Molarity is just another word for concentration. The formula for molarity
is equal to the moles of your solute, divided by the liters of your solution.
Now molarity can be used as a conversion factor because it has two different
units. Allowing you to go from one unit to the next. Alright, let's head over to
the blackboard for our example. It takes 12.5 milliliters of a 0.300
molar HCl solution to neutralize 285 milliliters of NaOH
solution. What is the concentration of the NaOH solution? And we're given our
balanced equation. Let's start with identifying our given and what we're
finding. So we're given 12.5 milliliters of HCl solution and then we're also
given our molarity of our HCl solution as well. Now whenever you see that
molarity, that's going to be used as your conversion factor. I want you to split
that up automatically to moles over liters. So the 0.300
will go with the moles on top and this will be per every one liter of HCl. We're
also given 285 milliliters of NaOH solution. Alright, now we're finding the
concentration of NaOH solution or really molarity. And once again molarity is
mole's over liters, so we first have to find the moles of NaOH and then we'll
divide that by the leaders of NaOH. A lot of times students don't know where to
start so my trick for you is, whatever you are not looking for, that's what
you're going to start with. So since we're looking for molarity of NaOH. We're
gonna start with HCl. So we're gonna start with the milliliters of HCl and
then make our way to moles of NaOH. So our first step is to find the moles
NaOH. And let's plan this out, we said we're gonna always start with the
compound or really, we're starting with whatever we're not looking for. So since
we're finding NaOH we're gonna start with the milliliters of HCl. Our plan
will be, starting with the milliliters of HCl then convert that to liters of HCl
using the metric system. From there we're going to go from liters of HCl to moles
of HCl using the molarity of HCl that was initially given and then we'll get
to moles of HCl then we'll use a mole to mole ratio from our balanced
equation and get to moles of NaOH. Okay let's set this up. So starting with our
12.5 milliliters of HCl let's align the milliliters of HCl so they can
cancel and our first step is to change this to litres of HCl using the metric
system. So I'm going to put a 1 milliliter on the bottom and 10 to the
negative third liters on top of HCl. From here our milliliters of HCl would then
cancel and we're at liters of HCl. Next step is to change this to moles of HCl
using the molarity that was provided. So I'll put the 0.300
moles on top of HCl and on the bottom would be one liter of HCl. The liters of
HCl would then cancel and we're now at moles of HCl. Now we want to find moles
of NaOH so we're going to use a mole to mole ratio and we'll take that from our
balanced equation. So note that going back to our balanced equation, everything
is a one-to-one relationship meaning, there is no coefficient in front there's
no number in front of that compound, so it's just a one mole relationship to
everything. So I'm going to put one mole of NaOH on top and one mole of HCl on
the bottom. Then once we do that our moles of HCl would then cancel and we're
finally left with moles of NaOH. Make sure to multiply straight across
and divide. Really in this case it's just divided by one. And our answer would be
0.00375 moles of NaOH. Our next step is to
convert the milliliters to liters of the NaOH that was provided, so in our given
we were given the two hundred and eighty five milliliters of NaOH. So since we're
finding molarity, this is going to be the second part where we have moles that we
just found in step one and now we're finding the liters of NaOH and at the
end we'll divide. So I'm going to take the milliliters of NaOH and align
milliliters and milliliters so they can cancel and then on top I'll place 10 to
the negative third liters of NaOH. So our milliliters and milliliters cancel
and now we're at liters of NaOH. There's another way to do this. So if you prefer
putting 1,000 milliliters on the bottom and a 1 liter on top that is completely
fine. Both of these will give you the same exact answer. So now we have 0.285
liters of NaOH. Our last step is to divide moles of NaOH that we found in
step one by the liters of NaOH that we found in step two. Because we're
finding molarity, which was the moles of NaOH divided by the liters of NaOH. This
is why we first have to start with finding our moles of NaOH, then finding
our liters of NaOH, dividing and that's our final answer. So I'm going to put the
moles of NaOH on top that we found in step one. And divide that by the liters
of NaOH that we found in step two. Once we do this, this would give us
0.0132 molar solution of NaOH. So by molar solution I really just
mean molarity. That capital M means molarity, you could also put moles over
liters as your final answer but you'll commonly see a capital M. Why don't you
try one out on your own, I placed a practice problem at the very top of the
description box and at the bottom you'll find the step-by-step
solution. Let me know how you do in the comments and guys my personal mission is
to help millions of students with chemistry so if I've helped you in any
way let me know by subscribing and liking this video. Alright I wish you the
absolute best, I'm gonna go and record the voiceover now. So I've been working
on these molarity notes for a week, I had to make sure to put how to find molarity,
different examples of solution stoichiometry,
I even covered dilutions. These notes are so much easier to understand than the
average textbook. You know I wish I had something like these notes because when
I was in class there wasn't enough time and I was always rushing to write
everything down. So check the link in the description, download the notes and
master molarity. Alright guys I'll see you in the next video