SOLUTIONS Definitions A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance • homogeneous mixture -- uniform appearance -- similar properties throughout mixture • The solvent is the dissolving agent -- i.e., the most abundant component of the solution • The solute is the component that is dissolved -- i.e., the least abundant component of the solution 1 2 Concentration of solutions Concentration units based on the number of moles of solute concentration -- the amount of solute dissolved in a given quantity of solvent or solution molarity -- number of moles of solute per liter of solution There are many di ff erent types of concentration units: molarity we will focus on this moles of solute Molarity = mass % liters solution volume % mass/volume % mol ( ) parts per million (ppm) molarity has units of moles per liter L parts per billion (ppb) mole fraction which can be abbreviated as M molality (not to be confused with molarity ) 3 4
Example: Preparation of a 1 molar solution of NaCl Example: Preparation of a 1 molar solution of NaCl Note: Dissolving 1 mole of solute to make 1 liter of solution is not the only 1 mole NaCl = water way to prepare a solution with a concentration of 1 M ( i.e., 1 mol / L ) 58.44 g NaCl 1.000 mol 0.250 mole NaCl 0.2500 mol Molarity = Molarity = = 14.61 g NaCl 1.000 L water 0.2500 L fill 1000.0-ml line = 1.000 M = 1.000 M flask fill line 250.0-ml flask Add 1 mole of NaCl Add water to Add water until the Add 0.250 mole of Add water to Add water until the to an empty 1-liter completely fill mark is reached NaCl to an empty completely fill mark is reached volumetric flask dissolve NaCl and mix thoroughly 250-ml volumetric flask dissolve NaCl and mix thoroughly 5 6 What is the molarity of a solution made by dissolving 2.00 g of Example: Preparation of a 0.5 molar solution of NaCl potassium chlorate in enough water to make 150. ml of solution? 0.5 mole NaCl = 29.22 g NaCl Step 1: Start with the definition of molarity: 0.5000 mol moles of solute Molarity = fill Molarity = 1000.0-ml 1.000 L line liters solution flask = 0.5000 M Step 2: Determine the number of moles of solute Molar mass of KClO 3 = 39.10 + 35.45 + 3(16.00) = 122.6 g / mol 0.250 mole NaCl 0.2500 mol = 14.61 g NaCl Molarity = 1 mole KClO 3 0.5000 L fill 2.00 g KClO 3 = 0.0163 moles KClO 3 line 122.6 g KClO 3 = 0.5000 M 500.0-ml flask 7 8
What is the molarity of a solution made by dissolving 2.00 g of What is the molarity of a solution made by dissolving 2.00 g of potassium chlorate in enough water to make 150. ml of solution? potassium chlorate in enough water to make 150. ml of solution? moles of solute moles of solute Molarity = Molarity = liters solution liters solution Step 4: Plug values into molarity equation Step 3: Determine the number of liters of solution 0.0163 moles KClO 3 Molarity = 1 liter 150. ml 0.150 L = 0.150 L 1000 ml Molarity = 0.109 moles KClO 3 / L = 0.109 M KClO 3 9 10 How many grams of potassium hydroxide are required to How many grams of potassium hydroxide are required to prepare 500. ml of 0.450 M KOH solution? prepare 500. ml of 0.450 M KOH solution? Step 1: Start with the definition of molarity: moles of solute Molarity = moles of solute liters solution Molarity = liters solution Step 3: Plug known values into molarity equation and solve for unknown (moles of solute) Step 2: Determine the number of liters of solution x moles of KOH = 0.450 M KOH 0.500 L 1 liter 500. ml = 0.500 L x moles of KOH moles KOH 1000 ml (0.500 L) 0.450 (0.500 L) = L 0.500 L 0.225 moles KOH = x 11 12
Calculate the number of moles of nitric acid in 325 ml of How many grams of potassium hydroxide are required to 16 M HNO 3 prepare 500. ml of 0.450 M KOH solution? Step 1: Start with the definition of molarity: moles of solute Molarity = moles of solute liters solution Molarity = liters solution Step 4: Convert moles KOH to grams KOH Step 2: Plug known values into molarity equation and solve for unknown (moles of solute) Molar mass of KOH = 39.10 + 16.00 + 1.008 = 56.11 g / mol x moles of HNO 3 16 M HNO 3 = 0.325 L 56.11 g KOH 0.225 moles KOH = 12.6 g KOH moles HNO 3 x moles of HNO 3 1 mole KOH (0.325 L) 16 = (0.325 L) 0.325 L L 5.2 moles HNO 3 = x 13 14 Concentrations of ions in aqueous solutions Concentrations of ions in aqueous solutions When an ionic compound dissolves, the concentrations of the When an ionic compound dissolves, the concentrations of the individual ions depend on the chemical formula of the compound individual ions depend on the chemical formula of the compound Example: What are the concentrations of Na + and Cl - ions in Example: What are the concentrations of Mg 2+ and Cl - ions a 1.0 M aqueous solution of NaCl? in a 1.0 M aqueous solution of MgCl 2 ? 1.0 mol Na + 1.0 mol Mg 2+ = 1.0 M Na + = 1.0 M Mg 2+ 1.0 mol NaCl 1.0 mol MgCl 2 1.0 M NaCl = L solution 1.0 M MgCl 2 = L solution L solution L solution 1.0 mol Cl - 2.0 mol Cl - = 1.0 M Cl - = 2.0 M Cl - Each NaCl molecular unit produces Each MgCl 2 molecular unit produces L solution L solution 1 Na + ion and 1 Cl – ion in solution 1 Mg 2+ ion and 2 Cl – ions in solution Na + Cl - Na + Cl - Cl - Cl - Cl - Cl - Mg 2+ Mg 2+ 15 16
