Chapter Four Reactions in Aqueous Solutions Chapter Four

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Chapter Four Reactions in Aqueous Solutions

Chapter Four Reactions in Aqueous Solutions

Chapter Four / Reactions in Aqueous Solutions and concentrations • Solution is a homogenous

Chapter Four / Reactions in Aqueous 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

Chapter Four / Reactions in Aqueous Solutions and concentrations • Steps to prepare a

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

Chapter Four / Reactions in Aqueous Solutions and concentrations Example 1: How many grams

Chapter Four / Reactions in Aqueous Solutions and concentrations Example 1: 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 m. L = 250/1000= 0. 25 L n=MXV = 2. 16 X 0. 25 = 0. 54 mol

Chapter Four / Reactions in Aqueous Solutions and concentrations n = mass / molar

Chapter Four / Reactions in Aqueous Solutions and concentrations n = mass / molar mass Molar mass of K 2 Cr 2 O 7 = 2 x 39. 1 + 2 x 52 + 7 x 16=294. 2 g/mol Mass = n x molar mass = 0. 54 x 294. 2 = 158. 9 g Mass (g) Mole (mol) Molar mass

Chapter Four / Reactions in Aqueous Solutions and concentrations Example 2: In a biochemical

Chapter Four / Reactions in Aqueous 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. 2 g/mol n = mass/molar mass = 3. 81 / 180. 2 = 0. 021 mol M = n/V V = n/M = 0. 021 / 2. 53 = 8. 30 X 10 -3 L = 8. 36 X 10 -3 X 103 L = 8. 30 m. L

Chapter Four / Reactions in Aqueous Solutions and concentrations FOR IONIC COMPOUNDS Na. Cl

Chapter Four / Reactions in Aqueous Solutions and concentrations FOR IONIC COMPOUNDS Na. Cl Na +Cl 1 mol Na. Cl , 1 mole Na+ , 1 mole Cl-1 1 M Na. Cl , 1 M Na+ , 1 M Cl-1 Ba (NO 3)2 Ba+2 + 2 NO 31 mole Ba (NO 3)2 , 1 mole Ba+2 , 2 mole NO 31 M Ba (NO 3)2 , 1 M Ba+2, 2 M 2 NO 3 -

Chapter Four / Reactions in Aqueous Solutions Dilution Add Solvent

Chapter Four / Reactions in Aqueous Solutions Dilution Add Solvent

Chapter Four / Reactions in Aqueous Solutions Dilution : is the procedure for preparing

Chapter Four / Reactions in Aqueous Solutions 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 102 m. L 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 X 102 M 1 V 1 = M 2 V 2 8. 61 X V 1 = 1. 75 X 5. 00 X 102 / 8. 61 = 101. 6 ml.

Chapter Four / Reactions in Aqueous Solutions Dilution Example 2: How would you prepare

Chapter Four / Reactions in Aqueous Solutions Dilution Example 2: How would you prepare 60. 0 m. L 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.