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## Chemistry library

### Course: Chemistry library > Unit 14

Lesson 3: Solubility equilibria- Dissolution and precipitation
- Common polyatomic ions
- Introduction to solubility equilibria
- Worked example: Calculating solubility from Kₛₚ
- 2015 AP Chemistry free response 4
- The common-ion effect
- pH and solubility
- Solubility and complex ion formation

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# Solubility and complex ion formation

Calculating the solubility of silver chloride in pure water and 3.0 M ammonia. Calculating new equilibrium constant to account for formation of complex ion. Created by Jay.

## Want to join the conversation?

- Was the reason why you decided to keep the 3-2x denominator term @11:37(instead of assuming it to be 3) due to the magnitude of the result?(36 votes)
- Generally if ksp < 1.0*10^-5 you can assume the x is small enough to disregard. In this problem ksp= 2.4*10^-3 so x is too large to disregard it.(33 votes)

- In part (b), why do we need 2 NH3 to interact with 1 Ag+? How can we determine how many NH3 to be used when we are doing the question?(19 votes)
- This is a bad example (I think) because there is no way you can know this other than with practice. This will often happen with solubilities but I can assure you that when you get questions on this, you will be given the formula of the solute.(16 votes)

- At8:30why do you multiply the Ksp and the Kf to get a new K for the net reaction? What is the reasoning behind this rule?(12 votes)
- You multiply them because you're trying to get the K for the net reaction. Multiply the K's of the individual reaction for your net reaction. If you're reversing one of the individual equations, you'd change the K to 1/K.(9 votes)

- At12:00, I don't understand when we can or can't assume x is small. If the K value was 2.9x10^-5 instead of 2.9x10^-3 can I assume x is small? What is the guideline for assuming x is small? Thanks.(6 votes)
- A good guideline for the "small x approximation" is the 5% rule. It's used by comparing the x value you get, to the approximation you make in the first steps.

Let's take the dissociation of 0.1M acetic acid (I'll call it HX for simplicity sake), a weak acid with Ka = 1.8x10^-5.

HX + H2O = X- + H3O+

0.1M 0 0

-x +x +x

0.1M-x x x*[the formatting messed up here, but it's a standard ICE table with HX dissociating]*

Then Ka = 1.8x10^-5 = x^2/(0.1M-x)

Now that 0.1M-x on the denominator of the right hand side is the pain. Go ahead and approximate it to 0.1, then solve for x and you get x = 0.0134*[this should have been 0.00134, please refer to Jo Ha's comment below]*

To check if your approximation of 0.1-x = 0.1 is valid, divide x by 0.1 and see if it falls within 5%:

0.0134/0.1 = 0.134 = 13.4%

Because we exceed 5%, our approximation was invalid and we need to return and solve this problem using quadratic. This would result in x = 0.00133, which is about 10x less than what we got for x when we approximated.

And if you analyze the math in this closely, you can conclude that the small x approximation is going to be invalidated when Ka is really large (i.e., a relatively strong weak acid) and the concentration is really small.(13 votes)

- How can one know that the silver ion will react with ammonia?(3 votes)
- You can't. But, now that you know, you will remember it in the future.(5 votes)

- at3:45part b I don't understand where the 2 is coming from when he writes NH3? Why is there a two in front of NH3?(4 votes)
- This is a bad example (I think) because there is no way you can know this other than with practice. This will often happen with solubilities but I can assure you that when you get questions on this, you will be given the formula of the solute.(3 votes)

- For example b), if Ag has a +1 ion charge in the Reactants, why does it have a +2 charge in Products? This confuses me because 2 NH3's are bound to Ag in the Products when it should only just bind 1 NH3. I dont understand why the balanced equation is not: Ag^+(aq) + NH3(aq) = Ag(NH3)^+.

- I understand that NH3 is neutral; this meaning it has no charge. But why are there 2 NH3's in the Reactants?(3 votes)- NH3 is a product by itself of 1 N and 3 H, so when it mixes with Ag there shouldn't be 2 of them to make a completely neutral atom, but in this case the atom ends up having a positive charge.(3 votes)

- What's the explanation, or at least intuition behind multiplying equilibrium constants when adding equations? Is it about finding a common probability by multiplying probabilities (individual equilibrium constants) of independent events?(3 votes)
- In this problem the equations add up to form the third equation, which has an equilibrium equivalent to the product of the first two equations. What if the products and solutions cancelled out by a different factor (for example the second equation was something like 2Ag + NH3 -> some product. Would we need to multiple the Ksp twice to get the new equilibrium constant?(2 votes)
- You would have had to square the K_sp.

The equations**might**have looked something like this:

2AgCl(s) ⇌ 2Ag⁺(aq) + 2Cl⁻(aq); K₁ = K_sp² = 3.2 × 10⁻²⁰

2Ag⁺(aq) + NH₃(aq) ⇌ Ag₂(NH₃)²⁺(aq); K_i = 1.6 × 10⁷

2AgCl(s) + NH₃(aq) ⇌ Ag₂(NH₃)²⁺(aq) + 2Cl⁻(aq);

K = K₁K_i = 3.2×10⁻²⁰ × 1.6×10⁷ = 5.2 × 10⁻¹³(3 votes)

- What does the f in Kf stand for?(1 vote)
- f stands for "formation".

