1. 8.4 Preparing Solutions

2. Review of Concepts Here mole-y mole-y!!

3. “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l)

4. Concentration (Molarity)

5. Concentration: Molarity Example If 0.435 g of KMnO 4 is dissolved in enough water to give 250 mL of solution, what is the molarity of KMnO 4 ? Step #2: Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters Step #1: Convert Mass to Moles n= m/MW n= m = 0.435 g KMnO 4 = 2.75 x 10-3 mol KMnO 4 MW 158.0 g/mol KMnO 4 Molarity KMnO 4 = 2.75 x 10-3 mol KMnO 4 = 1.10 x 10-2 M 0.250 L solution

8. Dilution When chemists purchase solutions, they generally purchase “stock solutions” which are extremely concentrated solutions This way a chemist can dilute the strong solution to any concentration that they wish. This stops the chemist from having to buy several concentrations

9. C 1 V 1 = C 2 V 2 Dilution Equation C 1 = Initial Concentration (mol/L) V 1 = Initial Volume (in mL or L) C 2 = Final Concentration (mol/L) V 2 = Final Volume (in mL or L)

10. Dilutions What is really happening? The amount of solute remains constant before and after the dilution just the volume is changing!!!

11. A volumetric flask A graduated pipette What you will need…

12. Sample Problem #1 Suppose we want to make 250 mL of a 0.10 M solution of CuSO4 and we have a stock solution of 1.0 M CuSO4. What volume would we need to prepare the solution? C 1 V 1 = C 2 V 2 (1.0 M) V 1 = (0.10M)(250mL) V 1 = 25 mL

13. How will it be done?

14. Let’s Get the Job Done? You know you want 250 mL So, go get a 250 mL volumetric flask You know you will need 25 mL of the 1.0 M solution to dilute it to 0.1 M Pipette 25 mL of the 1.0 M stock solution and put it in the volumetric flask. Dilute, by adding H 2 O to 250 mL mark And-shaka…shaka…shaka!!! You’re DONE!

15. Sample Problem #2 What is the final molar concentration if 360.0 mL of a 0.825 mol/L sodium dichromate (Na 2 Cr 2 O 7 ) is diluted to a new volume of 645 mL ?

16. C 1 V 1 = C 2 V 2 (0.825M)(360 mL)= C 2 (645mL) C 2 = 0.460 M Therefore the new concentration of Na 2 Cr 2 O 7 would be 0.460 M of diluted to 645 mL.

17. Sample Problem #3 You need to prepare a 4.0 M solution of NaOH for your lab experiment. Miss Botticelli provides you with the solute in solid form and a 1.0 L volumetric flask. A)How much solid will you need? B)You now need to dilute it to 0.25 M and provided with a 500 mL flask. What volume will you need to do this dilution?

18. A)Determine the number of moles C=n/V n (NaOH) = C x V = (4.0 M) (1L) = 4.0 mol of NaOH present m= nx MW = (4.0 mol of NaOH)(40.0g/mol) = 160 g of NaOH are needed!

19. B) Use the dilution formula C 1 V 1 = C 2 V 2 (4.0 M) V 1 = (0.25M)(500mL) V 1 = 31.25mL You will need to pipette 31 mL of the 4.0 M solution you made to dilute it in the 500 mL v flask to get a new concentration of 0.25 M