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