03 - Concentration - DILUTIONS

1. Chemistry 30 – Unit 2 – Solubility – Ch. 16 in Text 03 - Concentration - DILUTIONS

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: - m1 = initial moles of solute After Dilution: - m2 = 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. Final moles of solute Initial 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: M1V1 = M2V2 • M1V1 means the molarity and the volume of the original solution • M2V2 means the molarity and the volume of the second solution.

7. CAUTION! • You may also see this equation written as C1V1 = C2V2. • “C” stands for concentration. • We use the formula M1V1 = M2V2 when our concentration is expressed in molarity. • We would use C1V1 = C2V2 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 • M1v1 = M2v2 10. ( vi ) = 0.200 (.01) • v1 = 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 • M1v1 = M2v2 • 10M ( 10mL ) = ( M2 ) 100mL M2 = 1.0 M • Step 2: Now you have a 1.0 molar solution so redo the math • M1v1 = M2v2 • (1.0)v1 = 200 ( .01 ) v1 = 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 • M1v1 = M2v2 • (1.0) 10. = (M2) 100 M2 = .10 M • Step 2: Now you have a .10 molar solution so redo the math • M1v1 = M2v2 • (.10)v1 = 200 (.01) v1= 20 mL • 20 mL is even easier than 2 mL to measure accurately