DENA LEGGETT: Hi, and welcome to AP Daily practice
session, where we're going to explore some multiple choice
set questions.
My students call me Doc Dena, and I'm
grateful to help you now.
Remember, you have control over the Pause button.
If I'm going too fast, you need to pause.
All right.
A set question means that you have
information that applies to more than one
multiple-choice question.
So we have three reactions here.
And my suggestion is when it tells you the question numbers,
you circle those question numbers.
So this one said 34 to 36.
Circle them, and so you know which ones to look back
to this information.
This is a short set.
There's not a lot of information here.
The first part talks about precipitation.
We have a 1-gram sample added to a liter,
and that's when it precipitates.
Now, it's asking us when calcium carbide will precipitate.
And it gives us our Ksp value.
So if you're going to look at precipitation,
we have to compare Q and K. Q is going to be initial.
K is always at equilibrium.
And it's asking us if calcium hydroxide precipitates.
That's my reactant.
Ksp is always when a solid dissociates into ions.
We want to eliminate illogical possibilities.
If Q is greater than Ksp, we have too much product,
precipitate forms.
If Q is less than K, too little product, it's unsaturated,
so no precipitate forms.
So that's illogical, and this one is illogical.
So now, if you're stuck with guessing,
at least you've narrowed down what you need to do.
At this point, we actually have to do some calculations.
We need to find out how much calcium hydroxide
is theoretically present.
So mass to moles, moles to moles.
And it's in 1 liter, so I have 0.001 molar calcium hydroxide.
That means I have 0.001 calcium ion and 0.002 hydroxide.
It's a 2-to-1 mole ratio.
I can calculate Q now.
Q is given as 4 times 10 to minus 9.
It's definitely smaller than Ksp,
so no precipitate will form.
All right.
Now, we're looking at the other reaction with excess HCl
and some NaOCl and some chlorine.
So it's asking, what is the molarity of NaOCl that
will form?
So we have 100 milliliters, and we have 0.01 moles.
And from molarity, we're going to need moles of NaOCl,
and we need liters.
We've got the liters, so we've got to get to the moles.
So if I have 0.01 moles of chloride,
it's a 1-to-1 mole ratio of NaOCl to the chlorine,
if you look back at the reaction.
So I have 0.01 moles of NaOCl, moles divided by volume.
And I get 0.1 molar.
Always reference back to that original balanced equation
for your mole ratios.
Now, we flip from stoichiometry and equilibrium.
And now, we're going to do a hybridization.
So we want to look at the hybridization in reaction 3
and ask if it changes.
Draw, draw, draw.
If you want to know, you have to draw those structures.
So that first carbon, as a reactant,
it's got two things hanging off, 1, 2 sp.
So if I do the structure--
so I can cross off that sp2, is what
that line is for because I'm starting with sp.
For the second one, on the carbons,
I have three atoms or groups of atoms off of it.
And so that means I have sp2.
Three things, I've got an s, a p, and a p.
So I go to sp2, so my answer is A. Now,
let's look at the next set of questions, 22 to 25.
Refer to the following.
And I have an acid base neutralization.
And I'm titrating the acid.
And it has the reaction above, and I'm
monitoring with a pH meter.
We get that beautiful titration curve.
I love doing titration curves.
At S, we're at the half equivalence point.
So half of my base has been neutralized with my HCl.
At R, we have fully neutralized our acid.
NaOH and the HCl run out.
They are co-limiting, is what I like to call it.
And at Q and P, I have excess HCl.
So let's evaluate the questions here.
At point R on the pH curve, we want
to know what is present in the molarity of greater than 0.01.
So there's our curve again.
Now, we had our--
at R, we have our exact equivalence point,
and both acid and base have run out.
So I don't have excess H plus.
I don't have any more hydroxide.
Those were co-limiting, meaning the H
plus from the HCl and the OH minus from the NaOH
both ran out.
So all I have are those spectator ions.
Now, what would happen if I used a different molarity?
This time, I have a much stronger molarity of HCl.
How would it differ from the original curve?
So let's process.
It says the initial page would be 11 instead of 13.
Well, my initial pH depends on NaOH, and that didn't change.
At the initial, we haven't added any of the HCl.
So it's irrelevant what the HCl was.
So that's not going to change.
So that's out.
So as the pH at the equivalence point is 5 instead of 7.
Remember, R was the equivalence point.
And the only thing we had there was Na and Cl, neither of which
react with water.
So that's out.
It's not going to make it acidic or basic.
