Dilution Solution
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Transcript of Dilution Solution
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Dilution Solution
Solution Dilution
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What’s the “concentration” of red triangles?
500 mL
A. 10 MB. 5 M
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What’s the “concentration” of red triangles?
500 mL
1 g
1 g
1 g 1 g
1 g
A. 10B. 10
C. All of the above
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Concentration is…
…any statement of the relationship between the amount of stuff (“solute”) dissolved in a solvent/solution.
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Could be ANYTHING
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UNITS! UNITS! UNITS!
The units are your friend – ALWAYS! The units tell you how to measure your “stuff”.
So, if I’ve got Molarity (M), I probably want to measure the volume…
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𝐿𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑚𝑜𝑙𝑒𝑠𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒𝐿𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=𝑚𝑜𝑙𝑒𝑠𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
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If I have…
Then I want to measure…
GRAMS of solution!
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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.
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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.
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If I add 1 L of water
1 g
1 g
1 g 1 g
1 g
500 mL
1 L
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1 g
1 g
1 g 1 g
1 g
1500 mL
Volumes add.
Pick your favorite concentration unit:
Etc.
Chemistry is all about MOLES! MOLES! MOLES!So something like Molarity is usually our fav
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This is true whenever I mix things
1 g
1 g
1 g 1 g
1 g
500 mL 1 L
2g
2g 2g
2g
2g 2g2g
2g
10 8
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If I mix them…
1 g
1 g
1 g 1 g
1 g1500 mL
The concentration is…?
Etc.
2g
2g2g
2g
2g 2g2g
2g
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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.
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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!
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This is true whenever I mix things
50 mL 200 mL
0.215 M Na2SO4 0.5 M PbSO4
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It’s a question of how many there are…
250 mL
MOLES IS NUMBER!
But how many “moles” do I have?
0.01075 mol
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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.
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This is true whenever I mix things
50 mL
0.215 M Na2SO4
250 mL
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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.
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Plug and chug
This sometimes gets written in algebraic form:
M1V1= M2V2
Or sometimesC1V1=C2V2
It still boils down to:Moles in “solution 1” = Moles in “solution 2”
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Plug and chug
This sometimes gets written in algebraic form:
M1V1= M2V2
Notice that the Molarity of a solution is coupled with the volume of that solution.
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That’s why the 0.215 M goes with the 50 mL – that’s the solution that has that concentration.
50 mL
0.215 M Na2SO4
250 mL
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0.215 M Na2SO4 (50 mL) = M2 (250 mL)
M2=0.043 M Na2SO4
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That’s why the 0.215 M goes with the 50 mL – that’s the solution that has that concentration.
50 mL
0.215 M Na2SO4
250 mL
0.043 M Na2SO4
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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.
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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?
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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?
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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!
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ALWAYS A BALANCED EQUATION!
AgNO3 (aq) + Na2SO4(aq) → Ag+ + NO3- + Na+ + SO4
2-
AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 + NaNO3
2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 + 2 NaNO3
2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq)
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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?
These numbers don’t compare! Apples and oranges.It’s not how much you have, it’s how much you need.
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2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq)
These numbers compare!
The silver nitrate runs out first!
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2 AgNO3 (aq) + Na2SO4(aq) → Ag2SO4 (s) + 2 NaNO3 (aq)
This is the “theoretical yield” – what I get if everything goes perfectly. In this reaction, it didn’t!
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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?