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Ch 6.3b Dilution | Donald Hirsh | YouTubeToText
YouTube Transcript: Ch 6.3b Dilution
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Core Theme
Dilution is a fundamental chemical process used to reduce the concentration of a solution by adding more solvent. This process is governed by the principle that the amount of solute remains constant, allowing for the calculation of initial or final concentrations and volumes using the formula C1V1 = C2V2.
so here in the second half of section
6.3 molarity we're going to focus on
being able to perform dilution
so dilution is the process whereby the
concentration of a solution is lessened
by the addition of solvent and it's very
common in chemistry to have stock
solutions that are concentrated and then
dilute them for use at a later time so
as noted here dilution is a common means
of preparing Solutions of a desired
concentration and by adding solvent to a
measured portion of a more concentrated
stock solution we can achieve a
particular concentration that we desire
so um here's an example where we have
two solutions they both contain the same
mass of copper but the solution on the
right is more dilute because it has a
greater volume right the copper
nitrogate nitrate is dissolved in more solvent
so the molar amount of solute in a
solution is equal to the product of the
solution's molarity and its volume in
milliliters so that's what's shown in
this equation here number of moles
equals molarity times volume in liters
and we can see that that's the case by
looking at the units so molarity has
units of moles per liter and liters are
and volume in liters obviously has the
units of liters so that our leaders
could cancel out and we're left with
number of moles
so expression like expressions like
these may be written for a solution
before which is uh represented by the
number one and after represented by the
number two a dilution occurs okay
and furthermore since the dilution
process doesn't change the amount of
solute in the solution uh the number of
moles at the beginning and one is equal
to the number of moles of solute at the
end okay so again just to be clear the
number of moles that you begin with uh
of solute is the same at the beginning
of the process and at the end of the
dilution process
so what that means then is that N1
equals N2 and therefore we can set the
two equations that we had earlier equal
to each other so
um we have that the molarity in the
beginning times the volume in the
beginning is equal to the molarity at
the end times the volume in liters at
the end
and we can actually write a very general
equation for any type of concentration
units that we might want to use and for
any volume units that we want to use so
typically what we'll simply say is that
C1 V1 equals C2 V2 meaning the initial
concentration times the initial volume
is equal to the final concentration
times the final volume
so let's see how we can use this to
calculate a concentration or volume
so here we have a problem it says if
0.850 liters of a five molar solution of
copper nitrate is diluted to a volume of
1.8 liters by the addition of water what
is the molarity of the diluted solution
so what is the final molarity so the
final molarity that we're interested in
is C2 right so that's the thing we don't know
know
um and so we what we can do is solve for that
that
so dividing both sides by V2 we have
that C2 equals C1 V1 divided by V2
and then we can
put in our values the initial
concentration is 5 molar so that goes here
here
the initial volume
is 0.850 liters so that goes here
and the final volume is 1.80 liters so
that's V2 that goes in the bottom here
and then we simply work the math out
notice that liters will cancel out and
we'll be left with the concentration in
molarity which turns out to be 2.36
molar now just a final word on this is
we can use this equation to um to solve
for any one of these four quantities
right so as long as we know the other
three then we can solve for the fourth
so if you wanted to know for instance
what volume you needed to achieve a
certain final concentration we could
solve for v2 or if you wanted to know
um uh how much volume of a particular
stock solution say C1 that you should
use that would be V1 we can solve for V1
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