Transcript ppt
Solution Concentration Remember: Periodic Trends - Electronegativity • Electronegativity = a number that describes the ability of an atom to attract electrons – More electronegative = stronger pull on electrons being shared – Less electronegative = weaker pull on electrons being shared Trend: Electronegativity Increasing Increasing Difference in Electronegativity If the electronegativity difference is: • less than 0.4 = bond is non-polar covalent • is between 0.5 and 1.6 = bond is polar covalent • is greater than 1.7 = bond is ionic Types of Bonds • Non-Polar Covalent = the attractive forces between two atoms that results when electrons are equally shared by the atoms with similar electronegativities • Polar Covalent = a covalent bond formed between atoms with significantly different electronegativities resulting in unequal sharing of electrons • Ionic = a bond formed due a large difference in electronegativity between atoms resulting in a complete transfer of electrons Comparison… Non-Polar versus Polar Covalent Solution and Concentration How can we express the amount of solute to solvent in a solution? • This is called the concentration •a measure of the number of particles of the solute in the solvent •Concentration = Dilute Solutions Small quantity of solute per unit volume of solution. Concentrated Solutions Large quantity of solute per unit volume of solution. Ways to express concentration 3 ways of expressing concentration Percentages PPM or PPB concentrations molar concentrations and mass concentrations Percentage Concentration m ass solute( g ) m ass solution( g ) •% (w/w) = x 100 • W is weight same as mass •% (w/v) = m ass solute( g ) volum esolution(m L) x 100 •% (v/v) = volum esolute(m L) volum esolution(m L) x 100 Everyday Examples • Vinegar is 5% v/v acetic acid – 5 mL of CH3COOH in 100 mL vinegar • Hydrogen peroxide is 3% m/v – 3 g of H2O2 per 100 mL of solution • Yogurt is 2% m/m – 2 g of milk fat per 100 g yogurt Examples 1. % V/V = 4.1 L / 55 L = 7.5% V/V 2. % W/V = 16 g / 50 mL = 32% W/V 3. % W/W = 1.7 g / 35.0 g = 4.9% W/W More practice 1. What is the % W/W of copper in an alloy when 10 g of Cu is mixed with 250 g of Zn? 2. What is approximate % V/V if 30 mL of pure ethanol is added to 250 mL of water? 3. What is the % W/W if 8.0 g copper is added to enough zinc to produce 100 g of an alloy % Concentration: % Mass Example 3.5 g of CoCl2 is dissolved in 100mL solution. Assuming the density of the solution is 1.0 g/mL, what is concentration of the solution in % mass? %m = 3.5 g CoCl2 100g H2O = 3.5% (m/m) Low concentrations • For very dilute solutions, weight/weight (w/w) and weight/volume (w/v)concentrations are sometimes expressed in parts per million (ppm). – Example: • Toxic substances found in the environment • Chlorine in a swimming pool • Metals in drinking water • Parts per million also can be expressed – as milligrams per liter (mg/L) (w/v) – As milligram per kilogram (mg/Kg) (w/w) PPM learning check 1. Question: A solution has a concentration of 1.25g/L. What is its concentration in ppm? • • Convert the mass in grams to a mass in milligrams: 1.25g = 1.25 x 1000mg = 1250mg Re-write the concentration in mg/L = 1250mg/L = 1250ppm 2. Question: 150mL of an aqueous sodium chloride solution contains 0.0045g NaCl. Calculate the concentration of NaCl in parts per million (ppm). ppm = mass solute (mg) ÷ volume solution (L) mass NaCl = 0.0045g = 0.0045 x 1000mg = 4.5mg volume solution = 150mL = 150 ÷ 1000 = 0.150L concentration of NaCl = 4.5mg ÷ 0.150L = 30mg/L = 30ppm Molar concentration • is able to compare the amount of solute (moles) dissolved in a certain volume of solution. • Molar concentration (mol/L) or also called Molarity (M) = Moles of solute (n)/Volume of solution (V MUST be in litres) • is the number of moles of solute in one litre of a solution. We use "M" to denote molar concentration and it has the units of"mol/L". • C=n/V • n=CxV Concentration: Molarity Example If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO4? As is almost always the case, the first step is to convert the mass of material to moles. MM = 158.0 g/mol(39+55+(4X16) n= m/MM 0.435 g KMnO4 / 158.0 g/ mol = 0.00275 mol KMnO4 Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L . Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 Mol/L = M 0.250 L solution Molarity learning check 1. How many moles of H2SO4 are there in 250mL of a 0.8M sulphuric acid solution? 2. If 20g of NaOH is dissolved in sufficient water to produce 500 mL of solution, calculate the molar concentration in molarity. Questions 1. What is the molarity of the solution formed by dissolving 80 g of sodium hydroxide (NaOH) in 500 mLs of water? (Ar's: Na=23, O=16, H=1) Ans = 4 M 2. What is the molarity of the solution formed by dissolving 9.8 g of sulphuric acid (H2SO4) in 1000 cm3 of water? (Ar's: H=1, S=32, O=16) Ans = 0.1 M 3. What mass (g) of hydrogen chloride (HCl) is needed to make up 500cm3 of a solution of concentration 0.2mol/L? (Ar's: H=1, Cl=35.5) Ans = 3.65g Making Solutions… Solution Preparation • Standard Solution = a solution for which the precise concentration is known • Used in research laboratories and industrial processes • Used in chemical analysis and precise control of chemical reactions Preparing a Standard Solution Equipment: – Electronic balance precise measurement of solids – Pipets (pipettes) precise measurement of liquids – Volumetric flask calibrated to contain a precise volume at a particular temperature, used for precise dilutions and preparation of standard solutions Accurate Reading of a Volumetric Flask Bend down to see the meniscus Pipets and Bulb/Pump to transfer small quantities of liquid Volumetric Pipet Serological (Blow Out) Pipet Mohr (Graduated) Pipet Automatic Dispensers Micropipets – Dispense µL (microliters) Preparing a Solution by Dilution • Dilution = the process of decreasing the concentration of a solution, usually by adding more solvent • Stock Solution = a solution that is in stock or on the shelf (i.e., available); usually a concentrated solution Preparation of a Solution of Known Concentration by Diluting a Stock Solution Preparation of a Solution of Known Concentration Using a Solid Solute Calculating the New Concentration of the Diluted Solution • • • • C1 x V1 = C2 x V2 C1 = initial concentration V1 = initial volume C2 = final concentration V2 = final volume Sample Problem • Water is added to 0.200L of 2.40mol/L NH3(aq) cleaning solution, until the final volume is 1.000L. Find the molar concentration of the final, diluted solution.