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## Chemistry library

### Course: Chemistry library > Unit 19

Lesson 1: 2015 AP Chemistry free response questions- 2015 AP Chemistry free response 1a
- 2015 AP Chemistry free response 1b and c
- 2015 AP Chemistry free response 1d
- 2015 AP Chemistry free response 1e
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- 2015 AP Chemistry free response 2c
- 2015 AP Chemistry free response 2d and e
- 2015 AP Chemistry free response 2f
- 2015 AP Chemistry free response 3a
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- 2015 AP Chemistry free response 4
- 2015 AP Chemistry free response 5
- 2015 AP Chemistry free response 5a: Finding order of reaction
- 2015 AP Chemistry free response 6
- 2015 AP Chemistry free response 7

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# 2015 AP Chemistry free response 4

Calculating the molar solubility of calcium hydroxide in a 0.10 M calcium nitrate solution. From 2015 AP Chemistry free response 4.

## Want to join the conversation?

- Ca(NO₃)₂ is already in the solution. So how come we are not including the concentration of NO₃⁻ in the equation for Ksp?(4 votes)
- It is because Ksp only specifies the concentration relations between Ca2+ and (OH)- regardless of other substances.(8 votes)

- in writing the balanced equation I ended up with:

Ca(OH)2 + H2O ---- CaO + 2H20

why is it wrong?(3 votes)- Because they specifically asked for the dissolution reaction. When ionic compounds dissolve in water they turn into the ions that make it up, which in this case is Ca^2+ ions and OH^- ions.(4 votes)

- In part (a), why is it a double arrow? I thought that Ca(OH)2 is a strong base and would dissociate completely .(2 votes)
- Regardless of it being a strong base, Ksp expressions are written using double headed arrows(3 votes)

- what is molar solubility in b part(2 votes)
- The maximum concentration of a compound in a solution.

EDIT: Maybe even better would be the maximum amount of a compound that can dissolve in a solution, expressed as its final concentration ...

see: https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/copy-of-solubility-equilibria-mcat/v/solubility-product-constant-from-the-solubility(2 votes)

- around5:00, how did you already have 0.10M of Ca+2 ions?(0 votes)
- You are starting with 0.10 mol/L Ca(NO₃)₂.

This is a strong electrolyte. It dissociates completely in solution.

Thus, 0.10 mol Ca(NO₃)₂ forms 0.10 mol Ca²⁺ and 0.20 mol NO₃⁻.

Then you add the Ca(OH)₂, which produces more calcium ions.(6 votes)

- I wanted to calculate the molar solubility of CaCO3 in .050 M CaCl2 (Ksp for CaCO3 is 3.8 x 10^-9). I followed the steps in this video but with using the different numbers/exponents and got 2.8 x 10^-4, but the answer choices it gave were: a) 7.6 x 10^-8 b) 1.9 x 10^-7 c) 6.2 x 10^-5 d) 8.3 x 10^-5 e) 5.0 x 10^-2. What did I do wrong?(2 votes)
- I can't tell where you went wrong. You may simply have made a calculation error.

The correct answer is c) 6.2 × 10⁻⁵ mol/L.

CaCO₃ ⇌ Ca²⁺ + CO₃²⁻; Ksp = 3.8 × 10⁻⁹

_________x__x+0.050

Ksp = [Ca²⁺][CO₃²⁻] = 3.8 × 10⁻⁹

x(x+0.050) = 3.8×10⁻⁹

0.050/3.8×10⁻⁹ = 1.4 ×10⁷ >> 400; ∴ x ≪ 0.050

x² = 3.8×10⁻⁹

x = 6.2× 10⁻⁵

[CaCO₃] = [Ca²⁺] = x mol/L = 6.2 × 10⁻⁵ mol/L(1 vote)

- So Sal essentially neglects the x in (x+.1), but how did he know that the calcium from the calcium hydroxide would be significantly less than .1M, what if someone had put in 5 times as much Ca(OH)2 as they had Ca(NO3)2, where did he get the intuition that x<<.1M?(1 vote)
- A rule of thumb is that if x is less than 5% of the number from which it is added or subtracted then the "small x" approximation is valid. Otherwise, it's necessary to solve the quadratic equation. In this video, x was 1.8% of 0.1 so the approximation was valid.(1 vote)

- at2:56, why does Khan say "but that's going to be constant

because if there's any of the calcium hydroxide that gets dissolved, well the volume goes down,and if any gets formed, the volume goes up. So the concentration stays constant."

