Unit 4 Solubility

1. Unit 4 Solubility 4.2.1- Dilutions 4.2.2- MOLARITY 4.2.3 Concentration 4.2.4 Dilution of Solutions 4.2.5concentration of ions

2. 4.2.1 Dilute & Concentrated Solutions We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms - dilute and concentrated. A dilute solution is one in which there is a relatively small amount of solute dissolved in the solution. A concentrated solution contains a relatively large amount of solute.

3. 4.2.2 Molarity There are several different ways to indicate the amount of solute dissolved in a given volume of solution. The most common is called molarity. Molarity = moles solute litre solution Symbol for Molarity = M

4. As an example: The concentration of a solution might be expressed as a "0.50 M NaOH". You would read as "a 0.50 molar sodium hydroxide solution", meaning that there are 0.50 moles of NaOH dissolved per one litre of solution.

5. Sample Problem 1. Antifreeze is a solution of ethylene glycol, C2H6O2 in water. If 4.50 L of antifreeze contains 27.5 g of ethylene glycol, what is the concentration of the solution?

6. Solution: The question asks moles solute/litre solution Thus we will need to calculate the molar mass of C2H6O2 in order to convert between mass and moles. the molar mass of C2H6O2 to be 62.1 g·mol-1 arrange the information to end up with the desired unit: Mol = 27.5 g × 1 mol × 1__ L 62.1 g 4.5 L = 0.098 mol/L Our final answer: [C2H6O2 ] = 0.098 M

7. Practice Questions 4.2.2

8. 4.2.3 Other Measures of Concentration There are some other ways to express solution concentration. Here are some alternatives: Molality- Molality is defined as the number of moles of solute dissolved in one kilogram of solvent Note: Molarityuses mass rather than volume, and Molality uses solvent instead of solution.

9. Mass per unit volume. This unit is typically milligrams per milliliter (mg/mL) or milligrams per cubic centimeter (mg/cm3). A useful note is that 1 mL = 1 cm3 and that cm3 is sometimes referred to as a "cc" (cubic centimeter). Mass per unit volume is handy when discussing how soluble a material is in water or a particular solvent.

10. Percent by Mass. Also called weight percent or percent by weight. This is simply the mass of the solute divided by the total mass of the solution and multiplied by 100%: % mass = mass of component x 100% mass of solution

11. Parts per million (PPM). Parts per million works like percent by mass, but when there is only a small amount of solute present. PPM is defined as the mass of the component in solution divided by the total mass of the solution multiplied by 106 (one million): A solution with a concentration of 1 ppm has 1 gram of substance for every million grams of solution. In general, one ppm implies one mg of solute per liter of solution.

12. Example A sample of water contains 25 ppm of lead ions, Pb2+. Convert this concentration to molarity, mol·L-1.

13. Solution: Since the conversion involves mass, we will need to know the molar mass of Pb2+, which is 206.2 g·mol-1  1. change g to mL 2. change mL to L 3. change g to mol 25 ppm = 25 g Pb2+ = 25 g Pb2+= 1 mol Pb2+= 106mL H2O 103L H2O 206.2 g =1.2 × 10-4mol L

14. 4.2.4 Dilutions of Standard Solutions Imagine we have a salt water solution with a certain concentration (i.e a certain amount of salt dissolved in a certain volume of solution), to dilute this solution - we add more water

15. M1 = moles1  M2 =moles2 litre1 litre2 rearrange the equations to find moles: moles1= M1× litre1 moles2= M2 × litre2

16. By adding more water, we changed the volume of the solution and it's concentration. But the number of moles of solute did not change. So, moles1 = moles2 Therefore, M1 × litre1 = M2 × litre2 Which we will usually express as: M1V1= M2V2 Where M1 and M2 are the concentrations of the original and diluted solutions and V1 and V2 are the volumes of the two solutions * note C may be used for M *

17. 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?

18. Solution: In a dilution question there are 4 variables - M1, V1, M2 and V2. You will know three of these values and have to calculate the fourth. Write out what you know and what you are trying to find out M1 = 18.0 M M2= 0.150 M V1 = ?  V2 = 5.00 L Set up the formula, and rearrange to solve for the unknown, M1V1= M2V2 18.0 × V1= 0.150 × 5.00 V1= 0.150 × 5.00 18.0 V1= 0.0417 L= 41.7 mL

19. Example: Describe how you would prepare 500.0 mL of a 0.100 M standard solution of KNO3 Solution: We must first determine the mass of KNO3 that will be needed to prepare 500.0 mL (0.500 L) of a 0.100 M solution. The molar mass of KNO3 is 101.1 g·mol-1: g = 101.1 g × 0.100mol× 0.500 L= 5.05 g mol L To prepare the standard solution we will carefully measure out 5.05 g of KNO3 and dissolve it in some water. Then we add enough water to make 500 ml

20. Solution: We must first determine the mass of KNO3 that will be needed to prepare 500.0 mL (0.500 L) of a 0.100 M solution. The molar mass of KNO3 is 101.1 g·mol-1: g = 101.1 g × 0.100mol× 0.500 L= 5.05 g mol L To prepare the standard solution we measure out 5.05 g of KNO3 and dissolve it in some water. Then we add enough water to make 500 ml

21. Practice 4.2.4

22. 4.2.5 Concentration of Ions in Solution The solutions often involve ionic compounds and acids. Both of these produce electrolytic solutions, meaning they conduct an electrical current due to the production of ions in solution. It will often be important for us to be able to determine the concentration of these ions, not just the overall concentration of the solution.

23. It will be important for you to be able to write balanced reactions that show how these substances break down into ions. If you are given the formula of the compound (either ionic or acid) you will need to be able to determine: which ions are produced and in what mole ratio.

24. For example, sodium carbonate, Na2CO3 dissociates into ions as: Na2CO3 (s) → 2 Na+(aq)+ CO32-(aq) Notice that two molse of Na+ ions are produced for every one mole of Na2CO3 . The total volume of the solution remains unchanged. If we have a 0.20 M solution of Na2CO3 , what will be the concentration of our two ions, Na+ and CO32-? This can easily be determined from the coefficients in our balanced equation: [Na+] = 2 × [Na2CO3] = 2 × 0.20 M = 0.40 M [CO32-] = 1 × [Na2CO3] = 1 × 0.20 M = 0.20 M

25. Practice problems 4.2.5 Assignment 4.2.5