11 - Dilutions
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Dilution Solution Solution Dilution What’s the “concentration” of red triangles? 500 mL 5 A. πΏ 5 πππ π‘ππππππππ B. πΏππ‘ππ 10 πππ π‘ππππππππ C. πΏππ‘ππ D.10 M E. 5 M What’s the “concentration” of red triangles? 500 mL A. 1g 1g B. πππ π‘ππππππππ 10 πΏ π 10 πΏ C. 1% ππ¦ πππ π π D. 0.01 ππΏ 1g E. All of the above 1g 1g Concentration is… …any statement of the relationship between the amount of stuff (“solute”) dissolved in a solvent/solution. π πππ ππππ π’ππ ππ π πππ’π‘π π πππ ππππ π’ππ ππ π πππ’π‘πππ Could be ANYTHING π ππ π πππ’π‘π π ππ π πππ’π‘πππ ππΏ ππ π πππ’π‘π ππΏ ππ π πππ’π‘πππ πππππ ππ π πππ’π‘π πππππ ππ π πππ’π‘πππ πππππ ππ π πππ’π‘π πΏ ππ π πππ’π‘πππ UNITS! UNITS! UNITS! The units are your friend – ALWAYS! The units tell you how to measure your “stuff”. πππππ ππ π πππ’π‘π π= πΏ ππ π πππ’π‘πππ So, if I’ve got Molarity (M), I probably want to measure the volume… πππππ ππ π πππ’π‘π πΏ ππ π πππ’π‘πππ = πππππ ππ π πππ’π‘π πΏ ππ π πππ’π‘πππ If I have… π ππ π πππ’π‘π π ππ π πππ’π‘πππ Then I want to measure… GRAMS of solution! πππππ ππ π πππ’π‘π πππππ ππ π πππ’π‘πππ = πππππ ππ π πππ’π‘π πππππ ππ π πππ’π‘πππ Concentration is always just a conversion factor between the way you measured the solution and how much solute you’ve got! The SOLUTE is almost always the thing you care about. The solvent/solution is just the carrier. When you mix things… …you “dilute” them. But the concentration is still just the ratio of the amount of solute and the amount of solution. If I add 1 L of water 1g 1g 500 mL 1g 1g 1g 1L Volumes add. Pick your favorite concentration unit: 5 π ππ πππ π‘ππππππππ π ππ ππ‘ = 3.333 1.5 πΏ π πππ’π‘πππ πΏ 1500 mL 5 πππ π‘ππππππππ ππ‘ = 3.333 1.5 πΏ π πππ’π‘πππ πΏ 1g 1g 1g 1g 1g Etc. Chemistry is all about MOLES! MOLES! MOLES! So something like Molarity is usually our fav This is true whenever I mix things 2g 2g 1g 2g 1g 2g 2g 500 mL 1L 1g 2g 2g 1g 10 πππ π‘ππππππππ πΏ 1g πππππ π ππ’ππππ πΏ 8 2g If I mix them… 2g 2g 2g 2g 2g 2g The concentration is…? 2g 2g 1500 mL 1g 1g 5 πππ π‘ππππππππ ππ‘ = 3.33 1.5 πΏ πΏ 8 πππππ π ππ’ππππ ππ = 5.33 1.5 πΏ πΏ 16 π π. π . π π. π . = 10.67 1.5 πΏ πΏ 1g Etc. 1g 1g It’s tough when they are invisible… …so make them visible. A picture paints 1000 words (…so why can’t I paint you?). If you can draw it so you can see it, it makes pretty good common sense. 50 mL of 0.215 M Na2SO4 is mixed with 200 mL of 0.500 M PbSO4. What is the concentration of Na2SO4 after mixing? I think I’ll draw a picture! This is true whenever I mix things 200 50 mL 0.215 M Na2SO4 0.5 M PbSO4 It’s a question of how many there are… MOLES IS NUMBER! But how many “moles” do I have? 250 mL 0.215 πππ 0.215 π = πΏ 0.215 πππ 0.050 πΏ ππππ πΏ ππππ =0.01075 mol M is just a conversion factor Notice I multiply 50 mL by the concentration not 250 mL. Why? (you may ask) Because the concentration applied to THAT solution. This is true whenever I mix things 250 mL 50 mL 0.215 M Na2SO4 Conservation of moles… In general, there’s no such thing. But in a dilution there is… Moles at the beginning = moles at the end! It’s just the volumes that