A solution is made by dissolving 9.82 g of copper (II) chloride (CuCl 2 ) in enough A solution is made by dissolving 9.82 g of copper (II) chloride (CuCl 2 ) in enough water to make 600. mL of solution. What is the molarity of Cl – ions in solution? water to make 600. mL of solution. What is the molarity of Cl – ions in solution? Step 1: Determine the number of moles of solute Step 3: Determine the ion to solute ratio CuCl 2 dissociates to give one Cu 2+ ion and two Cl – ions Molar mass of CuCl 2 = 63.55 + 2(35.45) = 134.45 g / mol 2 moles Cl – ions CuCl 2 Cu 2+ + 2 Cl – 1 mole CuCl 2 9.82 g CuCl 2 1 mole CuCl 2 = 0.0730 moles CuCl 2 134.45 g CuCl 2 Step 4: Determine molarity of the ion Step 2: Determine molarity of solute 2 moles Cl – ions 0.122 M CuCl 2 = 0.244 M Cl – ions 1 mole CuCl 2 0.0730 moles CuCl 2 Molarity = = 0.122 M CuCl 2 0.600 L 17 18 Dilutions Dilution: Reducing the concentration of a solution by adding more solvent to the solution NO 3- • More solvent is added: Na + -- volume of the solution increases • No additional solute is added Moles = 1.0 mol -- number of moles of solute stays the same Volume = 1.0 L 1.0 mol Net result: The molarity of the solution decreases Molarity = 1.0 L (unchanged) moles of solute = 1.0 M Molarity = NaNO 3 solution liters solution 19 20
Calculate the molarity of a solution prepared by diluting 125 ml of 0.400 M HCl to a final solution volume of 1.00 L. For any dilution problem, remember that the number of moles of NO 3- solute remains the same : Na + moles of solute (before) = moles of solute (after) Moles = 1.0 mol Based on the definition of molarity, this can be expressed as: Volume = 2.0 L M 1 V 1 = M 2 V 2 1.0 mol Molarity = 2.0 L Where M 1 is the molarity of the original solution • Solution volume is doubled = 0.5 M V 1 is the volume of the original solution • Moles of solute remain the same M 2 is the molarity of the diluted solution • Solution concentration is halved V 2 is the volume of the diluted solution 21 22 Calculate the molarity of a solution prepared by diluting Solubility 125 ml of 0.400 M HCl to a final solution volume of 1.00 L. Solubility refers to the ability of a compound to dissolve in a solvent M 1 V 1 = M 2 V 2 -- different compounds will dissolve to different extents in a given solvent When water is the solvent: M 2 = ? M 1 = 0.400 M V 2 = 1.00 L If a solute dissolves readily in water, it is said to be soluble in water V 1 = 125 ml = 0.125 L Examples: solute dissolves in NaCl water to form solution AgNO 3 (0.400 M ) (0.125 L) = (M 2 ) (1.00 L) (NH 4 ) 2 CO 3 1.00 L 1.00 L If a solute will not dissolve in water, it is said to be insoluble in water Examples: 0.0500 M = M 2 solute does not Fe(OH) 3 dissolve in water PbCl 2 CaCO 3 23 24
Solubility rules Solubility rules AN IONIC COMPOUND AN IONIC COMPOUND IS IS SOLUBLE IN WATER NOT SOLUBLE IN WATER EXCEPTIONS EXCEPTIONS IF IT CONTAINS THE IF IT CONTAINS THE FOLLOWING IONS: FOLLOWING IONS: Carbonate (CO 32- ) Ammonium ion (NH 4+ ) none Compounds containing Li + , Na + , K + , NH 4+ (soluble) Phosphate (PO 43- ) Alkali metal (Group IA) ions none Compounds containing (Li + , Na + , K + ) Li + , Na + , K + , NH 4+ (soluble) Hydroxide (OH - ) Nitrate (NO 3- ) none Compounds containing Acetate (C 2 H 3 O 2- ) Ca 2+ , Ba 2+ , Sr 2+ (slightly soluble) Compounds containing Halides (Cl - , Br - , I - ) Compounds containing Ag + , Pb 2+ , Hg 22+ Li + , Na + , K + , NH 4+ (soluble) Sulfide (S 2- ) Compounds containing Compounds containing Sulfate (SO 42- ) Ca 2+ , Ba 2+ , Sr 2+ (soluble) Ag + , Pb 2+ , Ca 2+ , Sr 2+ , Ba 2+ 25 26 Double-displacement reactions Sample problems double-displacement reaction -- two ionic compounds exchange Are the following compounds soluble or insoluble in water? partners ( i.e., cations and anions) to form two different compounds soluble NaCl General form: AB + CD AD + BC (NH 4 ) 3 PO 4 soluble cations two new CaCO 3 insoluble switch compounds places are formed MgSO 4 soluble Precipitation reactions are a type of double-displacement reaction BaSO 4 insoluble 27 28
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