Solution Concentration

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2 Agenda Day 66 Concentration Lesson: PPT, Handouts: 1. Concentration& Dilution Handout. 2. Concentration of Solutions Worksheet Text: 1. P Concentration ( %, ppm) HW: 1. Worksheets, P. 400 # 1-4, P.402 # 3-10 %, ppm

3 Solution Concentration Concentration = quantity of solute quantity 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/V Like 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/l 3 g H 2 O 2 /100 ml solution 3 g H 2 O 2 /100 g solution

4 Solution and Concentration Other ways of expressing concentration Molarity(M): moles solute / Liter solution Mass percent: (mass solute / mass of solution) * 100 Molality* (m) - moles solute / Kg solvent Mole 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.

5 % Concentration % (w/w) = mass solute mass solution x 100 [ 3% w/w = 3 g/100 g] % (w/v) = % (v/v) = mass solute volume solution volume solute volume solution x 100 [ 3% w/v = 3 g/100 ml] x 100 [ 3% v/v = 3 ml/100 ml]

6 Units of Concentrations amount of solute per amount of solvent or solution Percent (by mass) = g solute g solution x 100 = g solute x 100 g solute + g solvent Molarity (M) = moles of solute volume in liters of solution moles = M x V L

7 Solution Concentration Expressing concentrations in parts per million (ppm) requires the unit on top to be 1,000,000 times smaller than the unit on the bottom E.g. 1 mg/kg or g/g Notice that any units expressed as a volume must be referring to a water solution (1L = 1kg)- density of water For parts per billion (ppb), the top unit would have to be 1,000,000,000 times smaller 1 ppm = 1 g/10 6 ml Cppm = m solute m solution x 10 6 = 1 g/ 1000 L = 1 mg/l = 1 mg/kg = 1 g/g 1g = 1000mg 1000mg/1000L 1mg = 1000 g 1000 g/1000g

8 Molarity C = n V Molar 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 1. What is the % W/W of copper in an alloy when 10 g of Cu is mixed with 250 g of Zn? 10 g / 260 g = 3.8 % W/W 2. What is approximate % V/V if 30 ml of pure ethanol is added to 250 ml of water? 30 ml / 280 ml = 11% V/V (in reality may be off) 3. What is the % W/W if 8.0 g copper is added to enough zinc to produce 100 g of an alloy? 8.0 g / 100 g = 8% W/W

10 Concentration: Molarity Example If g of KMnO 4 is dissolved in enough water to give 250. ml of solution, what is the molarity of KMnO 4? As is almost always the case, the first step is to convert the mass of material to moles g KMnO 4 x 1 mol KMnO 4 = mol KMnO g KMnO 4 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 L. Molarity KMnO 4 = mol KMnO 4 = M L solution

11 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 ml of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity [NiCl 2 6 H 2 O ] = M

12 Concentration: Mixed Example A solution of H 2 O 2 is 3% (w/v). a) Calculate the mass of H 2 O 2 in ml of solution. b) Calculate the mass of H 2 O 2 in 1L of solution. c) Calculate the number of moles of H 2 O 2 in 1L of solution. d) State the molar concentration of the solution e) Calculate the ppm of H 2 O 2.

13 Concentration: Mixed Example Answers a) b) c) d) e) mass = 250mL mass =1000mL 3g 100ml = 7.50g 3g 100ml = 30g n = 30g 1mol 34g = 0.88mol M = 0.88 mol L 3g Cppm = 100mL x 106 = 30000ppm

14 Making Molar Solutions From Solids

15 Preparation of Solutions

16 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 L x 0.10 mol NaCl = 0.75 mol 1 L But 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 L x 0.10 mol NaCl 1 L x g NaCl 1 mol NaCl =43.83 g

19 More Practice Questions 1. How many grams of nitric acid are present in 1.0 L of a 1.0 M HNO 3 solution? 63 g 2. Calculate the number of grams needed to produce 1.00 L of these solutions: a) 1.00 M KNO g b) 1.85 M H 2 SO g c) 0.67 M KClO 3 82 g 3. Calculate the # of grams needed to produce each:1 a) 0.20 L of 1.5 M KCl b) L of M HCl c) 0.20 L of 0.09 mol/l AgNO 3 d) 250 ml of 3.1 mol/l BaCl 2 4. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)h 2 SO 4 b)ca(oh) 2 5. Describe how 100 ml of a 0.10 mol/l NaOH solution would be made. a) 22 g b) 1.75 g c) 3 g d) 0.16 kg a) mol/l b) mol/l

