1. Unit 12 Solutions And you

2. OBJECTIVE To gain informed insights into reactions that take place in aqueous environments you need to have a solid conceptual understanding about the ions that are present. Solutions are important to chemists because much of the chemistry in the lab, the environment, and within living organisms, occurs in solutions.

3. A brief review of terms: Mixture = two or more substances physically combined. Homogenous mixtures have similar properties throughout Heterogenous mixtures’ properties vary.

4. Solution =a homogenous mixture. Solution Solute Solvent Salt Water Carbonated Water Gold Jewelry Air Solute is present in ____________ amount. Solvent is present in ___________ amount salt water CO 2(g) water CuAu CO 2, O 2,...N2N2 smaller larger

5. Solution = Solute + Solvent in other words: The solute is dissolved in the solvent!

6. Concentration and Molarity (Sections 15.2, 15.4) For solutions, chemists use the idea of ___________ to describe the amount of solute in a given amount of solution. Concentrations are always RATIOS: A solution that is 1.0 molar (= 1.0M) contains 1.0 mol. of solute per liter of solution Molarity amount of solute amount of solution

7. Molarity Molarity” is a common method of expressing the concentration of a solute dissolved in a solution. Molarity, abbreviated by M, is defined as: M = #mol. solute L of solution

8. Sample Molarity Calculation A: Making solutions from solids What is the molarity (M) of a 250. mL solution containing 9.46 g of CsBr? M = moles solute = L solution moles solute = 9.46g (1 mol./212.81g) =.04445 mol.04445 mol.250 L =.178 Molar

9. Sample Molarity Problem Just For You Calculate the number of grams of NaCl needed to make 125 mL of a 3.78 M solution.

10. M = mol/L Therefore: #mol = M (L) = 3.78 (.125) =.4725 mol NaCl But ? grams????????? Moles to grams conversion:.4725 mol. NaCl ( 58.44g/1 mol NaCl) = 27.6 g NaCl

11. Please work on molarity questions at this point – thank you ever so much.

12. A brief comment on ions in solution When FeCl 3 dissolves it produces ions as follows: FeCl 3(s)  Fe +3 (aq) + 3 Cl - (aq) Therefore: 1 mol FeCl 3 yields 1 mol. Fe +3 and 3 moles Cl - ions. Therefore: A 1M FeCl 3 solution contains 1M Fe +3 ions and 3M Cl - ions.

13. What is the of ions in the following solutions?? 0.10 M Na 2 CO 3 ? 0.010 M Al 2 (SO 4 ) 3 ? 0.20 M Na + 0.10 M CO 3 -2 0.020 M Al +3 0.030 M SO 4 -2

14. Now for a more challenging question How many moles of Ag + ions are present in 25 mL of a 0.75 M AgNO 3 solution? Need a hint?????? 1) Calculate how many moles of AgNO 3 are present 2) Use the mole ratio of Ions to compound to determine the number of moles of ion are present.

15. Ag + ions are present in 25 mL of a 0.75M AgNO 3 solution? M = # mol/L so #mol = M(L) # mol AgNO 3 = 0.75 (.025L) =.019 mol AgNO 3 1 mol AgNO 3 / 1 mol Ag +.019 mol AgNO 3 /.019 mol Ag +

16. Dilutions (Section 15.5) One of the most common lab techniques is the dilution of a more concentrated solution to make a less concentrated solution. The basic idea is for the procedure is to remove a “small” portion of the concentrated solution (often referred to as the “stock”), place it into a volumetric flask, and then fill the flask to a mark with water. The addition of the water “dilutes” the concentrated stock to the new molarity.

17. To calculate the amount of stock to remove and the volume of your new solution, use the equation M C V C = M D V D where M = molarity, V = volume, C = concentrated, and D = dilute Therefore: Molarity (con.) X Volume (con.) = Molarity (dilute ) X Volume (dilute)

18. Calculate the volume of solution needed to prepare 500 mL of a 3.0 M HCl solution from the 12.0 M stock. Start by defining what you have been given: M C = larger molarity = 12.0V C = ? M D = smaller molarity = 3.0V D = 500. mL (doesn’t have to be in L – just consistent units) M C V C = M D V D V C = M D x V D = 3.0 x 500 = 125 mL M C 12.0

19. But how do you make the dilute solution? M C V C = M D V D V C = 125 ml Pipet 125 mL from stock (in 500 mL volumetric flask) into ~ 200 mL water (Add Acid) Fill to the 500 mL mark with water.

20. Note how the dilution factor of 4 (from 12M to 3M) is the same as the factor of water addition: 125 mL stock x 4 will give 500 mL solution total! NO WAY!!! WAY!!!!!

21. Calculate the preparation of 100. mL of 2.0 M HNO 3 solution from a 16 M HNO 3 stock. V C M C = V D M D so V C = M D V D / M C V C = 2.0 (100 mL) / 16 = 12.5 mL So you would add 12.5 mL of the concentrated acid to ~ 60 mL of distilled water (in a 100 mL volumetric flask) then top to the 100 mL mark on the flask with distilled water.

22. NOW Get to work on the problem set questions that focus on dilutions please.