1. Dilution Adding water to a solution

2. Dilution The number of moles of solute doesn’t change if you add more solvent! The # moles before = the # moles after M 1 x V 1 = M 2 x V 2 M 1 and V 1 are the starting concentration and volume. M 2 and V 2 are the final concentration and volume. Stock solutions are pre-made to known Molarity

3. Ex:1 - 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make? M 1 V 1 = M 2 V 2 (1.50 mol/L)(53.4 mL) = (0.800 mol/L) (x) (0.800 mol/L) (0.800 mol/L) 100. mL = x

4. Ex:2 - 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results? M 1 V 1 = M 2 V 2 (2.500 mol/L)(100.0 mL)=(0.5500mol/L)(x) 454.55 mL = x 454.6 mL = x

5. Ex: 3 - 53.4 mL of a 1.50 M (mol/l) solution of NaCl is on hand, but you need some 0.500 M solution. How many ml of 0.500 M can you make? M 1 V 2 = M 2 V 2 (1.50 mol/L) (53.4 ml) = (0.500 mol/L) (x) 160. ml = x

6. C b V b = C a V b This is the same as the dilution equation, concentration before and volume before is equal to the concentration after and the volume after.

7. Practice a. 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? b. You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make? c. Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?

8. Questions P 942#365,370,371,372 Answers: 365: a) 0.100M, b) 38ml, c)0.986M, d)1.7L, e) 50.ml 366: 50.6 ml 368: 0.44 ml 370: 6.35 ml 371: 348 ml 372: 0.558 M