Units of Concentration A solution is a homogeneous mixture of one substance (the solute) dissolved in another substance (the solvent). Concentration is a ratio of the amount of solute to the amount of solvent. A common task in a biotechnology lab is preparing solutions. What exactly is a solution? A solution is a mixture of what is dissolved (the solute) and the dissolving medium (the solvent). For example, if you are making koolaid the colored sugar is the solute and the pitcher of water is the solvent. The concentration of a solution is the ratio of the amount of solute to solvent. It is necessary to prepare solutions with the correct concentration or you can destroy months of hard work in a biotechnology lab.

Units of Concentration Percent volume % volume = volume solute (ml) x 100 volume solution (ml) Percent mass % mass = mass solute (g) x 100 mass solution (g) Solution = solvent + solute

Units of Concentration Example 1: What is the percent by volume concentration of a solution in which 75.0 ml of ethanol is diluted to a volume of 250.0 ml? 75.0 ml x 100 = 30.0% 250.0 ml

Units of Concentration Example 2: What volume of acetic acid is present in a bottle containing 350.0 ml of a solution which measures 5.00% concentration? x = 0.05 350.0 ml x = 17.5 ml

Units of Concentration Example 3: Find the percent by mass in which 41.0 g of NaCl is dissolved in 331 grams of water. 41 g x 100 = 11.0% 372 g

Units of Concentration Molarity (M) is the most common unit of concentration Molarity is an expression of moles/Liter of the solute. Molarity is the most common unit of concentration. You will often be asked to prepare a given molarity of solution. What does molarity mean? Well, it tells you the number of moles of solute in a liter of solvent.

Units of Concentration A mole is the SI unit of number of particles and can be used as an expression of the molecular weight of a substance. So your next question is probably “what is a mole”? Well, it’s not the furry animal wearing sunglasses in the previous slide, it is actually the SI unit of number of particles. It is a chemistry term that can be used to calculate the formula weight of a substance. In fact formula weight is often called molar mass. The formula weight of an element is recorded on the periodic table. For example, the formula weight of sodium is 22.990 grams/mole. The formula weight of an element is expressed as grams/mole

Units of Concentration The molar mass of a compound can be calculated by adding the molar mass of the individual elements. To calculate the molar mass of a compound, you simply add the formula weights of the individual elements. So in this example you add the formula weight of sodium (22.990) with the formula weight of chlorine (35.453) to tell you that the molar mass of salt (NaCl) is 58.443 grams/mole. 22.99 + 35.45 = 58.44 g/mol

Making Solutions g = M x L x molar mass You just calculated the molar mass of sodium chloride to be 58.44 g/mol. To determine how to make a stock solution of sodium chloride, use the formula: You have to know the formula weight of a compound if you are going to prepare the correct concentration of a solution with that compound. The formula (grams = molarity x liters x molar mass) is used to tell you how much of a compound to use to make a solution with a specific concentration. g = M x L x molar mass

Making Solutions g = M x L x molar mass How many grams of NaCl would you need to prepare 200.0 mL of a 5 M solution? g = M x L x molar mass g = (5mol/L) (0.2L) (58.44g/mol) g = 58.44 g For example, if I wanted to make a 200 milliliters of sodium chloride with a concentration of 5 M, how much of the compound would I use? Remember that Molarity actually stands for moles per Liter. That means I’m going to have to convert 200 milliliters to liters. On our metric line, there were three places between milli and the base unit, so we will have to move our decimal three places to the left. That means 200 milliters is 0.2 L. Now we can plug our numbers into our formula. According to my formula, the number of grams I need will be 5M x 0.2L x 58.44 g/mol. After canceling units and multiplying these three numbers together, I get the answer 58.44g. My balance goes to 2 decimal places, that means that I will weigh out 58.44 grams of NaCl and add 0.2 L or 200 millilters of water. When making a stock solution, we usually dissolve the solid in about two-thirds volume of water. So that means we would dissolve our 58.44 g of NaCl in only about 150 mls of water and stir. When it is completely dissolved, we would transfer the solution to a graduated cylinder and bring it to the final volume of 200 milliliters.

Diluting Solutions Often once you have made a stock solution, you need to dilute it to a working concentration. To determine how to dilute the stock solution, use the formula: Great! Now that you’ve made your one molar stock solution of sodium chloride, you may need to dilute it to a different concentration as a working solution. The formula C1V1 = C2V2 is used to figure out this dilution. C2 is the concentration of the new solution you want to make and V2 is the volume of that solution you want to make. C1 is the concentration of your stock solution. The question is “how much of that stock do you need to make your dilution”? That’s why you will usually be solving for V1. C1 – concentration of stock C2 - concentration of diluted solution V1 – volume needed of stock V2 – final volume of dilution C1V1 = C2V2

Diluting Solutions Example 5: C1 V1 = C2 V2 (5) V1 = (0.4)(100) How many milliliters of a 5 M stock solution of NaCl are needed to prepare 100 ml of a 0.4 M solution? C1 V1 = C2 V2 (5) V1 = (0.4)(100) V1= 8 ml If you want to make 100 milliliters (that’s your V2) of a 0.4 M sodium chloride solution (that’s your C2). The concentration of the stock solution is 5M (that’s your C1) How much of that stock solution will you need (that’s your V1)? So to solve for V1, let’s plug our numbers in to the formula and see. Solving for V1 means we have to multiply 0.4 by 100 and divide by 5. That gives us an answer of 8 milliliters. That means we will use 8 milliliters of our stock solution. Can you guess how much water we add to that 8 mls to dilute it? That’s right! We wanted a final volume of 100 mls, so 100-8 means we will add 92 mls of water to produce our final product of a 0.05 M solution of sodium chloride. Alright!

Diluting Solutions Serial Dilutions are dilutions made in series (for example, if you needed to make solutions that were 2M, 1M, 0.5M, and 0.25 M) The formula for serial dilutions is: Dilution Factor = (V1 + V2) V1 V1 – volume of solution being diluted V2 – volume of solvent Great! Now that you’ve made your one molar stock solution of sodium chloride, you may need to dilute it to a different concentration as a working solution. The formula C1V1 = C2V2 is used to figure out this dilution. C2 is the concentration of the new solution you want to make and V2 is the volume of that solution you want to make. C1 is the concentration of your stock solution. The question is “how much of that stock do you need to make your dilution”? That’s why you will usually be solving for V1.

Units of Concentration Example 6: Propose a method to prepare 100 ml of a 0.5 M glucose solution from a 5 M glucose solution. 10 = (v1 + 100) v1 10v1 = v1 + 100 -v1 -v1 9v1 = 100 9 9 v1 = 11.1 ml of 5 M glucose + 100 ml H2O

Diluting Solutions Dilutions tutorial Click on the Dilutions tutorial for more practicing diluting solutions.