I don't think the logic holds up here: clearly the concentration is fluctuating, but then Khan says it stays constant.(1 vote)- The reactions where calcium hydroxide is dissolving and solidifying are happening simultaneously. So any amount of solid calcium hydroxide which dissociates into ions is immediately replaced by ions in solution solidifying back onto the block of sodium hydroxide. Therefore the concentration of the solid calcium hydroxide remains constant because the two rates are equal.

Hope that helps.(1 vote)

- On part C (around9:10) is it necessary to label the H's and O's or the + and - sides of H2O, or can you get full credit without doing that?(1 vote)
- So why did we have to include the 0.1M Ca(NO3)2 in our calculation in b? Does the fact that we already have some Ca in the solution decrease the solubility of our Ca(OH)2 ?(1 vote)
- What you say is right. This is an example of Le Chatelier's Principle which applies to reversible reactions (https://www.khanacademy.org/science/chemistry/chemical-equilibrium/factors-that-affect-chemical-equilibrium/v/le-chatelier-s-principle). The partial dissolution of an insoluble salt is reversible and if you add more of a common ion, it pushes the equilibrium to the left, meaning that less of the salt dissolves.(1 vote)

## Video transcript

- [Voiceover] Answer
the following questions about the solubility of calcium hydroxide, and they give us the solubility product. Write a balanced chemical equation for the dissolution of
solid calcium hydroxide in pure water. So we're going to start
off with calcium hydroxide. And it is solid, and it is going to be in equilibrium. When it's dissolved, you have calcium ions that have a positive two charge. Writing aqueous here since it's dissolved in our solution, a solution of water, and then plus you have
your hydroxide ions, also dissolved in the water. For every calcium, you
have two hydroxides. We see that we are balanced, one calcium, one calcium, two hydroxides, two hydroxides. All right, let's do the next part. Calculate the molar solubility
of calcium hydroxide, not in pure water, but in a solution that already has a concentration of another salt, calcium nitrate. It's a .1 molar concentration, .10 molar concentration
of calcium nitrate, which you might recognize is a very soluble salt. It's very soluble, it's a nitrate salt, which are very soluble. So how do we think about this? How do we think about the molar solubility of calcium hydroxide? How much incremental calcium can be dissolved when you already have some calcium? And for whatever the
concentration of calcium that gets dissolved, you're gonna have twice the concentration of
hydroxide that gets dissolved. You might have guessed,
or intuited, or have known that this will involve
the solubility product. So the solubility product is equal to the concentration of the product of the concentrations of the two things that
actually get dissolved. So you have your calcium ions and you have your hydroxide ions, hydroxide ions, and since you two have moles of this for every mole of the calcium, we square it just like that. Now some of you might be saying, hey, this looks kind of like an equilibrium constant, but if we were dealing with
an equilibrium constant, we would divide by the concentration of the reactants. And we could have done an
equilibrium constant here, used an equilibrium constant, but the concentration of the reactant, this is in its solid state. If we imagine what's
happening right over here, we have a block, you can imagine having a block of solid calcium hydroxide here inside the water, and we've been saturated. So we've saturated as
much of the ions as we can into our solution. And so every time you have
some calcium hydroxide that gets dissolved, you have an equal amount that forms from the solution. And so if you were doing
a equilibrium constant, you would divide by the concentration of the solid calcium hydroxide. We're not used to thinking
about concentrations of solids in terms of molarity and things like that, but that's going to be constant because if there's any
of the calcium hydroxide that gets dissolved, well
the volume goes down, and if any gets formed,
the volume goes up. So the concentration stays constant. The convention is you multiply both sides by that to get a solubility product, and that's where the
solubility product comes from. I'll do another video that goes into much more depth than that. But anyway, how do we use this? Well, we already have, let me write this down, we already have some calcium ions. So we already have 0.10 molar calcium ions dissolved. I don't even have to