change. Plug and chug This sometimes gets written in algebraic form: M1 V1= M2 V2 Or sometimes C1V1=C2V2 It still boils down to: Moles in “solution 1” = Moles in “solution 2” Plug and chug This sometimes gets written in algebraic form: M1 V1= M2 V2 πππππ ππ 1 πππππ ππ 2 πΏ ππ 1 = πΏ ππ 2 πΏ ππ 1 πΏ ππ 2 Notice that the Molarity of a solution is coupled with the volume of that solution. That’s why the 0.215 M goes with the 50 mL – that’s the solution that has that concentration. 250 mL 50 mL 0.215 M Na2SO4 0.215 M Na2SO4 (50 mL) = M2 (250 mL) M2=0.043 M Na2SO4 That’s why the 0.215 M goes with the 50 mL – that’s the solution that has that concentration. 250 mL 50 mL 0.215 M Na2SO4 0.043 M Na2SO4 The units must cancel… …but it doesn’t matter what they are… C1V1 = C2V2 As long as the “C”s are in the SAME units and the “V”s are in the SAME units, everything is fine. πππππππ‘π¦ × ππΏ = πππππππ‘π¦ × ππΏ % ππ¦ πππ π × πΏ = % ππ¦ πππ π × πΏ πππππππ‘π¦ × πππππππ = πππππππ‘π¦ × πππππππ Question When 50.00 mL of 0.125 M silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. If I recover 0.813 g of solid, what is the yield of the reaction? Limiting reactant problem How do I know? I have limited amounts of each reactant. When 50.00 mL of 0.125 M silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. If I recover 0.813 g of solid, what is the yield of the reaction? Where do I start? ALWAYS A BALANCED EQUATION! When 50.00 mL of 0.125 M silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. AgNO3 (aq) + Na2SO4(aq) → ??? What type of reaction is this? Double replacement – two ionic reactants! ALWAYS A BALANCED EQUATION! AgNO3 (aq) + Na2SO4(aq) → Ag+ + NO3- + Na+ + SO42- AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 + NaNO3 2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 + 2 NaNO3 2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq) 2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq) When 50.00 mL of 0.125 M silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. If I recover 0.813 g of solid, what is the yield of the reaction? 0.050 πΏ π΄πππ3 π πππ’π‘πππ 0.050 πΏ ππ2 ππ4 π ππ 0.125 πππ π΄πππ3 = 6.25 × 10−3 πππ π΄πππ3 πΏ π πππ’π‘πππ 0.250 πππππ2 ππ4 = 1.25 × 10−2 πππ ππ2 ππ4 πΏ π πππ’π‘πππ These numbers don’t compare! Apples and oranges. It’s not how much you have, it’s how much you need. 2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq) 1 πππ π΄π2 ππ4 6.25 × 10 πππ π΄πππ3 = 3.125 × 10−3 ππππ΄π2 ππ4 2 πππ π΄πππ3 −3 1.25 × 10−2 1 πππ π΄π2 ππ4 πππ ππ2 ππ4 = 1.25 × 10−2 πππ π΄π2 ππ4 1 πππππ2 ππ4 These numbers compare! The silver nitrate runs out first! 2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq) 3.125 × 10−3 ππππ΄π2 ππ4 311.80 π π΄π2 ππ4 = 0.974 ππ΄π2 ππ4 ππππ΄π2 ππ4 This is the “theoretical yield” – what I get if everything goes perfectly. In this reaction, it didn’t! I only got 0.813 g of product! When 50.00 mL of 0.125 M silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. If I recover 0.813 g of solid, what is the yield of the reaction? πππ‘π’ππ π π‘π’ππ πππππ = × 100 π‘βπππππ‘ππππ π π‘π’ππ 0.813 π πππππ£ππππ πππππ = × 100 0.974 π π‘βπππππ‘ππππ πππππ = 83.4%
