1. Chapter 16 – Solutions Mr.Yeung

2. Lesson 5 - Objectives Take up questions Dilutions (Super important)!

3. Dilutions When you buy a can of juice concentrate from the supermarket you are expected to dilute the juice concentrate. What would a spoonful of concentrate taste like compared to a spoonful of the diluted juice? Yuck? –The concentrate would be a lot stronger

4. Think of the orange concentrate are molecules ConcentrateDiluted

5. What if … We think 1 can of concentrate = 1 mol And we need to make 1L of orange juice? 1 can / 1 L = 1M What about 0.5L? 1 can / 0.5L = 2M What about 0.75L? 1 can / 0.75L = 1.33M More water = less concentrated

6. Now.. Say we took the –2M concentrated orange juice we made (1 can / 0.5L) (Stock solution) –We used 2L of this concentration (2M) and we want to dilute this solution to make 6L of orange juice –What would you do?

7. Dilution Before dilution After dilution 2M of 2L Total volume of 6L 2L is 3 times less than 6L so the 2M will be 3 times as less as well = 0.67M

8. Formula Basically the formula for dilution –C1 V1 = C2 V2 C1 = initial concentration (mol/L) V1 = initial volume (L) C2 = final concentration (mol/L) V2 = final volume (L) From last example –2mol/L * 2L = C2 * 6L –C2 = 2/3 mol/L

9. Examples Example: What volume of concentrated sulfuric acid, 18.0 M, is required to prepare 5.00 L of 0.150 M solution by dilution with water? –C1 = 18M C2 = 0.150M –V1 = ?(*required)V2 = 5.0L 18M * (?) = 0.150M * 5.0L (0.150M * 5.0L) / 18M = 0.042L –Does that make sense? 18.0M is A LOT more concentrated than 0.150M so using a little amount is expected!

10. Example How much in grams would you need to prepare 500.0 mL of a 0.100 M standard solution of KNO3? –Find mols 0.5L * 0.1mol/L = 0.05mol –Find mol mass 101.1g/mol –Find grams 101.1g/mol * 0.05mol = 5.05g

11. Example If I have 340 mL of a 0.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it? –C1V1 = C2V2 V1 = 340ml or.34L C1 = 0.5M V2 = 560ml or.56L C2 = ? 0.34L * 0.5M = 0.56L * (?) C2 =.30M

12. Real life example How would you prepare 500 ml of 3 M HCl using 6 M HCl from the stockroom. In other words how much water and how much 6 M HCl would you mix to accomplish this dilution? –Determine the volume of 6M HCl to use applying the dilution equation 6 M ( Volume of 6M) = 3 M HCl ( 500 ml) –Volume of 6 M HCl = 3 M HCl ( 500 ml) / 6 M HCl = 250 ml 6M HCl So what is the 250ml 6M HCL? –That is the amount you need but you still need to dilute this… To what volume? –Total = 500ml – 250ml

13. Example in chem labs Example 1 Let us consider 5.00 mL of a solution labeled “6.00 M HCl” and we now add enough water to give a total volume of 14.00 mL. What is the concentration of the new solution?

14. Summary Dilution Problems