write dissolved there. We already have that in the solution. We are going to add x molar, I guess I could say, calcium ions through dissolving the calcium hydroxide. And that means that we
would have two x molar concentration, I could write it this way, concentration of hydroxide because for every,
whatever the concentration of calcium ions, you're gonna have twice the concentration of hydroxide because for every calcium molecule, you're gonna have two hydroxide molecules. So how do we use that with this solubility product? Well, you already have
this right over here, you already have 0.10 already, and then you're going to add x. So that's that right over there. I think this notation that I wrote here might be a little bit confusing. So bear with me. So maybe not putting the brackets there will make it a little
bit easier to understand because there's clearly already a concentration. All right, so this is going
to be the concentration of calcium ions, the ones that you already had plus the ones that you add. And then your concentration of hydroxide, well, that's going to be two x. For every calcium you add, you're gonna add twice as many hydroxides. And of course, you square that. That�s going to be equal
to the solubility product, which they tell us is
1.3 times 10 to the -6th. So we have 1.3 times 10 to the -6th. Now if you try to solve this outright, it's going to get quite complicated. You're gonna end up, this is going to be four x squared, and you multiply it times this, you're gonna get an x squared term, and then you�re getting
an extra third term. Not that easy to solve. A simplifying assumption that we make when we solve these, especially because this nitrate salt is so soluble, much more soluble than this salt right over here, this hydroxide salt, and so you can assume this is, this part right over here is approximately 0.1 since x is much smaller than 0.10. That's the assumption you make. You have to know that okay, well first of all, this is going to be hard to solve the way it is, but then the fact that we're adding a much less soluble thing to something that is nowhere close to saturation, this is a
very high concentration. We can go back fill to see if we make this assumption, whatever we solve for x, if it gets us still pretty close to this solubility product. So if we make that assumption, we get 1.3 times 10 to the -6th is equal to 0.10 times four x squared, or we could get x squared is equal to 1.3 times 10 to the -6th divided by 0.10, and then times four, or we could say divided by 0.40, or we could say that x is going to be the square root of this. So x is going to be equal to, let's get a calculator out, so we have 1.3 EE times 10 to the -6th divided by .4 is equal to that, and now we want to take
the square root of that. And let's see, we're dealing with how many significant figures are over here? Well looks like it's about two. So 0018. So approximately, let me write that a little bit neater, approximately 0.0018 molar. Let me see if that's, is that what I got? Yeah, .0018 molar. And so that's our answer. If you want to feel better about it, you can substitute this back into our original equation here and say, okay, this does still get me something close to 1.3 times 10 to the -6th. And if you try that, you will see that that
is actually the case. Now let's do this last part. In the box below, complete the particle
representation diagram that includes four water molecules with proper orientation around the calcium ions. Well, the water molecules we know are polar. If this is the oxygen here and this is a hydrogen, this is a hydrogen, we have our lone pairs of electrons here. So you have a partially
negative charge there, and you have partially positive charges at the hydrogens that get
their electrons hogged, and so you're gonna orient it. So they just want us to draw four. So the oxygen end is going to be oriented towards
the positive calcium because you have a
partially negative charge being attracted to the positive charge. So that's one, that is two, that is three, and that is four. They told us to draw only
four right over there. If you want to be clear, you can say those are oxygens right over there, and these are the
hydrogens right over here. The hydrogens have a partially, that end of the molecule has
a partially positive charge so it's going to be repelled from the calcium ion, and then the other side
that has the lone pairs, well, that's gonna have a
partially negative charge so it's going to be attracted to the positive charge. And there you go.