution. Express the desired unit of concentration (mol/l) using the information given as appropriate factors. Calculate the GMW of the compound. HNO 3 H 1.0 x 1 = 1.0 N 14.0 x 1 = 14.0 O 16.0 x 3 = g/mol Determine the grams per liter of HNO g ml X X = 994 g/l HNO 3 ml 100 L Multiply the gram per liter concentration of the acid times the grams per mole, using the factors appropriately. 994 g mol X = 15.8 mol/l L 63.0 g MD

9 c. Conversion From g/ml to Eq/L. Normality is based on equivalents of solute per liter of solution. For example: What is the Eq/L concentration of a H 2 SO 4 solution with an S.G. of 1.84 and an assay of 95.0%? Solution. Express the desired unit of concentration (Eq/L) using the information given as appropriate factors. Calculate the GEW of the compound. H 2 SO 4 H 1.0 X 2 = 2.0 S 32.1 X 1 = 32.1 O 16.0 X 4 = g/mol 98.1 g mol X = 49.0 g/eq mol 2 Eq Determine the grams per liter of H 2 SO g ml X X = 1748 g/l ml 100 L Multiply the gram per liter concentration of the acid times the grams per equivalent, using the factors appropriately g Eq X = 35.7 Eq/L L 49.0 g Continue with Exercises MD

10 EXERCISES, LESSON 8 INSTRUCTIONS: Answer the following exercises by writing the answer in the space provided at the end of the question. After you have completed all the exercises, turn to "Solutions to Exercises" at the end of the lesson and check your answers. For each exercise answered incorrectly, reread the material referenced with the solution. 1. How many ml of phosphoric acid (H 3 PO 4 ), 1.69 S.G., 85.0% purity would be needed to make 500 ml of a 1.00 Eq/L H3PO4 solution? 2. A commercial HCl solution has a S.G. of 1.15 and 33.0% purity. a. Find the mol/l concentration of the solution. b. How much concentrated solution would be required to make 1.00 L of a 2.00 Eq/L solution? 3. How much NH 4 OH with a specific gravity of and an assay of 26.0% is required to make 750 ml of 4.00 Eq/L ammonium hydroxide solution? 4. What is the g/dl concentration of the concentrated solution in exercise 3? MD

11 5. What volume of acetic acid (CH 3 COOH), S.G and assay of 99.7% is required to prepare 250 ml of a mol/l acetic acid solution? 6. What is the mass of H 2 SO 4 present in 150 ml of sulfuric acid solution with a specific gravity of 1.86 and a purity of 93.0%. 7. What is the mol/l concentration of the sulfuric acid solution in exercise 6? 8. How much of the sulfuric acid solution in exercise 6 would be required to prepare 1000 ml of a 2.50 mol/l sulfuric acid solution? 9. A HCl solution of 500 ml was prepared by using 25.0 ml of commercial reagent, S.G and assay of 34.0%. What is the mol/l concentration of the solution that was prepared? 10. A H 3 PO 4 solution was prepared by adding 12.5 ml of concentrated acid, S.G and 87.0% purity, to a one liter flask and Q.S. to the mark. Assuming that the acid ionizes completely, what is the normal concentration of the phosphoric acid that was prepared? Check Your Answers on Next Page MD

12 SOLUTIONS TO EXERCISES, LESSON ml (para 8-5) 2. a mol/l (para 8-6b) b. 192 ml (para 8-4) ml (para 8-5) g/dl (para 8-6a) ml (para 8-5) g (para 8-4) mol/l (para 8-6b) ml (para 8-5) mol/l (para 8-5) Eq/L (para 8-5) End of Lesson 8 MD