SOLUTIONS. Definitions. Solvation. Hydration. Energy changes involved in solutions

Transcription

1 1 SOLUTIONS Definitions Solvation Hydration Energy changes involved in solutions

2 2 Solubility Definition Unsaturated Saturated supersaturated Factors affecting solubility Interactions of solute with solvent

3 3 Effect of pressure -Henry s law Effect of temperature Colligative properties of solutions Physical properties of solutions differ from that of the pure solvent. Physical properties that depend on the quantity and not on the kind of solute are called colligative properties. e.g. Lowering of vapor pressure (Raoults law)

4 4 Elevation of boiling point Depression of freezing point Osmotic pressure

5 5 Solution Stoichiometry A solution is a homogenous mixture of two or more substances. It is made up of a solute (solid) dissolved in a solvent (liquid). When water is the solvent, the solution is named aqueous solutions. Molarity is the amount of moles of solute dissolved in a liter of solution. To prepare a solution of known molarity from a solute e.g. 1M sodium hydroxide How many moles of solute is needed? Molarity = no of moles of solute/vol of solution Rewrite eqn No of moles of solute = molarity x vol of solution = =1mol/L x 1L = 1 mole 1M = 1 mole of solute in 1 L of solution For NaOH molar mass is 40g = 1 mole So dissolving 40g NaOH in 1 L of solution gives us 1M 2) To prepare a 0.25 M HCl How many moles of solute is needed? Molarity = no of moles of solute/vol of solution Rewrite eqn No of moles of solute = molarity x vol of solution =0.25mol/L x 1L =0.25 mole For HCl molar mass is 36.5g = 1 mole 0.25 mol HCl x 36.5g HCl/1mol HCl = 9.125g HCl

6 6 So dissolving 9.125g HCl in 1 L of solution gives us 0.25 M 3. Calculate the molarity of a solution that contains mol NH 4 Cl in 400ml of solution. Molarity = no of moles of solute/vol of solution in L =0.0345mol/0.4L = 0.09M Diluting a more concentrated solution Use the equation MaVa =MbVb Ma = molarity of original solution Va =Volume to be taken from original solution Mb= molarity of solution to be prepared Vb = volume of solution to be prepared 1. To prepare a 0.1M solution in a 100mL flask from a 5M solution MaVa =MbVb Va= MbVb /Ma = 0.1M x 100ml/5M =2ml Take 2 ml from original solution and dilute to 100mls. 2. To prepare a 1M solution in a 100mL flask from a 10M solution Dilute it 10 times. If final vol of solution to be prepared is 100

7 7 Take 10ml from original solution and dilute to 100ml (10x) 3. To prepare a 1M solution in a 50mL flask from a 10M solution MaVa =MbVb Va= MbVb /Ma = 1M x 50ml/10M =5ml Take 5ml and dilute to 50ml (10x). 4. Calculate the molarity of a solution prepared by diluting 125 ml of 0.400M K 2 Cr 2 O 7 with 875 ml of water. Conversions 1. Calculate the number of moles of solutes in 25 ml of 3M H 2 SO 4 2. What volume of 0.256M KCl will contain 20g KCl? 3. Calculate the grams of solute in 150L of 1.0M NaCl. 4. How many grams of calcium nitrate are required to make 0.5L of 0.1M Ca(NO 3 ) 2? Mass percentage Mass % of component = mass of component in solution/total mass of solution x 100 Parts per million (ppm)= mass of component in solution/total mass of solution x 10 6 Parts per billion (ppb) = mass of component in solution/total mass of solution x 10 9 Mole fraction of component, X = moles of component/total moles of all component

8 8 Molality, m = moles of solute/kg of solvent Molarity depends on volume while molality depends on the mass of solvent. Titrations Titrations are used to determine the concentration of a particular solute in a solution. This involves combining a solution of known concentration with that of an unknown solution e.g in a neutralization reaction between an acid and a base. At the end point equivalent quantities are present stoichiometrically. In order to identify or determine the end point of titration, an indicator is added before titrating. Indicators have different colors in acidic, basic and neutral mediums. The color change from acidic to neutral to basic indicates that all the acid has been consumed and is the end point. Hence care has to be taken to ensure that the last drop is not in excess. Example 1. If 45.7ml of 0.500M H 2 SO 4 is required to neutralize a 20.0 ml sample of NaOH solution. What is the concentration of the NaOH solution? Use the equation CaVa =a/b CbVb Ca = concentration of acid Va =Volume of acid Cb= concentration of base Vb = volume of base a/b = mole ratio of acid to base=1/2

9 9 H 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O Rearrange equation Cb= 2 x CaVa/Vb Cb= 2 x 0.5M x 45.7ml / 20ml Cb= 2.28M 2. How many grams of Ca (OH) 2 are needed to neutralize 25ml of 0.100M HNO 3 Molarity = no of moles/vol in L Rearrange No of moles = molarity x vol in L = 0.1mol/L x 0.025L = mols 2HNO 3 (aq) + Ca(OH) 2 (s) mols x 1mol Ca(OH) 2 / 2mols HNO 3 x 74.1g Ca(OH) 2 /1 mol Ca(OH) 2 =0.0926g Ca(OH) 2 3. What is the molarity of a NaOH solution if 48 ml is needed to neutralize 35 ml of M H 2 SO 4? Cb= 2 x CaVa/Vb = 2 x 0.144M x 35ml/ 48ml