Idea Transcript
Lab 3
Solutions & pH
Background The preparation of solutions of known concentration is a common procedure in the biology laboratory. There are several ways to make solutions and express their concentrations based on what the starting material is. 1. Weight-to-Weight (w/w) In the preparation of a weight-to-weight solution, a measured weight of solute is dissolved in a measured weight of solvent to make the solution. The w/w ratio is the weight of solute dissolved in a certain weight of solution. To make a 1:10 w/w aqueous solution of salt, for example, would entail dissolving 1 gram of salt in 9 grams of water to give ten grams of solution. Weight-to-weight is rarely used in biology. 2. Weight-to-Volume (w/v) The method of solution preparation preferred by most biologists is the weight-to-volume system. Such a preparation is usually made by dissolving a certain weight of solute in a volume of solvent less than the total volume required. Once dissolved, more solvent is added to the final volume desired. 3. Volume-to-Volume (v/v) In situations where both the solute and the solvent are liquids, a volume-to-volume solution is made. The usual method is to add a volume of solute to a volume of solvent, resulting in a solution.
Concentrations Percent A way of designating concentration is percent (%) which means parts per hundred. Percent is frequently used by biologists to indicate concentration. If you were preparing a w/v or a v/v percent solution, the actual amounts of the solute would be different for each solution. As a result, it is very important to indicate which way the percent solution is made. To make a percent solution always think in terms of 100, such as grams per 100 ml or milliliters per 100 ml.
g / 100 ml = % To make a 20% (w/v) SDS solution, you would Lab 3
weigh out 20 g of SDS (be careful not to inhale the powder). Then dissolve the SDS in a volume of laboratory grade water that is less than 100 ml. Once all the SDS is dissolved, add water up to 100 ml total. To make a 70% (v/v) EtOH solution, you would measure out 70 ml of EtOH and then add water (30 ml) to 100 ml total.
Molarity Besides percent, biologists often express concentration in terms of molarity (M). Since different substances have different molecular (or formula) weights, the same weight of the two substances will contain different amounts of molecules. To correct for this difference molarity is used. Molarity is expressed in grams of molecular weight (mw) per liter. It is a weight per volume type of solution (w/v). A one molar solution is the molecular weight in grams of the compound dissolved in one liter of solution. It is prepared by weighing out in grams the amount needed of a compound and dissolving it in a solvent to a total volume of one liter. The molecular weight is usually found on the label of the bottle which contains the substance or it can be looked up in the Handbook of Chemistry and Physics. For example, the molecular weight of KCl is 74.56. Therefore, a 1M (molar) concentration of KCl contains 74.56 grams/liter. To illustrate the utility of molarity let’s compare the one molar solution of KCl above with a 1M solution of acetic acid (mw = 60.05). A 1M solution of acetic acid contains 60.05 g of acetic acid per liter. But a gram molecular weight of any substance contains the same number of molecules of that substance (6.02 × 1023 Avogadro’s number). Consequently, a volume of 1M KCl contains the same number of molecules of KCl as does the same volume of 1M acetic acid contains acetic acid molecules. This is true even though the number of grams of each substance differs. By definition a 1 M solution = the molecular weight (mw) in grams dissolved in 1 liter.
mw g / l = 1 M To make 1 liter of 1 M Tris, you would first find the molecular weight of Tris on the bottle Page 1
(121.12). Then weight out 121.12 grams and dissolve in less than 1 liter of water. Once dissolved add water up to 1 liter. If you want to 100 ml of a 1 liter KCl solution, again first look up the molecular weight of KCl (74.56). Then. . . 74.56 g / l = 1 M Since 1 liter = 1000 ml, substitute 1000 ml for l 74.56 g / 1000 ml = 1 M Since you want 100 ml divide the left side of the equation by 10 (or move both decimal points one place to the left). 7.456 g / 100 ml = 1 M Now weight out 7.456 g of KCl and dissolve in less than 100 ml of water. Once dissolved bring the volume up to 100 ml with water. At this time, take special note of the fact that molarity is expressed as part per liter which means parts per 1000 since a liter is 1000 ml. Since percent means per 100, it should be obvious that conversion of molarity to percent involves dividing the number of moles by ten.
