Lecture Ch. 9: Solution Concentration Units When working with solutions, it is convenient to use units that express concentration – that is to say, how much solute there is per a set amount of solution or solvent. Highly concentrated solutions have a lot of solute dissolved. Weakly concentrated solutions have very little solute dissolved. Some solutions will have as little as one part solute/ 10,000,000,000 parts solvent ! For some mixtures, particularly liquids dissolved in liquids, dissolving in any proportion will lead to a solution – we call these solutions miscible in all proportions. For others, only a maximum amount of solute can be dissolved in solution before the rest will start to precipitate out. At that point the solution is saturated . For solutions in liquid solvents, there are a number of units we can use to express concentration. Some are better for extremely dilute concentrations, and some are better used for relatively concentrated solutions. Some will be used by chemists, some by biologists, and some by laboratory technicians. As professionals in the health field, you should be comfortable working with all of the concentration units we’ll cover in this chapter. I. Molarity Molarity is most often used by chemists in the practical lab. You may already have encountered it during your own lab classes here . Molarity stands for Moles of Solute / Liters of Solution. The most common mistake I see students make is to think that Molarity, or its shorthand symbol, M, stands for moles. This is not true. It is moles PER liter of solution. The shorthand for moles is mol, there is no single letter shorthand for moles. When doing math with Molarity, you need to remember that it includes both the solute and the volume of the solution. Example: Calculate the concentration of 0.034 moles of HCl in 1.5 L of solution in units of Molarity. Dimensional analysis tells us that we want moles of the solute and liters of the solution, and that we divide the one by the other. Looking at the problem, we have 0.034 moles of HCl – that is our solute. We also have 1.5 L of solution. When we divide one by the other, we get the following: 0.034 moles HCl = 0.0227 M HCl 1.5 L solution If instead I had given you grams or mL, you would first have had to convert to moles or to Liters in order to get the right units for Molarity! II. Weight/Volume percent concentration Weight percent is often used when solutions are relatively concentrated. Weight percent is given by dividing the mass of the solute in grams by the volume of the solution in mL, and multiplying by 100%. We express the result as % (w/v). For instance, for a solution of 15 g glucose in 100 mL solution, the weight percent would be found as follows: (15.0 g glucose) * 100% = 15% (w/v) (100 mL solution) If I give you the solute in moles or the solution in Liters, you will have to convert first to grams and mL to obtain the correct units. Alternately, if I give you the % concentration (w/v) and one of the other factors, you should be able to calculate the other – just treat this expression as a mathematical formula. III. Volume/volume percent concentration Volume percent is also often used when two solutions are relatively concentrated – and it is specifically used when the solvent and the solute are both liquid. Examples of this include the concentration units used to express the concentration of alcohol or hydrogen peroxide in the bottles you buy at the drugstore - both of which contain a substantial amount of water as well as ‘active ingredient’. Volume percent is given by dividing the volume of the solute by the volume of the solution and multiplying by 100%. We express the results as % (v/v). While generally the units used are mL, since both the volume of the solution and the solute must be given in the same units and they cancel out, any volume units should work. For example, if we have a solution of 4.5 mL of alcohol in 6.5 mL of solution, our volume percent would be found as follows: (4.5 mL alcohol) * 100% = 69.2% (v/v) (6.5 mL solution) IV. ppm and ppb Ppm and ppb are concentration units often used when the concentration of a solute is so small as to be nearly undetectable. ppm and ppb stand for “parts per million” and “parts per billion”, respectively . While it is tempting to think of these as parts in terms of molecules, the truth is ppm and ppb both refer to a ‘part’ as a unit of mass. So a solute that is present in concentrations of 1 ppm contributes 1 g for every 1 milllion grams of that solution. You will often see these concentration units on your water utility statements every year or so when they are required to provide you with information pertaining to the pollutants and drinkability of the local tap water. Items such as arsenic and lead in the water supply, as well as pharmaceuticals, will be expressed in terms of ppm and ppb. To calculate concentration in units of ppm, simply divide the mass of the solute by the mass of the solution (both in the same units, such as grams), then multiply be 10 6 (that is, a million). For example, if tap water is fluoridated by the addition of NaF at 32 mg NaF for every 20 kg of solution, what is its concentration in ppm? First, we need to convert both the NaF and the solution to the same mass units. I can either convert 32 mg NaF to kg, or 20 kg of solution to mg. Or I can convert them both to grams. Then I need to divide the mass of NaF by the mass of the solution and multiply by 10 6. 32 mg NaF = 32x10 -3 g NaF 20 kg solution = 20x10 3 g solution -3 32x10 g NaF 6 ppm = 20x10 3 g solution * 1x10 = 0.16 ppm NaF To calculate concentration in units of ppb, simply divide the mass of the solute by the mass of the solution (both in the same units, such as grams), then multiply be 10 9 (that is, a billion). For example, if the concentration of chloroform in drinking water is 100 ppb, how many grams of chloroform will be in a glass that contains 400 g of water? For this problem we will have to work backwards, but we can just use the same formula and plug it in: X g chloroform * 1x10 9 = 100 ppb Chloroform ppb = 400 g water When we solve for X, we get the following equation: 100 ppb chloroform*400 g water 9 = X g Chloroform 1x10 We solve for X and find 0.00004 g of chloroform in the glass of drinking water. I have included a great informative link to a page on understanding ppm and ppb in the weblinks section – if you have trouble with this concept I’d highly recommend checking it out. But consider: 100 ppb is such a small amount – it would be like selecting 700 people on the planet out of everybody alive today V. Dilutions Once we have mastered using using concentration units, we find it is easy to calculate concentrations of diluted solutions. If we multiply the concentration by the volume of the solution, the volume dimensions cancel out and we are left with the amount of solute. Since diluting a solution by adding liquid does not change the amount of solute present, the product of the concentration and the volume will be the same for the initial solution as well as the final. For example, if we did this with Molarity it would look like the following: M1V1 = M 2V2 Where M 1 and M 2 are the initial and final Molarity concentrations of our solution respectively, and V 1 and V 2 are the initial and final volumes of our solutions. For example, if I have a 0.500 L sample of 0.10 M NaOH, and I add enough water to dilute the solution volume to 2.500 L, what will the final concentration be? In this problem, M 1 = 0.10 M NaOH, V 1 = 0.500 L, and V 2 = 2.500 L. When we plug them into the equation and solve for M 2, we get the following: (0.10 M NaOH)*(0.500 L) = M = 0.020 M NaOH (2.500 L) 2 Again, this method will work with any of the concentration units discussed previously. Watch your units! .