We're focusing on an example of solution stoichiometry. So our example states,
according to the label on a bottle of concentrated hydrochloric acid, the
contents are 36% HCl by mass and have a density of 1.18 grams per
milliliter. We're asked three different questions here ,our first one is, "What is
the molarity of the concentrated HCl?" So let's start off with that one. Identify
what we're given and what we're finding as always, so we're given 36% of HCl by
mass. Now whenever you see this percentage by mass, that's gonna act as
your conversion factor. So what we're gonna do is we're gonna split this apart
and 36 grams of HCl will go on top and then on the bottom would be a
100 grams of HCl solution. Next we're also given our density of 1.18
grams per milliliter. And this is specifically of the HCl solution. Next
we're asked to find the molarity and remember, molarity is mole's divided by
liters. So some way somehow we're gonna get to moles of HCl and divide that by
the liters of our HCl solution. So what we're gonna do first is, taking that
36 grams per every 100 gram of HCl solution, we're gonna put that and
multiply that by the density that was provided because our grams of HCl and
grams of HCl solution are going to then cancel and then now we have grams
of HC divided by the milliliters of our HCl solution. And then so we're at this
current step right now. After we've multiplied already we have to keep going
because remember we're trying to find the molarity, so that means we want our
moles of HCl on top divided by liters of our HCl solution.
So taking what we just found that 0.42 for 8 grams of HCl, then I'm gonna change
this to moles using the molar mass of HCl. So aligning our grams of HCl
and remember grams of HCl how we find that, how we found the molar mass is just taking
our mass of each individual element, we have hydrogen that has 1.01
chlorine which has 35.45 and we're going to add those
two together to get our molar mass of HCl which would be 36.46
grams per mole. So I'm going to put the grams on the bottom and the moles on
top since we're trying to find moles on top. And then from here our grams of HCl
and grams of HCl will then cancel and then now we have moles of HCl on top and
milliliters on the bottom. So we're almost done. The next portion here is to
change milliliters on the bottom to liters. So what we're going to use is
1,000 milliliters, put that on top so our units would then cancel and we always
want those units across from each other so they can cancel. And then next our
liters of HCl is finally on the bottom. So what we're going to see is we're
finally going to have our unit of moles of HCl divided by the leaders of our
HCl solution. And then from there our final answer would give us
11.7 molarity of our HCl solution, so that's the molarity of HCl. Let's move
on to Part B. Part B states, "What volume of it would you need to prepare 283
milliliters of a 2.20 molar solution of HCl so we're gonna use our
previous portion, our previous molarity in Part A that we found which was that
11.7 molarity of HCl, so that is gonna be our given, one of our
givens in this question. Now we're asked to find another volume, so what I want
you to notice is the fact that we started offer in Part A we had
11.7 molarity of HCl but this time we decreased our molarity, so
anytime we're decreasing that molarity and we're dealing with molarity and
volume that's a dilution. So we have to use our dilution formula of
M1V1 those two are multiplied and set that equal to M2 x V2. Let's identify what
our M 1 and V 1 and M 2 and V 2 are. So our M 1 is going to be what we found
initially, what we found in Part A, and then what we're given our molarity to is
that 2.20 molarity of HCl. Next were asked what volume of it, meaning what
volume of our previous molarity, so our initial molarity. So we're finding V 1
and they give us V 2 is 283 milliliters. Okay now that we've identified
everything here, let's just plug in. So we'll plug in 11.7 for our
M 1 and then V 1 remember that's what we're solving for and then M 2
we'll plug in 2.20 and V 2 we'll plug in 283 and then from there all
we have to do is to get V 1 by itself so we'll divide over that eleven point
seven to either side and then from there these two would cancel we'll solve for V
1 after we do the math so multiply the numerator, first what's on top and divide
it by 11.7, we would get 53.2 milliliters of HCl.
One more question let's go to Part C. Part C states,
"What mass of sodium bicarbonate would be needed to neutralize the spill if a
bottle containing 1.75 liters of concentrated HCl dropped on a lab floor
and broke open? So once again let's identify what are we given, what are we
finding? So we're given that 1.75 liters of HCl and then remember we're
still going to use our concentrated HCl molarity that we found in Part A which
was that 11.7 molarity of HCl. Now we're asked to find grams of a completely
different compound and in this case its sodium bicarbonate. So we're
gonna do a little bit of stoichiometry here. So before I move on to
stoichiometry, one thing to note is we need our balanced equation. Because we're
going from moles or really one compound of HCl to a totally different compound
of sodium bicarbonate, we have to do a mole to mole ratio and where do we get a
mole to mole ratio, our balanced equation, those coefficients in front. So our
balanced equation would be HCl, plus sodium bicarbonate and what's gonna
happen is that sodium or NA would then combine with Cl so we'll see that here
these two would combine to form that NaCl and then the next portion here is a
little bit different because we're gonna form two additional products and this is
from that hydrogen combining with that bicarbonate portion. So when hydrogen
gets combined with bicarbonate it forms water and carbon dioxide. Okay now if we
were to add everything up we'll see that this is already balanced, how nice. So
everything has a 1, every single compound has a 1 in front. So it's all a
one-to-one relationship. Now let's start this problem off with our given. So our
given was that 1.75 liters of HCl. Remember molarity is usually used as a
conversion factor which is exactly what we're gonna use it for with this question.
Now we're gonna have 1.75 liters of HCl, align our units so they can cancel. So
1 liter of HCl and where I'm taking this by the way, I can rewrite this as
11.7 moles of HCl divided by 1 liter of HCl. So that's what you
want to do whenever you're looking at molarity, split it apart into moles
divided by liters. Now we're at moles of HCl on top and then I want those moles
of HCl to cancel to then get to moles of a completely different compound or
moles really of our sodium bicarbonate. So this is where our balanced equation
will come in. We're gonna do a mole to mole ratio again
and then remember everything had a 1 in front of the compound so I'm just
gonna do 1 mole of HCl on top would be 1 mole of sodium bicarbonate. So moles
of HCl would then cancel and then finally we're asked to find the grams of
sodium bicarbonate, so we need to use the molar mass of sodium bicarbonate. That's
what we're gonna calculate right now. So sodium has a mass of 22.99. Hydrogen
has a mass of 1.01. Carbon 12.01 and oxygen is 16 however there are 3
oxygen within this compound so I'm gonna multiply that by 3 and we'll get 48.
Now we'll add everything together, so adding all of these masses that we
found together we will get the molar mass of 84.01 grams per mole
which we have to place on top. Since we're looking for grams, that's why
replacing grams on top and then remember we want our units to cancel so moles of
sodium bicarbonate, let's align those across from each other so they can
cancel and finally we have our grams of sodium bicarbonate. Multiply straight
across since really we're dividing by 1 and then from here our answer would
be 1,720 grams of sodium bicarbonate. And that was our pact
example of a solution stoichiometry question. If this video helped you out
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