2 Agenda Day 66 – Concentration Lesson: PPT, Handouts: 1. Concentration& Dilution Handout. 2. Concentration of Solutions WorksheetText: 1. P Concentration ( %, ppm)HW: 1. Worksheets, P. 400 # 1-4, P.402 # %, ppm
3 Solution Concentration Concentration = quantity of solutequantity of solution (not solvent)There are 3 basic ways to express concentration: 1) percentages, 2) very low concentrations, and 3) molar concentrations% concentration can be in V/V, W/W, or W/VLike most %s, V/V and W/W need to have the same units on top and bottom.W/V is sort of in the same units; V is mostly water and water’s density is 1 g/mL or 1 kg/L3 g H2O2/100 mL solution 3 g H2O2/100 g solution
4 Solution and Concentration Other ways of expressing concentrationMolarity(M): moles solute / Liter solutionMass percent: (mass solute / mass of solution) * 100Molality* (m) - moles solute / Kg solventMole Fraction(A) - moles solute / total moles solution* Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.
6 Units of Concentrations amount of solute per amount of solvent or solutiong soluteg solutionx 100g soluteg solute + g solventx 100=Percent (by mass) =moles of solutevolume in liters of solutionMolarity (M) =moles = M x VL
7 Solution Concentration Expressing concentrations in parts per million (ppm) requiresthe unit on top to be 1,000,000 times smaller than the uniton the bottomE.g. 1 mg/kg or g/gNotice that any units expressed as a volume must be referring to a water solution (1L = 1kg)- density of waterFor parts per billion (ppb), the top unit would have to be 1,000,000,000 times smaller1 ppm = 1 g/106 mL= 1 g/ 1000 L= 1 mg/L= 1 mg/kg= 1 g/g1g = 1000mg1000mg/1000L1mg = 1000 g1000g/1000g
8 MolarityMolar concentration is the most commonly used in chemistry. amount of solute (in moles) Molar concentration = volume of solution (in litres) UNITS: ( mol/L) or (mol .L-1) or M
9 Concentration: Percentage Examples 10 g / 260 g = 3.8 % W/W30 mL / 280 mL = 11% V/V (in reality may be off)8.0 g / 100 g = 8% W/WWhat is the % W/W of copper in an alloy when 10 g of Cu is mixed with 250 g of Zn?What is approximate % V/V if 30 mL of pure ethanol is added to 250 mL of water?What is the % W/W if 8.0 g copper is added to enough zinc to produce 100 g of an alloy?
10 Concentration: Molarity Example If 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.0.435 g KMnO4 x 1 mol KMnO4 = mol KMnO4158.0 g KMnO4Now 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 L .Molarity KMnO4 = mol KMnO4 = M0.250 L solution
11 Step 1: Calculate moles of NiCl2•6H2O PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity.Step 1: Calculate moles of NiCl2•6H2OStep 2: Calculate Molarity[NiCl2•6 H2O ] = M
12 Concentration: Mixed Example A solution of H2O2 is 3% (w/v).Calculate the mass of H2O2 in mL of solution.Calculate the mass of H2O2in 1L of solution.Calculate the number of moles of H2O2 in 1L of solution.State the molar concentration of the solutionCalculate the ppm of H2O2 .
13 Concentration: Mixed Example Answers a) b) c) d) e)
16 1. 0 L of water was used to make 1. 0 L of solution 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.
17 What are molar solutions? A molar solution is one that expresses “concentration” in moles per volume. Usually the units are in mol/L mol/L can be abbreviated as M Molar solutions are prepared using: a balance to weigh moles (as grams) a volumetric flask to measure litres L refers to entire volume, not water! Because the units are mol/L, we can use the equation M = n/L Alternatively, we can use the factor label method.
