1. How many grams of sodium sulfate would you need to make 100. mL of a 3.5 M solution?

2. Making Solutions - Dilutions

3. Adding water to a solution of high concentration to create a larger volume of a less concentrated solution. The # of moles of substance stays the same.

4. Why are dilutions used? If it is impossible to weigh out the amount of solid for a low concentration. If you need a concentration but you don’t have solid, you have higher concentrated solution. (ie HCl)

5. M1V1 = M2V2 (or C1V1 = C2V2) This can only be used for dilutions!! NEVER NEVER NEVER use it for a stoichiometry or titration problem.

6. If you use 50.0 mL of a 6.00 M NaOH solution and dilute it to 150. mL, what is your new concentration? M1V1 = M2V2 (6.00 M) (50.0 mL) = (M2) ( 150. mL) M2 = (6.00 M) (50.0 mL) = 2.00 M NaOH (150 mL)

7. If you use 100.0 mL of a 3.00 M HCl solution and dilute it to 0.750 M, what is your new volume? M1V1 = M2V2 (3.00 M) (100.0 mL) = (0.750 M) ( V2) V2 = (3.00 M) (100.0 mL) = 400. mL HCl (0.750 M)

8. If you have 250. mL of 6.00 M HCl, how much 1.00 M HCl can you make? M1V1 = M2V2 (6.00 M) (250.0 mL) = (1.00 M) ( V2) V2 = (6.00 M) (250.0 mL) = 1500. mL HCl = 1.50 L (1.00 M)

9. How much volume of a 0.0500 M solution of copper sulfate is needed to create 1000. mL of a 0.0000500 M solution of CuSO4? M1V1 = M2V2 (0.0500 M) (V1) = (0.0000500 M) ( 1000 mL) V1 = (0.00005000 M) (1000.0 mL) = 1 mL CuSO4 (0.0500 M)

10. If you have 3.00 M NaOH and you want to make 2.00 L of 1.25 M NaOH, how much stock solution do you need to mix with how much water? M1V1 = M2V2 (3.00 M) (V1) = (1.25 M) (2.00 L) V2 = (1.25 M) (2.00 L) = 0.833 L NaOH = 833 mL (3.00 M) This is the amount of stock solution (concentrated NaOH) I would need. To get the amount of water I subtract this from 2000 mL. I would need 1167 mL water.