1 / 11

DILUTING A SOLUTION

DILUTING A SOLUTION. The objectives of this video are twofold : Using the dilution formula Establishing the steps to follow in order to realize the dilution. How to prepare a solution of 100 mL :. Pour around half of your solvent , in this case 50 mL.

vida
Download Presentation

DILUTING A SOLUTION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. DILUTING A SOLUTION The objectives of thisvideo are twofold: Using the dilution formula Establishing the steps to follow in order to realize the dilution

  2. How to prepare a solution of 100 mL: • Pour around half of your solvent , in this case 50 mL 2. Put all of your solute into the half solution, making sure that no solute is lost. 3. Stir (dissolve) the solute completely 4. Complete the solution by topping it off with solvent to reach the 100 mL line. Remember, that it is 100 mL of solution not 100 mL of solvent. Na2SO4 5. Stir again 50 mL

  3. We have a 100 mL solution of Na2SO4, a salt solution, dissolvedin water. The concentration is 0.1 M (moles/ liter).Let’ssaywewant to diluteall of this solution to 0.01 M. How much water must weadd? The formula is: C1V1 = C2V2 where C1 = initial concentration V1 = initial volume , or volume to use C2 = final concentration V2 = final volume, or volume to make

  4. Let us first examine the relationshipbetween volume and concentration. Here are someeasynumbers: If C1 = 10 V1 = 4 C2 = 5 and C1 x V1 must alwaysequal C2 x V2, Then10 x 4 = 5 x ? 10 x 4 / 5 = 8 Weseethat as C decreases ( 2 times less), 10x4=5x8 V must converselyincrease ( 2 times more) 40=40

  5. In our case,C1 = 0.1 M ( 0.1 molar or 0.1 moles/liter)V1 = 100 mLC2 = 0.01 MV2 = ?There are 3 variables for whichwe have information.Wealso know that, as C decreases, V must increase. Applying the formula, weget: C1 X V1 = V2 C2 0.1M x 100 mL V2 0.01M 1000 mL = V2

  6. The question thatremainsis: How much water do weneed to add?In the lab, dilutingalwaysinvolvesaddingmore solvant.We have found V2, the volume to make; not thevolume to add. This requires an additionnalstep: V2 – V1 = Volume to add Thus: 1000 mL – 100 mL = 900 mL Weneed to make 1000 mL of solution by adding 900 mL of solvant, water.

  7. Dilution: Dilution lowers the concentration of a substance by adding a solvent. Let’s look at our previous example: 100 mL solution a) Volume to use: V1: 100 mL First concentration: C1: 0.1 M b) Final volume: V2: 1000 mL Second concentration: C2: 0.01 M C1 V1 C2 V2 Adding 900 mL of solvent

  8. Before C1 = n1 V1 n1 = C1 V1 After C2 = n2 V2 n2 = C2 V2 Now with this formula, we can predict concentration and dilutions of solutions. Laws of concentration: C1V1 = C2V2 Now, since: n1 = n2 So C1V1 = C2V2

  9. Hereis a different situation. What if westillwant to dilutefrom 0.1M to 0.01M but wewant to make a small volume of diluted solution, say 50 mL. Westill have the initial solution atourdisposal, 100 mL. C1 = 0.1 M V1 = 100mL C2 = 0.01 M V2 = 50mL As C decreasestenfold, V must increasetenfold. If weadd water to end with 50, wecan’t use all of our initial 100mL!

  10. So, wewillonlytakesome of our initial solution.How much? This meansthatwe are actuallylooking for V1. V1 = C2 x V2 C1 V1 = 0.01M x 50 mL 0.1 M V1 = 5 mL

  11. Weagainneed to find how much water to add: Volume to add = V2 – V150 mL – 5 mL = 45 mL In conclusion, weuse 5 mL of our initial solution and add 45 mL to make 50 mL of solution (final).

More Related