We've got a strong acid, strong base.
We have non-reactive conjugates.
It's pH 7.
Says the pH far beyond the equivalence point
would be higher.
Well, at Q and P, that was dictated by the excess HCl.
Now, I have a stronger acid.
That means I have more H plus present.
And that would make it a lower pH, not a higher pH.
So that's out.
So this says that the pH far beyond the equivalence point
would be lower.
So Q and P would be at a lower pH
to represent an excess of a higher molarity of HCl.
Well, now, we want to take a look at enthalpy.
Oh, my gosh, so much content, so little time.
So this time, we're going to add these together,
and we're going to look at the heat that is either
given off or released.
I'm not going to give that away yet.
So let's look at what the table says
and what we want to look at.
All right.
So we have a volume of 0.1 molar, is 50/50 in trial 1.
And in trial 2, we double our acid.
And we have still the amount of NaOH.
Now, this time, the question is really nice
because it says that we're looking
at the amount of energy released,
meaning this is an exothermic process.
Now, in the first one, if we have the same volume
and the same molarity in trial 1,
we have the same number of moles.
Again, we would say that is a co-limiting situation.
In trial 2, if we double the volume of HCl,
yeah, we have more moles of HCl, but we
have the same moles of NaOH that we're neutralizing.
So if we increase the moles of HCl and keep NaOH the same,
we don't expect Y to be double because we have the same amount
of limiting reagent.
And it's not going to be 2x over 3,
which is implying something about the changes
in overall volume.
100 plus 50 is 150, so that's not
going to tell us that there are more particles distributed.
That's kind of cool, but that's about entropy.
That's not about enthalpy.
The relationship can be predicted
because the amount of heat that's going to be released,
just like the amount of product that's going to be formed,
is totally dictated by that limiting reagent.
And the limiting reactant is the same in both.
It's NaOH.
So the answer is B.
Now, we're going to have to do a calculation.
We're mixing samples.
It's somewhat different than what was in the table.
But you notice, that's still all relates to that original amount
of information given for this set involving these reactions.
So you want to circle or underline all of those values.
It's asking us for heat.
Now, when we measure heat, we measure the surroundings.
The system is where bonds are breaking, bonds are forming.
That's where we're breaking the attraction between NaOH and OH
minus or H plus and Cl minus.
That's my system.
I can't get a thermometer down in there.
When I'm measuring with a thermometer,
I am measuring my surroundings.
So we can measure the heat of our surroundings by using MCAT.
So the total mass of surroundings
is the sum of the volumes.
And it tells us to assume that the density is
the same as water.
So that means for every milliliter, we have a gram.
So we know our mass for MCAT.
So mass is going to be that-- or excuse me,
heat is going to be mass times the heat capacity.
You notice I have that circle.
It's given.
Look at those units.
It's joules per gram-Kelvin.
Well, I want to tell you a little clue here.
The change in temperature is the same in Kelvin
as it is in Celsius.
So it's OK to keep that in Celsius.
And in fact, it'll save us time if you
recognize that a delta T doesn't depend on the unit here.
So my surroundings was calculated by that formula.
But my system is equal in magnitude but opposite in sign
of what my surroundings are.
So I have to change the sign.
My surroundings was plus 500 kilojoules.
So my system, which remember, is exothermic is minus 500 joules.
And when we convert those joules to kilojoules,
we get 0.50 kilojoules.
I want to take a breath here because A,
I don't see that answer anywhere,
and B, my units from Q aren't matching up
with what's in my choices here.
I have pure kilojoules, and it's asking me
for kilojoules per mole of reaction.
So I can divide this by moles of reaction.
So in this situation, you [INAUDIBLE]
to moles Volume times molarity is moles.
So my moles of HCl were equal to my moles of NaOH.
Volume times molarity is moles.
So that gives me 0.010 moles.
But I need to convert that to moles of the reaction.
So I have 0.01 moles of NaOH.
But for every 1 mole of reaction, in other words,
for every 6.02 times 10 to the 23rd times
this reaction occurs, I'm going to consume 1 mole of NaOH.
So in this case, it's a 1-to-1 ratio.
So I have 0.01 moles of reaction.
So now, I can divide that heat released
by the moles of reaction.
And I can finally find my answer as that 0.5.
Negative 50, it was A.
OK, I rushed.
That was fast.
If you need more information, longer videos, more help,
go to the College Board YouTube channel.
I've given you the QR code for the chemistry videos.
And you'll get so much help there.
In the meantime, Thanks for letting me help you.