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Concentration

Concentration

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Concentration

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  1. Concentration

  2. Agenda • Day 66 – Concentration • Lesson: PPT, • Handouts: 1. Concentration& Dilution Handout. 2. Concentration of Solutions Worksheet • Text: 1. P. 398-401 - 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 H2O2/100 mL solution  3 g H2O2/100 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) = [ 3% w/w= 3 g/100 g] • % (w/v) = [ 3% w/v = 3 g/100 mL] • % (v/v) = [ 3% v/v = 3 mL/100 mL]

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

  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/106 mL = 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 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 10 g / 260 g = 3.8 % W/W 30 mL / 280 mL = 11% V/V (in reality may be off) 8.0 g / 100 g = 8% W/W • What 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 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. 0.435 g KMnO4x 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g 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 M 0.250 L solution

  11. 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•6H2O Step 2: Calculate Molarity [NiCl2•6 H2O] = 0.0841 M

  12. Concentration: Mixed Example A solution of H2O2is 3% (w/v). • Calculate the mass of H2O2 in 250.0 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 solution • Calculate the ppm of H2O2 .

  13. Concentration: Mixed Example Answers a) b) c) d) e)

  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. x 0.10 mol NaCl x 0.10 mol NaCl x 58.44 g NaCl 1 mol NaCl 1 L 1 L 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 = 0.75 mol 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 =43.83 g

  19. More Practice Questions 63 g • 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 KNO3 b) 1.85 M H2SO4 c) 0.67 M KClO3 3. Calculate the # of grams needed to produce each:1 a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HCl c) 0.20 L of 0.09 mol/L AgNO3 d) 250 mL of 3.1 mol/L BaCl2 4. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2 101 g 181 g 82 g a) 22 g b) 1.75 g c) 3 g d)0.16kg 5. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made. a) 0.041 mol/L b) 0.054 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 20 - 30 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? 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. 403-411- • HW: 1. Worksheets, P. 405 # 2-7, P. 410 # 1-10, P.416-421- Review questions

  23. Dilution of Solutions

  24. Dilution of solutions • Since moles are constant, the new concentration may be found using the following formula: n 1 = n 2 C1V1 = C2V2 Initial concentration Final volume Final concentration Initial volume

  25. Dilution Example #1 A stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a 100.0 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 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 solution which is 12M, and the two values for the diluted solution which are C2=1M and V2=2L. C1V1 = C2V2

  27. Dilution Example #3 Calculate the final concentration if 2L of 3M of NaCland 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 pipette Identify each volume to two decimal places (values tell you how much you have expelled) 4.48 - 4.50 4.86 - 4.87 5.00

  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 mLof 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L? 3. 100 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 1.5M. Whatisthe totalvolumeofthenew solution? 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out,whatistheconcentrationoftheremaining 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 = 0.03125 L = 31.25 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

  33. Dilution problems (4 - 6) 4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L) = 0.8 mol + 1.8 mol = 2.6 mol # L = 0.4 L + 0.6 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 + 0.4 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 M1V1 = M2V2, M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

  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 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.