K_f is the equilibrium constant for the formation of the complex.(5 votes)

## Video transcript

- First let's calculate
the molar solubility of silver chloride in pure water, and then we're going to compare the solubility that we get in part A to the solubility that we're going to get in part B. And in part B, it's no longer pure water. We're going to have some ammonia present. So we take some solid silver chloride, so that's this over here, and we put it in some pure water. And some of the silver
chloride is going to dissolve, but most of it's going to
stay in the solid form. So silver chloride is only a slightly soluble compound. So there's my little pile of undissolved silver chloride. But some of it does dissolve, and so we get a saturated solution of silver chloride. So there are silver plus-one
cations in solution, and there are also chloride
anions in solution. And we have our solubility equilibrium. To find the molar solubility
of silver chloride, let's go ahead and write our ICE table. So our initial concentrations, our change, and finally our
equilibrium concentrations. Before we put the silver chloride in, we didn't have any ions in solution. So the initial
concentrations of our product are zero. We have zero for the
concentration of our ions. Some of the silver chloride
that we put in dissolved, so we lost a certain concentration of our silver chloride. We're going to call that X. So we're losing a certain concentration of our silver chloride. If we look at our mole ratios, for every one mole of silver
chloride that dissolves, we get one mole of silver
plus-one cations in solution. So if we're losing X over here, we're gaining X. We're gaining X with a concentration of silver cations in solution. And since our mole ratios
are still one to one, if we dissolve one mole
of silver chloride, we gain one mole of chloride anions. So over here we get a plus X. So at equilibrium, our concentration of silver cations is X, and our concentration
of chloride anions is X. So we write our equilibrium expression, and since this is a
solubility equilibrium, we're going to write Ksp, our solubility product constant, is equal to concentration of our products, so that's concentration of Ag plus, raised to the power of the coefficient, so raised to the first power, times the concentration
of chloride anions, Cl minus, raised to the
power of the coefficient, to the first power. So we can plug in our
solubility product constant at 25 degrees. It's 1.8 times 10 to the negative 10, so we put in here 1.8 times
10 to the negative 10. And this is equal to, this would be X times X. So this is equal to X squared. So we can get out the calculator and solve for X. So we have the square root of 1.8 times 10 to the negative 10. So we take the square root of that, and X is equal to 1.3 times
10 to the negative five, if we round that. So X is equal to 1.3 times 10 to the negative five molar. So that's the concentration of, let's say our silver
plus-one cations in solution. That's also the molar
solubility of silver chloride. That's how much we can dissolve. So it's a very small molar solubility, so silver chloride is only a slightly soluble compound. Let's see what happens when we have ammonia present instead of pure water. So we have a three molar, all right? We have some ammonia present now. What happens when ammonia
is around silver cations? We get this reaction. So silver cations plus
two molecules of ammonia give us what's called a
complex ion over here. Let's show what's happening. So if we have a silver plus-one cation, we have two molecules of ammonia, so let me go ahead and draw in my two molecules of ammonia. And this is certainly not drawn to scale, so I'll draw one in on that side, and I'll draw another one in on this side. Remember your definitions for Lewis acid, Lewis base. A Lewis base is an electron pair donor, and that's what's going to happen here. The nitrogen has a lone pair of electrons, and each ammonia molecule is going to donate a pair of electrons to our silver plus-one cation, all right? So ammonia is donating,
is an electron pair donor, so this must be our Lewis base, and the silver plus-one cation is accepting a pair of electrons, so this must be our Lewis acid. So the formation of a complex ion is a Lewis acid-base reaction. And notice the equilibrium
constant for this. 1.6 times 10 to the seventh. Kf is called the formation constant. This is a very high value for the equilibrium constant. So the equilibrium lies to the right, and this a stable complex ion. So any silver ions that are in solution, if you have ammonia present, ammonia is going to
pick up that silver ion and essentially remove the
silver ions from solution. So if we go back up here and think about our equilibrium, if we have ammonia present, we're going to be removing
silver cations from solution. So we're decreasing the concentration of silver cations. We're decreasing the concentration of one of our products. And Le Chatelier's principle says if we're decreasing the concentration of one of our products, our equilibrium's going to
shift to the right to make more, therefore increasing the
solubility of silver chloride. More is going to dissolve, because the ammonia is present because of the formation
of our complex ion. So if we added some ammonia, if we added a solution of ammonia here, we could get our silver
chloride to dissolve, because it's more soluble, because of the formation of a complex ion. So let's say we add
enough ammonia solution where we get our silver
chloride to dissolve here. So that's the idea. Adding our ammonia caused formation of a complex ion which increased the solubility of our silver chloride. So we would expect an
increased solubility here of our slightly soluble ionic compound, silver chloride. And next we're going
to do the calculation. We're going to find the
solubility of silver chloride in our three-molar ammonia solution. Now we've look at two reactions. We've looked at the solubility equilibrium for our slightly soluble compound, and we've also looked at the formation of our complex ion. Now let's add those two reactions together to get our net reaction
for what's happening. So we're going to add our
