Section 6.2—Concentration

1. Section 6.2—Concentration How do we indicate how much of the electrolytes are in the drink? • Objectives: • Calculate concentrations in percent concentration and molarity • Calculate the concentration of electrolytes in solutions of ionic compounds

2. Concentrated versus Dilute solvent solute Lower concentration Not as many solute particles Higher concentration More solute particles

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

4. Percent Mass/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. %(M/V) Example Example: If you dissolve 12 g of sugar in 150 mL of water, what percent mass/volume is the solution?

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

7. Practice Problems • Determine the % M/V of a carbohydrate in a sports drink if it contains 14 g in 240 mL; if it contains 20 g in 240 mL. • A certain sports drink contains 0.050 g sodium chloride in 240 mL. What is its concentration in % M/V? What if it had 0.100 g in 240 mL?

8. 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

9. Mole Review • The molar mass of a substance is found by adding up all the atomic masses in the substance • Example: The molar mass of NaCl is 58.5 g • The molar mass of a substance in grams represents 1 mole of that substance. • 1 mole of NaCl has a mass of 58.5 g

10. Practice • Determine the molar mass of each of the following substances. • Ca(OH)2 • Mg • C2H3OH • Cu(NO3)2

11. Review: Calculating # of Moles • If you are given the mass of a substance, you use the molar mass to determine the number of moles. • Example how many moles are in 35 g of Mg? • 35 g x 1 mole = 1.4 moles 24.3 g

12. Practice: How many moles are in 25.5 g NaCl?

13. Na 1  23.0 g/mole 23.0 g/mole =  + 35.5 g/mole Cl 1 35.5 g/mole = 58.5 g/mole Example: How many moles are in 25.5 g NaCl? 1 mole NaCl molecules = 58.5 g 25.5 g NaCl mole NaCl 1 = _______ mole NaCl 0.436 58.5 g NaCl

14. More Practice • Determine the number of moles in each of the following: • 68.0 g CaCl2 • 9.2 g CH4 • 3. 70.1 g Al(OH)3

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

16. Molarity Example Example: If you dissolve 12 g of NaCl to make 150 mL of solution, what is the molarity?

17. Na 1  23.0 g/mole 23.0 g/mole =  + 35.5 g/mole Cl 1 35.5 g/mole = 58.5 g/mole Molarity Example Example: If you dissolve 12 g of NaCl to make 150 mL of solution, what is the molarity? 1 mole NaCl molecules = 58.5 g 12 g NaCl mole NaCl 1 = _______ mole NaCl 0.21 58.5 g NaCl Next, remember to change mL to L! Divide by 1000. 150 mL of water = 0.150 L = 1.4 M NaCl

18. Practice Problems • Determine the molarity of 2.5 L of solution made from 125 g of AgNO3. • What is the molarity of a solution produced when 145 g of NaCl is dissolved in enough water to make 2.75 L of solution? • What is the molarity of a solution in which 85.6 g of HCl are dissolved in enough water to make 385 mL of solution? • 8.77 g of KCl is dissolved in sufficient water to make 4.75 L of solution. Determine the molarity of the solution.

19. 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

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

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

22. 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 CaCl2 Ca+2 + 2 Cl-1 For every 1 CaCl2 unit that dissolves, you will produce 1 Ca+2 ion and 2 Cl-1 ions If the concentration of CaCl2 is 0.25 M, the concentration of Ca+2 is 0.25 M and Cl-1 is 0.50 M

23. 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?

24. 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

25. Let’s Practice #2 Example: What is the %(W/V) of a 500. mL sample of a 0.25 M CaCl2 solution?

26. Ca 1  40.08 g/mole 40.08 g/mole =  + 70.90 g/mole Cl 2 35.45 g/mole = 110.98 g/mole Let’s Practice #2 Example: What is the %(W/V) of a 500. mL sample of a 0.25 M CaCl2 solution? ? = 0.125 moles CaCl2 1 mole CaCl2 molecules = 110.98 g 0.125 moles CaCl2 g CaCl2 110.98 = _______ g CaCl2 13.9 1 mole CaCl2 2.8 %(W/V) CaCl2

27. Let’s Practice #3 Example: What are the molarities of the ions made in a 0.75 M solution of Ca(NO3)2

28. Let’s Practice #3 Example: What are the molarities of the ions made in a 0.75 M solution of Ca(NO3)2 Ca(NO3)2 Ca+2 + 2 NO3-1 For every 1 Ca(NO3)2, there will be 1 Ca+2 and 2 NO3-1 ions Ca+2 = 0.75 M NO3-1 = 1.5 M