Urgent Help: Dilution & C1V1=C2V2 (1 Viewer)

[green_appletini]

Hey guys!

I am really confused and i hope someone will be able to help me cos my trial exam's on tuesday, and i hafta study for drama + ancient history today because my trials for them are tomorrow, so i'm only left with monday to cram for chemistry!!!>__<

Basically, i was wondering if someone can totally explain the C1V1=C2V2 formular for me? I got 55% in my last prac assessment b/c i had a mental blank and totally forgot how to use it (like which original concentration goes where, and which volume goes on which side and what the unknown formula i calculated actually MEANT! I'm so confused...does anyone know how it can explain this to me? And i also get confused about using mL and Litres in it..i'm soo lost!!

One question i really don't get is in "Conquering Chemistry HSC Course" page 151. I don't understand what they've done in the solution where it says "this is the molarity of acetic acid in the diluted vinegar" and how they worked it out from there onwards. Can someone please help me?

Thanks very much!

-marilia- xoxo

 

[Xayma]

C₁V₁=C₂V₂

Comes from: n=CV
ie if n₁=n₂

subbing n=CV gives

C₁V₁=C₂V₂

 

[CM_Tutor]

Firstly, C₁V₁ = C₂ V₂ should be used only for dilutions, and not for the mole calculation part of a titration. Looking at the example from CC_HSC, p. 151, we have two steps for the calculation:

Step 1: The titration - this covers the reaction of 25.00 mL of the diluted solution of vinegar, and titrated it against 0.105 molL^-1 sodium hydroxide, requiring 15.9 mL to reach the end point.

This is a regular mole calculation, given [CH₃COOH] = 0.0668 molL^-1

Step 2: The dilution - In the original question, the Chemist took 10.0 mL of vinegar, and diluted it to 100.0 mL. (In other words, 10.0 mL of vinegar was pipetted into a 100.0 mL volumetric flask, and then filled (with shaking) to the graduation mark with distilled water.) As a result, the solution in this volumetric flask now has a concentration 1/10th of that of the original vinegar. It was from this 100.0 mL that the 25.0 mL used in Step 1 was drawn, and so the 0.0668 molL^-1 concentration found above is the concentration in the 100.0 mL volumetric flask (ie. its the diluted vinegar), and not the conentration in the original 10.0 mL of vinegar, which is 0.668 molL^-1.

This can also be found by using the dilution formula, with

C₁ = concentration of the original, undiluted vinegar, which is what we seek.
V₁ = the volume of the original, undiluted vinegar used = 10.0 mL = 0.0100 L
C₂ = concentration of the diluted vinegar, shown in Step 1 to be 0.0668 molL^-1
V₂ = the volume to which the vinegar was diluted = 100.0 mL = 0.100 L

Now, C₁V₁ = C₂V₂
C₁ * 0.0100 = 0.0668 * 0.100
C₁ = 0.0668 * 0.100 / 0.0100
C₁ = 0.668 molL^-1

As a general comment, calculations are usually done in the opposite order to the experimental steps. Here, the chemist diluted then titrated, so the calculation should handle the titration then the dilution.

 

[green_appletini]

thank you so much for the help!
Thanks everyone!!!

-marilia- xoxo