two reactions together to get a net reaction for what happens when you put silver chloride in ammonia. We have silver cations
on our reactant's side and on our product's side, so we can take those out. And so for our reactants,
for our net reaction, we have solid silver chlorides plus ammonia. And for our product's side, we would have our complex ion and our chloride anion. So we have our complex ion, and we also have chloride anions. So Cl minus. So that's the net reaction. And if we added those
two reactions together to get our net reaction, we have a new equilibrium constant, which I will call K. And if you add two reactions together to get your new equilibrium constant, you multiply the equilibrium constants of those two reactions. So K is equal to Ksp times Kf. So if you did that calculation, 1.8 times 10 to the negatives 10 times 1.6 times 10 to the seventh, you'll get your new equilibrium constant. And I won't do it here
just to save some time, but if you do that, you'll get 2.9 times 10
to the negative third. So that's a nine. 2.9. All right, our goal is to find the molar solubility of silver
chloride in our ammonia. So we set up our ICE tables. So our initial concentrations, our change, and then our equilibrium concentrations. And let's pretend like we're starting out with our solution of ammonia, and we haven't added
any silver chloride yet. So let's go back up here and let's see what our concentration was. Well, we started with a three-molar solution of ammonia, so let me put that down here. So our initial concentration for ammonia is three molar. And if we haven't added
any of our silver chloride, then we don't have any
of our complex ions, so the initial concentration of our complex ion is zero, and we also wouldn't have any of our chloride anion either, so that concentration would be zero, too. So now we add our silver chloride to our solution of ammonia, and some of our silver chloride is going to dissolve. So we make that concentration X. So we're losing a certain concentration of silver chloride. So we write minus X. Next we look at our mole ratios. So the mole ratio of solid silver chloride to ammonia is one to two. So for every one mole of silver chloride that dissolves, we're also going to lose two moles of ammonia. So we write minus two X here. For our products, if one mole of silver chloride dissolves, we make one mole of our complex ion. So if we're losing X over here, we would be gaining X over here for the concentration of our complex ion. And the same thing for the chloride anion, because our coefficient is a one. So we're going to gain X for the concentration of chloride anion. So at equilibrium, our
concentration of ammonia is three minus two X, our concentration of complex ion is X, and our concentration of
chloride anion is also X. Next, we write our equilibrium expression. So K is equal to concentration of products over reactants. So I look at my complex ion over here. So concentration of our complex ion. And we raise the concentration to the power of the coefficient. So to the first power. This is times the concentration of our chloride anion, so times the concentration
of the chloride anion raised to the power of the coefficient, so raised to the first power. We also have some reactants here. So we leave silver chloride out because that's a solid, but we have ammonia, so this is all over the concentration of our reactants, so this over the concentration of ammonia, and notice our coefficient is a two. So we need to have a two here. Next we plug in our
equilibrium concentrations. So our equilibrium
concentration of ammonia is three minus two X. Our equilibrium concentration of a complex ion is X, and our equilibrium concentration of chloride anion is X. So on the right side, we would have X to the first power times X to the, let me take that out, times X to the first power, all right, all over three minus two
X to the second power, so squared. And then we can plug in our equilibrium constant K, which is 2.9 times 10
to the negative third, so 2.9 times 10 to the negative three is equal to, on the right side, we would have X squared, so let me write this out. We would have X squared over three minus two X squared is equal to 2.9 times 10 to the negative third. And so what we could do here, these are both X squared, all right, these are both X squared, so we could just say that this is all squared on the right side. And then we could take the square root of both sides. So take the square root of 2.9 times 10 to the negative third, and take the square root
of what's on the right. So the square root of what's on the right would be X over three minus two X, and let's take out the calculator to do the other one. So square root of 2.9 times 10 to the negative third is equal to, let's round that to .054. So that just makes it
easier here to round that. So .054. Our goal is to solve for X. Our goal is to solve for
the molar solubility. So we can multiply both sides by three minus two X, so when we do that we would have this. So we need to multiply .054 by three, and if you round that, you're going to get about .16, and .054 times two is approximately .11. so we have .16 minus .11X is equal to X. So we have .16 is equal to, we put both Xs on one side, this is one X, and this is .11X. So one X plus .11X is 1.11X. And so now we need to solve for X. So we've done our math here. So we have .16, and we divide that by 1.11, and that is equal to .14. So X is equal to .14, which is only approximate because of all the
rounding that we've done. So this, this is concentration. So this is our molar solubility. If we go back up here, X is equal to .14 molar, so that's our concentration of complex ion in solution at equilibrium. That's our molar solubility
of silver chlorides. So it's .14 molar. Let's go way back up to the beginning and compare the molar solubilities. So in pure water, the molar solubility of silver chloride is 1.3 times 10 to the
negative five molar, which isn't very soluble at all. But now we have ammonia present. We've increased the solubility, and we just calculated
the molar solubility is .14 molar. And obviously, .14 molar is a lot larger than 1.3 times 10 to the negative five. So the formation of a
complex ion increased the solubility of our
slightly soluble compound, silver chloride.