1. Chapter Four Reactions in Aqueous Solutions

2. Chapter Four / Reactions in Aqueous Solutions Solutions and concentrations Solution is a homogenous mixture of two or more substances. When water is the solvent, we called the solution aqueous solution. Concentration of a solution is the amount of solute present in a given amount of solvent. The concentration of a solution can be expressed in many different ways. MOLARITY (M): is the number of moles of solute per liter of solution. Unit of molarity is mol/L

3. Chapter Four / Reactions in Aqueous Solutions Solutions and concentrations Steps to prepare a solution of known molarity :

4. Chapter Four / Reactions in Aqueous Solutions Solutions and concentrations Example1: How many grams of potassium dichromate (K 2 Cr 2 O 7 ) are required to prepare a 250 ml solution whose concentration is 2.16 M.? M =2.16, V =250 mL = 250/1000= 0.25 L n = M X V = 2.16 X 0.25 = 0.54 mol n = mass / molar mass Molar mass of K 2 Cr 2 O 7 = 294.2 g/mol Mass = n x molar mass = 0.54 x 294.2 = 158.9 g

5. Chapter Four / Reactions in Aqueous Solutions Solutions and concentrations Example 2: In a biochemical assay, a chemist needs to add 3.81 g of glucose (C 6 H 12 O 6 ) to a reaction mixture. Calculate the volume in milliteres of a 2.53 M glucose solution he should use for the addition. M = n/V Molar mass of C 6 H 12 O 6 =180 g/mol n = mass/molar mass = 3.81 / 180 = 0.021 mol M = n/V V = n/M = 0.021 / 2.53 = 8.30 X 10 -3 L = 8.36 X10 -3 X 10 3 L = 8.30 mL

6. Chapter Four / Reactions in Aqueous Solutions Solutions and concentrations FOR IONIC COMPOUNDS NaCl Na +Cl 1 mol NaCl, 1 mole Na +, 1mole Cl -1 1 M NaCl, 1 M Na +, 1 M Cl -1 Ba (NO 3 ) 2 Ba +2 + 2NO 3 - 1 mole Ba (NO 3 ) 2, 1 mole Ba +2, 2 mole NO 3 - 1M Ba (NO 3 ) 2, 1 M Ba +2, 2 M NO 3 -

7. Chapter Four / Reactions in Aqueous Solutions Dilution Add Solvent

8. Chapter Four / Reactions in Aqueous Solutions Dilution Dilution : is the procedure for preparing a less concentrated solution from a more concentrated one. M 1 V 1 = M 2 V 2 BEFORE AFTER Example 1 : How you would prepare 5.00 x 10 2 mL of a 1.75 M H 2 SO 4 solution, starting with an 8.61 M stock solution of H 2 SO 4 ? M 1 = 8.61, V 1 = ?, M 2 = 1.75, V 2 = 5.00 X10 2 M 1 V 1 = M 2 V 2 8.61 X V 1 = 1.75 X 5.00 X 10 2 V 1 = 1.75 X 5.00 X 10 2 / 8.61 = 101.6 ml.

9. Chapter Four / Reactions in Aqueous Solutions Dilution Example 2: How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M 1 = 4, V 1 = ?, M 2 = 0,2, V 2 = 60 M 1 V 1 = M 2 V 2 4 X V 1 = 0.2 x 60 V 1 = 0.2 X 60 / 4 = 3 ml.