18 Calculations with molar solutions Q: How many moles of NaCl are required to make 7.5 L of a 0.10 M solution?M=n/L, n = 0.10 M x 7.5 L = 0.75 mol# mol NaCl =7.5 Lx 0.10 mol NaCl1 L= 0.75 molBut in the lab we weigh grams not moles, so …Q: How many grams of NaCl are required to make 7.5 L of a 0.10 M solution?# g NaCl =7.5 Lx 0.10 mol NaCl1 Lx g NaCl1 mol NaCl=43.83 g
19 More Practice Questions How many grams of nitric acid are present in 1.0 L of a 1.0 M HNO3 solution?2. Calculate the number of grams needed to produce 1.00 L of these solutions: a) 1.00 M KNO3b) 1.85 M H2SO c) 0.67 M KClO33. Calculate the # of grams needed to produce each:1a) 0.20 L of 1.5 M KCl b) L of M HClc) 0.20 L of 0.09 mol/L AgNO3d) 250 mL of 3.1 mol/L BaCl24. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)263 g101 g181 g82 ga) 22 gb) 1.75 gc) 3 gd) 0.16 kg5. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made.a) mol/Lb) mol/L
20 Practice making molar solutions Calculate # of grams required to make 100 mL of a 0.10 M solution of NaOH (see above).Get volumetric flask, plastic bottle, 100 mL beaker, eyedropper. Rinse all with tap water.Fill a beaker with distilled water.Pour mL of H2O from beaker into flask.Weigh NaOH. Add it to flask. Do step 5 quickly.Mix (by swirling) until the NaOH is dissolved.Add distilled H2O to just below the colored line.Add distilled H2O to the line using eyedropper.Place solution in a bottle. Place label (tape) on bottle (name, date, chemical, molarity). Place bottle at front. Rinse & return equipment.
21 Concentration and Dilution How can a solution be made less concentrated?More solvent can be added.What is this process called?DilutionThis process is used extensively in chemistry... the concentration decreases in dilution, BUT what happens to the moles of the solute?Do they increase?Decrease?Stay the same?
22 Agenda Day 67 – Dilutions Lesson: PPT, Handouts: 1. Concentration& DilutionHandoutText: 1. PHW: 1. Worksheets, P. 405 # 2-7, P. 410 # 1-10, P Review questions
24 Dilution of solutionsSince moles are constant, the new concentration may be found using the following formula:n 1 = n 2C1V1 = C2V2Initial concentrationFinal volumeInitial volumeFinal concentration
25 Dilution Example #1A stock solution of 1.00M of NaCl is available. How manymilliliters are needed to make a mL of 0.750M?What we know: the molarity of the stock solution which is1.00 M, and the two components of the diluted solutionwhich are C2= 0.750M and V2= 100 mL.C1V1 = C2V2
26 Dilution Example #2 Concentrated HCl is 12M. What volume is needed to Make 2L of a 1M solution?What we know: the molarity of the stock solutionwhich is 12M, and the two values for the dilutedsolution which are C2=1M and V2=2L.C1V1 = C2V2
27 Dilution Example #3Calculate the final concentration if 2L of 3M of NaCl and 4L of 1.50M of NaCl are mixed. Assume there is no volume contraction upon mixing. For this, you must use the equation:
28 Making Molar Solutions From Liquids (More accurately, from stock solutions)
29 Making molar solutions from liquids Not all compounds are in a solid form Acids are purchased as liquids (“stock solutions”). Yet, we still need a way to make molar solutions of these compounds. The Procedure is similar: Use pipette to measure moles (via volume) Use volumetric flask to measure volume Now we use the equation C1V1 = C2V2
30 Reading a pipetteIdentify each volume to two decimal places(values tell you how much you have expelled)5.00
31 More Practice1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? 2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L? mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M? 4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl? 5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl? 6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water? 7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution? 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out, what is the concentration of the remaining HF?
32 Dilution problems (1-6)1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = L = mL 2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L) M2 = 1.2 M 3. M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M) V2 = 0.4 L or 400 mL
34 Dilution problems (7, 8)7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M) V2 = 16 L 8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.
35 Practice making molar solutions Calculate # of mL of 1 M HCl required to make 100 mL of a 0.1 M solution of HClGet volumetric flask, pipette, plastic bottle, mL beaker, 50 mL beaker, eyedropper. Rinse all with tap water. Dry 50 mL beakerPlace about 20 mL of 1 M HCl in 50 mL beakerRinse pipette, with small amount of acidFill flask about 1/4 full with distilled waterAdd correct amount of acid with pipette. Mix.Add water to line (use eyedropper at the end)Place solution in plastic bottleLabel bottle. Place at front of the room.Rinse and return all other equipment.