20 Practice making molar solutions 1. Calculate # of grams required to make 100 ml of a 0.10 M solution of NaOH (see above). 2. Get volumetric flask, plastic bottle, 100 ml beaker, eyedropper. Rinse all with tap water. 3. Fill a beaker with distilled water. 4. Pour ml of H 2 O from beaker into flask. 5. Weigh NaOH. Add it to flask. Do step 5 quickly. 6. Mix (by swirling) until the NaOH is dissolved. 7. Add distilled H 2 O to just below the colored line. 8. Add distilled H 2 O to the line using eyedropper. 9. 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? Dilution This 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& DilutionHandout Text: 1. P HW: 1. Worksheets, P. 405 # 2-7, P. 410 # 1-10, P Review questions

23 Dilution of Solutions

24 Dilution of solutions Since moles are constant, the new concentration may be found using the following formula: Initial concentration n 1 = n 2 C 1 V 1 = C 2 V 2 Final volume Initial volume Final concentration

25 Dilution Example #1 A stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a ml of 0.750M? What we know: the molarity of the stock solution which is 1.00 M, and the two components of the diluted solution which are C 2 = 0.750M and V 2 = 100 ml. C 1 V 1 = C 2 V 2

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 solution which is 12M, and the two values for the diluted solution which are C 2 =1M and V 2 =2L. C 1 V 1 = C 2 V 2

27 Dilution Example #3 Calculate 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: C total = n total V total

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 C 1 V 1 = C 2 V 2

30 Reading a pipette Identify each volume to two decimal places (values tell you how much you have expelled)

31 More Practice 1. 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 CuSO 4 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. M 1 = 14 M, V 1 =?, M 2 = 1.75 M, V 2 = 250 ml V 1 = M 2 V 2 / M 1 = (1.75 M)(0.250 L) / (14 M) V 1 = L = ml 2. M 1 = 6 M, V 1 = 0.2 L, M 2 =?, V 2 = 1 L M 2 = M 1 V 1 / V 2 = (6 M)(0.2 L) / (1 L) M 2 = 1.2 M 3. M 1 = 6 M, V 1 = 100 ml, M 2 = 1.5 M, V 2 =? V 2 = M 1 V 1 / M 2 = (6 M)(0.100 L) / (1.5 M) V 2 = 0.4 L or 400 ml

33 Dilution problems (4-6) 4. # mol = (2.0 mol/l)(0.4 L) + (3.0 mol/l)(0.6 L) = 0.8 mol mol = 2.6 mol # L = 0.4 L L # mol/l = 2.6 mol / 1 L = 2.6 mol/l 5. # mol = (0.5 mol/l)(3 L) + (0.2 mol/l)(2 L) = 1.5 mol mol = 1.9 mol # mol/l = 1.9 mol / 5 L = 0.38 mol/l 6. # mol = (0.5 mol/l)(3 L) + (0 mol/l)(2 L) = 1.5 mol + 0 mol = 1.5 mol # mol/l = 1.5 mol / 5 L = 0.3 mol/l Or, using M 1 V 1 = M 2 V 2, M 1 = 0.5 M, V 1 = 3 L, M 2 =?, V 2 = 5 L

34 7. Dilution problems (7, 8) M 1 = 6 M, V 1 = 4 L, M 2 = 1.5 M, V 2 =? V 2 = M 1 V 1 / M 2 = (6 M)(4 L) / (1.5 M) V 2 = 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 HCl Get volumetric flask, pipette, plastic bottle, 100 ml beaker, 50 ml beaker, eyedropper. Rinse all with tap water. Dry 50 ml beaker Place about 20 ml of 1 M HCl in 50 ml beaker Rinse pipette, with small amount of acid Fill flask about 1/4 full with distilled water Add correct amount of acid with pipette. Mix. Add water to line (use eyedropper at the end) Place solution in plastic bottle Label bottle. Place at front of the room. Rinse and return all other equipment.