1. 03 - CONCENTRATION - DILUTIONS CHEMISTRY 30 – UNIT 2 – SOLUBILITY – CH. 16 IN TEXT

2. STOCK SOLUTIONS A common situation in a chemistry laboratory occurs when a solution of a particular concentration is desired, but only a solution of greater concentration is available: this is called a stock solution.

3. WHAT IS A DILUTION? To dilute a solution means to add more solvent without the addition of more solute. You may have heard it referred to as “watering down” This statement is somewhat accurate, as long as your solvent actually is water Of course, the resulting solution is thoroughly mixed so as to ensure that all parts of the solution are homogenous.

4. WE CAN DILUTE A SOLUTION BY ADDING WATER TO IT: BEFORE DILUTION: - M 1 = INITIAL MOLES OF SOLUTE AFTER DILUTION: - M 2 = FINAL MOLES OF SOLUTE

5. FROM PREVIOUS SLIDE… NOTICE THAT WE DID NOT ADD ANY MORE SOLUTE AT ALL THE NUMBER OF MOLES WILL NOT CHANGE. THE VOLUME CHANGES, AND ALSO THE CONCENTRATION. Initial moles of solute Final moles of solute …is equal to

6. DILUTION EQUATION: The fact that the solute amount stays constant allows us to develop calculation techniques. First, we write: moles before dilution = moles after dilution From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can substitute MV (molarity times volume) into the above equation, like this: M 1 V 1 = M 2 V 2 M 1 V 1 means the molarity and the volume of the original solution M 2 V 2 means the molarity and the volume of the second solution.

7. CAUTION! You may also see this equation written as C 1 V 1 = C 2 V 2. “C” stands for concentration. We use the formula M 1 V 1 = M 2 V 2 when our concentration is expressed in molarity. We would use C 1 V 1 = C 2 V 2 if our concentration was expressed in something else, like g/L, or ppm, etc. ***More on those later**.

8. EXAMPLE #1 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make? Using the dilution equation, we write: (1.50 mol/L) (0.0534 L) = (0.800 mol/L) (x) Solving the equation gives an answer of 0.100 L. Notice that the volumes need not be converted to liters. Any old volume measurement is fine, just so long as the same one is used on each side.

9. EXAMPLE #2 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results? Placing the proper values into the dilution equation gives: (2.500 mol/L) (0.1000 L) = (0.5500 mol/L) (x) x = 0.4545 L

10. SERIAL DILUTIONS The dilution equation is fairly easy to understand – we shouldn’t have any trouble calculating new or desired concentrations or volumes Sometimes however, we will calculate out a value that is too small or unrealistic – this makes it difficult to measure in a practical lab-like setting

11. SERIAL DILUTIONS - EXAMPLE How could we prepare 200. mL of.01 M HCl from 10. M HCl? Calculate the missing value using the dilution equation M 1 v 1 = M 2 v 2 10. ( v i ) = 0.200 (.01) v 1 = 0.00020 L or 0.20 mL It is unrealistic for us to measure.20 mL so we do a series of dilutions so we can maintain our accuracy

12. SERIAL DILUTIONS - EXAMPLE How do we solve this? Step 1: First pick a practical dilution into an appropriately small volume. For example use 10. ml into 100. ml M 1 v 1 = M 2 v 2 10M ( 10mL ) = ( M 2 ) 100mL M 2 = 1.0 M Step 2: Now you have a 1.0 molar solution so redo the math M 1 v 1 = M 2 v 2 (1.0)v 1 = 200 (.01 ) v 1 = 2 mL 2 mL is a much more reasonable amount to measure than 0.2 mL

13. SERIAL DILUTIONS - EXAMPLE If 2 mL is still too small and you don't have the equipment to do it, continue to dilute: Step 1: Do the 10. ml into 100. ml again; this time your initial concentration will be 1.0 M M 1 v 1 = M 2 v 2 (1.0) 10. = (M 2 ) 100 M 2 =.10 M Step 2: Now you have a.10 molar solution so redo the math M 1 v 1 = M 2 v 2 (.10)v 1 = 200 (.01) v 1 = 20 mL 20 mL is even easier than 2 mL to measure accurately