1. Section 6.2—Concentration How do we indicate how much of the electrolytes are in the drink?

2. Concentrated versus Dilute solute solvent Lower concentration Not as many solute (what’s being dissolved) particles Higher concentration More solute (what’s being dissolved) particles

3. Concentration Concentration gives the ratio of amount dissolved to total amount There are several ways to show concentration

4. Percent Weight/Volume This is a method of showing concentration that is not used as often in chemistry However, it’s used often in the food and drink industry  For example, your diet drink can might say you have less than 0.035 g of salt in 240 mL.  That would give you a concentration of 0.035 g / 240 mL, which is 0.015% solution

5. %(W/V) Example Example: If you dissolve 12 g of sugar in 150 mL of water, what percent weight/volume is the solution?

6. %(W/V) Example Example: If you dissolve 12 g of sugar in 150 mL of water, what percent weight/volume is the solution? 8.0% (W/V)

7. Concentration using # of molecules When working with chemistry and molecules, it’s more convenient to have a concentration that represents the number of molecules of solute rather than the mass (since they all have different masses) Remember, we use moles as a way of counting molecules in large numbers

8. Quick Mole Review 1 mole = 6.02 × 10 23 molecules The molecular mass of a molecule is found by adding up all the atomic masses in the atom Molecular mass in grams = 1 mole of that molecule

9. Quick Mole Example Example: How many moles are in 25.5 g NaCl?

10. Quick Mole Example 25.5 g NaCl = _______ mole NaCl g NaCl mole NaCl 1 58.44 0.44 1 mole NaCl molecules = 58.44 g Na Cl 1 1 22.99 g/mole 35.45 g/mole  = 22.99 g/mole = 35.45 g/mole + 58.44 g/mole  Example: How many moles are in 25.5 g NaCl?

11. Molarity Molarity (M) is a concentration unit that uses moles of the solute instead of the mass of the solute

12. Molarity Example Example: If you dissolve 12 g of NaCl in 150 mL of water, what is the molarity?

13. Molarity Example Example: If you dissolve 12 g of NaCl in 150 mL of water, what is the molarity? 1.4 M NaCl 12 g NaCl = _______ mole NaCl g NaCl mole NaCl 1 58.44 0.21 1 mole NaCl molecules = 58.44 g Na Cl 1 1 22.99 g/mole 35.45 g/mole  = 22.99 g/mole = 35.45 g/mole + 58.44 g/mole  Remember to change mL to L! 150 mL of water = 0.150 L

14. Converting between the two If you know the %(W/V), you know the mass of the solute You can convert that mass into moles using molecular mass You can then use the moles solute to find molarity

15. Converting from % to M Example Example: What molarity is a 250 mL sample of 7.0 %(W/V) NaCl?

16. Converting from % to M Example Example: What molarity is a 250 mL sample of 7.0 %(W/V) NaCl? 1.2 M NaCl 17.5 g NaCl = _______ mole NaCl g NaCl mole NaCl 1 58.44 0.30 1 mole NaCl molecules = 58.44 g Na Cl 1 1 22.99 g/mole 35.45 g/mole  = 22.99 g/mole = 35.45 g/mole + 58.44 g/mole  Remember to change mL to L! 250 mL of water = 0.250 L ? = 17.5 g NaCl

17. Concentration of Electrolytes An electrolyte breaks up into ions when dissolved in water You have to take into account how the compound breaks up to determine the concentration of the ions CaCl 2  Ca +2 + 2 Cl -1 For every 1 CaCl 2 unit that dissolves, you will produce 1 Ca +2 ion and 2 Cl -1 ions If the concentration of CaCl 2 is 0.25 M, the concentration of Ca +2 is 0.25 M and Cl -1 is 0.50 M

18. Let’s Practice #1 Example: You want to make 200 mL of a 15% (W/V) solution of sugar. What mass of sugar do you need to add to the water?

19. Let’s Practice #1 Example: You want to make 200 mL of a 15% (W/V) solution of sugar. What mass of sugar do you need to add to the water? 30 g of sugar

20. Let’s Practice #2 Example: What is the %(W/V) of a 500. mL sample of a 0.25 M CaCl 2 solution?

21. Let’s Practice #2 Example: What is the %(W/V) of a 500. mL sample of a 0.25 M CaCl 2 solution? 2.8 %(W/V) CaCl 2 0.125 moles CaCl 2 = _______ g CaCl 2 mole CaCl 2 g CaCl 2 110.98 1 13.9 1 mole CaCl 2 molecules = 110.98 g Ca Cl 1 2 40.08 g/mole 35.45 g/mole  = 40.08 g/mole = 70.90 g/mole + 110.98 g/mole  ? = 0.125 moles CaCl 2

22. Let’s Practice #3 Example: What are the molarities of the ions made in a 0.75 M solution of Ca(NO 3 ) 2

23. Let’s Practice #3 Example: What are the molarities of the ions made in a 0.75 M solution of Ca(NO 3 ) 2 Ca(NO 3 ) 2  Ca +2 + 2 NO 3 -1 For every 1 Ca(NO 3 ) 2, there will be 1 Ca +2 and 2 NO 3 -1 ions Ca +2 = 0.75 M NO 3 -1 = 1.5 M