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Chapter 16 – Solutions

Chapter 16 – Solutions . Mr.Yeung. Lesson 5 - Objectives. Take up questions Dilutions (Super important)!. Dilutions. When you buy a can of juice concentrate from the supermarket you are expected to dilute the juice concentrate.

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Chapter 16 – Solutions

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  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 Concentrate Diluted

  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 Total volume of 6L 2M of 2L 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

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