Solution Conversions It is often necessary to convert from one concentration to another. For example, in the laboratory the biologist will often make a “stock” solution and keep it on the shelf. For the experiment, however, the experiment requires a
concentration of the solution less than the stock solution concentration. Therefore it is essential that you be able to convert from one concentration to another. The first step is to list what you have and what you want. Next make sure everything is in the same units. The easiest way to convert from one concentration to another is to use the formula:
CHave × VTransfer = CWant × VWant C stands for concentration and V stands for volume. You have a stock solution of 90% ethanol. You want 100 ml of 70% ethanol for your experiment. You set up the equation as: (90%) (V) = (70%) (100 ml) Now solving for V gives you 77.8 ml. Now transfer 77.8 ml of 90% stock solution to a graduated beaker and add water up to 100 ml to obtain 100 ml of 70% alcohol. If you have a 5 M stock solution and you need 20 ml of a 1 M solution, you would set up the equation as: (5 M) (V) = (1 M) (20 ml) Now solve for V which will tell you how much of your 5 M stock solution you need to transfer to make a 20 ml 1M solution. (V = 4 ml)
Exercise 3.1 - Making solutions 1. Using the formula CV = CV calculate how to make the solutions on page 5. 2. Have your calculations checked by the instructor 3. Make the solutions in 15 or 50 ml blue capped centrifuge tubes.
Example Problem # 1 You need 0.03 l of 2 % SDS in the experiment. You have a bottle of 20% (w/v) SDS on the shelf. Step One: Express in Common Terms This applies to both the concentrations and the volumes. In the example problem both concentrations are expressed in percent so we are good there. For the volumes you need to transfer milliliters (ml) but need 0.03 l. So first convert the 0.03 l into ml. 0.03 l = 30 ml Step Two: Determine what you Have and what you Want You have a stock solution of 20 %. You want a final concentration of 2 % in 30 ml. Step Three: Calculate the Dilution Factor by filling in the equation and solving for VTransfer 20 % × VT = 2 % × 30 ml Now solve for VT which = 3 ml. Step Four: Use the results Transfer 3 ml of your 20 % (w/v) SDS stock solution and add water (27 ml) up to 30 ml total for a final concentration of 2 % SDS. Lab 3
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Example problem #2 You need 100 ml of 0.5 M EDTA (Disodium Ethylenediamine Tetraacetate). formula weight = 372.24. Based on the molar formula given above, 372.24 g / l = 1 M EDTA If you needed 1M you would be done with the math but you need 0.5 M, so divide both sides of the equation by two resulting in 186.12 g / l = 0.5 M EDTA. Now you want 100 ml not 1 liter. You could make up 1 l and then just use 100 ml but that would be a waste of material so do the math. Since 1 liter = 1000 ml replace the l in the formula with 1000 ml, to get 186.12 g / 1000 ml = 0.5 M EDTA. Now divide both the numerator and the denominator by 10 (or just move the decimal point to the left one digit), to get 18.612 g / 100 ml = 0.5 M EDTA. Now, at the lab bench weigh out 18.612 g of EDTA and add water (less than 100 ml) to dissolve the EDTA. Once dissolved bring the volume up to 100 ml to obtain a 0.5 M solution of EDTA.
pH Compounds that release hydrogen ions (H+) when dissolved in water are called acids. On the other hand, compounds that release hydroxide ions (OH-) in water are called bases. The released hydroxide ions will combine with any free hydrogen ions and form water thereby reducing the amount of free hydrogen ions in the solution.
H+ concentration
The concentration of hydrogen ions is measured on the pH scale. The scale ranges from 0 (the most acidic) to 14 (the most basic). The midpoint of the scale is 7 which represents a neutral pH. pH 7 is where most living systems function best. Your stomach is very acidic. The acidity aids in digestion and kills bacteria you may swallow to protect you against infection. When you suffer from acid indigestion, the acid level in your stomach is too high. To alleviate the uncomfortable condition you can take an antiacid which will add OH- ions to your stomach and neutralize the excess H+. pH can be measured with a pH meter or with pH indicator paper.
(moles/liter)
pH value 1 M hydrochloric acid (HCl) stomach acid lime juice lemon juice
10-0
0
10-1
1
10-2
2
10-3
3
10-4
4
vinegar, cola, tomatoes, orange juice beer
10-5
5
black coffee, tea
10-6
6
rain (5.6) urine (5-7) saliva, pure water, blood, sweat
10-7
neutral (H+ = OH-)
7
10-8
8
seawater
10-9
9
baking soda
10-10
10
10-11
11
phosphate detergents chlorine bleach household ammonia
10-12
12
10-13
13
oven cleaner
10-14
14
1 M sodium hydroxide (NaOH)
Exercise 3.2 pH solutions 1. Determine the pH of the solutions listed in Exercise 3.2 Report using a pH meter. (You can use pH paper to measure the pH of saliva.) 2. Record your results in the table below and answer if the solution tested is an acid, a base, or a neutral solution. Lab 3
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Normality Often the concentration of acids and bases is expressed in terms of normality. A one normal solution (abbreviated 1 N) is defined as one gram - equivalent weight per liter. It is a weight-to-volume solution also. A gram equivalent weight of a solute is its gram molecular weight divided by the number of potentially replaceable hydrogens per solute molecule. The following table illustrates this for acids and bases: Mol. Wt. HCl H2SO4
36.45 98.0
# of Groups can donate 1 2
H3PO4
102.99
3
34.33
NaOH Ca(OH)2
40 74.09
1 2
40 37.045
Exercise 3.3
Equiv. Wt.
Thus, a 1.0 N solution of HCl would be its gram-molecular weight divided by the number of hydrogens or potential hydrogens, which would be 36.45 g/l divided by 1, or 36.45 g/l. Similarly, a 1.0 N solution of H2SO4 would be 98 g/l divided by 2, or 49 g/l; for NaOH it would be 40 g/l divided by 1, or 40 g/l. In the case of NaOH, the number of potential hydrogens would be equivalent to its valence, since one hydrogen will combine with OH- to form H2O. Once normality is calculated, its conversion to percent concentration is done the same as converting molarity to percent.
36.45 49.0
Citric Acid
1. First calculate (show on report) how much citric acid to weigh out to prepare 25 ml of 1 M citric acid. (m.w. = 192.1) 2. Using the structural formula of citric acid below determine how many H+s are released by citric acid when it is dissolved in water. Write your answer on the report. 3. Next calculate (show on report) how much citric acid to weigh out to prepare 25 ml of 1 N citric acid. 4. Checked your calculations with your instructor. 5. Prepare 25 ml of 1 N citric acid (assume complete dissociation) from solid. Label and store the solution in a lab bench drawer.
Citric acid HC—COOH | HO—C—COOH | 2HC—COOH 2
Lab 3
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Exercise 3.1 Report
Name _______________
Prepare to following solutions: (Show calculations below.) 1. 20 ml of 0.005M KMnO4 (Potassium permanganate) from a stock solution of 0.05 M.
2. 10 ml of 1N hydrochloric acid (HCl) from a stock solution of 2 N.
3. 25 ml of 2% w/v glucose from a 5% (w/v) stock solution.
Lab 3
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Exercise 3.2 Report
Name _______________
pH of various solutions. Solution
pH
Acid / Base / Neutral
1% bleach saliva tap water laboratory grade water baking soda vinegar liquid soap 2 N hydrochloric acid (from this lab) 1 N hydrochloric acid (from this lab) 1 N citric acid (from exercise 3.3 of this lab) 1. Which solution tested had a pH closest to neutral?
2. Was there a difference in pH between the 1N HCl and the 1N citric acid? If so why?
3. Which of the solution(s) above would you use to neutralize excess acid in your stomach?
Exercise 3.3 Citric Acid Solution 1. 1 M Citric Acid solution calculations.
2. How many H+s can be released from citric acid? 3. 1 N Citric Acid solution calculations.
Lab 3
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