Nuclear Bombs

The amount of energy associated with nuclear reactions is massive compared to the amount of energy associated with chemical reactions. Although this is true in power plants and engines, it is much more obvious in explosions. A chemical explosion can destroy a large building, but even a little nuclear explosion can destroy a small city.

In the bomb that the United States dropped on Hiroshima Japan, the reaction was the fission of Uranium.  In this reaction, a neutron added to a uranium atom causes the uranium to split, spitting out additional neutrons. Each uranium atom, has the potential to "set off" three additional uranium atoms. As a result a few uranium atoms splitting can very quickly turn into a massive reaction which releases a tremendous amount of energy.

Nuclear reactions, unlike chemical reactions are single atom events governed by probability. This means, that an individual atom of uranium could decompose and spit out several neutrons. Those neutrons could hit other uranium atoms and start a chain reaction. However, this is unlikely in a small amount of uranium, of if the uranium is spread out enough, because the neutrons are more likely to be lost than to set off a chain reaction. A sample of uranium is called a critical mass if it is large enough and packed tightly enough that neutrons are likely to start a chain reaction. A smaller or less packed sample is called sub-critical.

 The trick, then, is to build a bomb of sub-critical samples and then to make them suddenly critical at the moment that you want the bomb to explode. This can be done several ways. 

One way to make a sample suddenly critical is to have several sub-critical masses suddenly pushed together. Imagine you had an incomplete sphere of uranium that was just short of critical. The "missing piece" of the sphere is separated along a tube. At the correct moment, that piece is "shot" into the sphere, reaching a critical mass.

Another way to do it is to have a critical mass, but to arrange it in a sphere that is not tightly packed enough to be critical. The entire sphere is wrapped in coventional explosives. Setting off those explosives, forces the uranium into a tighter sphere, reaching criticality.

Of course, the problem is, that uranium can emit a neutron on its own, so that if you had a large enough pile of uranium, it might go off on its own, when you didn't want it to. That "enough" is called a critical mass, and bomb makers are careful to never get a critical mass together while building their bomb.

 

Will a Precipitate Form?

We’re going to solve two different problems on this page. The first is relatively straight-forward, the second is the same sort of problem, but has an added issue that we’ll need to address.

A simple Problem 

 Let’s start by solving this problem:

A solution is made that is 0.0035 M \(Pb(NO_3)_2\) and 0.070 M \(NaCl\). Will a precipitate form? The \(K_{sp} ~of~ PbCl_2\) = \(1.7 \cdot 10^{-5}\).

There are several things we need to think about before we can start the problem:

• Since we are given the \(K_{sp}\) of \(PbCl_2\) then the relevant reaction must be the dissolving of \(PbCl_2\) even if that seems to come from out of nowhere.
• All nitrates are soluble (check your solubility table). That means that we don’t need to care about the \(NO_3^{-1}\) ions, on the \(Pb^{+2}\) ions, which have a concentration of 0.0035 M.
• All sodium compounds are soluble. So, like the nitrates, we can ignore the \(Na^{+1}\) ions and the \([Cl^{-1}]\) = 0.070 M.

Here is our relevant reaction and the current ICE table:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

It is tempting to think that this reaction must run backwards, since no reactant is present, but this is a solubility problem. The reactant is a solid, is not included in the math and therefore is NOT relevant.

What we actually need to do is to figure out what the question actually is. “Will a precipitate form? means “will a solid appear in the container?” For that to happen, the ions in solution need to stick together and form the solid compound. In other words:

“Will a precipitate form?” = “Does the reaction run LEFT?”

That question, of course, is really asking: Is \(K_{sp} \lt Q_{sp}\)?

And that is simple math.

\(K_{sp}\) > or = or < \(Q_{sp}\)

\(1.7 \cdot 10^{-5}\) ? \([Pb^{+2}] \cdot [Cl^{-1}]^2\)

\(1.7 \cdot 10^{-5}\) ? \([0.0035] \cdot [0.070]^2\)

\(1.7 \cdot 10^{-5} \lt 1.72 \cdot 10^{-5}\)

Since \(K \lt Q\), the reaction will run to the left and the answer is YES (a precipitate will form).

A More Complex Problem

Now, let's try a more complex problem:

A solution is made by mixing 100 mL of a 0.0035 M \(Pb(NO_3)_2\) solution and 200.0 mL of a 0.070 M \(NaCl\) solution. Will a precipitate form? The \(K_{sp} of PbCl_2 is 1.7 \cdot 10^{-5}\).

In the last problem, we were given the concentrations of the ions in the solution which had already been made. In this problem we are given volumes and concentrations of the two solutions that are being mixed. Since each compound will end up in a larger volume than it had been before, the concentrations of each must be determined before we can do the check for precipitation.

Here is the work for the lead II nitrate. We’ll do a FULL version of the math and then find the quicker and easier way.

The full version:

\(100 ~mL ~Pb(NO_3)_2 \cdot \frac{1 L}{1000 mL} \cdot \frac{0.0035~moles}{1 L~solution}= 3.5 \cdot 10^{-4} mols~PbCl_2\)

Then, calculating the new molarity of the \(Pb(NO_3)_2\) in the mixed solution with its volume of 300 mL (0.300 L) requires this math:

\(M~=\frac{moles}{L}=\frac{3.5 \cdot 10^{-4} mols~PbCl_2}{0.300~L}= 0.00117 M\)

There is a simpler and faster way to get the same value. We need to notice that the \(Pb(NO_3)_2\) started as 100 mL and ended up in 300 mL. We can, therefore, do the following math:

\(0.0035 \frac{mols}{L}\cdot \frac{100~mL}{300 ~mL} = 0.00117~M\)

or even more simply:

\(0.0035 \frac{mols}{L}\cdot \frac{1}{3} = 0.00117~M\)

The NaCl goes from 200 mL to 300 mL, so we can calculate that molarity:

\(0.070~M~NaCl \cdot \frac{2}{3}= 0.0467~M~NaCl\)

Finally, we can address the actual question, which was “Will a precipitate form?” which we know means: is \(K_{sp} \lt Q_{sp}\)?

So, with our newly determined concentrations, we can solve the problem: 

\(K_{sp} \gt ~or~ = ~or~ \lt Q_{sp}\)

\(1.7 \cdot 10^{-5}\) ? \([Pb^{+2}] \cdot [Cl^{-1}]^2\) 

\(1.7 \cdot 10^{-5}\) ? \([0.00117] \cdot [0.0467]^2\) 

\(1.7 \cdot 10^{-5} \gt 2.55 \cdot 10^{-6}\)

So, in this case, the answer is NO (a precipitate will not form).

Finding Solublity in g/100 mL

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the solubility of the compound in \(\frac{g}{100~mL}\)?

This problem exists because we don’t have devices that measure moles directly. Therefore, knowing the molar solubility is not all that helpful. In the end, this problem is about turning molar solubility into a usable value.

We start this problem by finding the molar solubility of the compound. Given that we solved this problem on this page, we’ll jump right to the answer for that portion:

Molar solubility = “x” = 0.016 M.

The first thing we need to do is fix the unit moles, by converting it to grams. That’s a relatively simple problem using molar mass:

\(0.016 \frac{moles~PbCl_2}{L} \cdot \frac{278.10~g~PbCl_2}{1~mole~PbCl_2}=4.45 \frac{g}{L}\)

This is now a useful and workable amount. It tells us that we can dissolve, at most, 4.45 g of \(PbCl_2\) in a liter of solution.

For some reason, that is beyond the reasoning of this author, it was decided that solubilities should be recorded in terms of the grams that will dissolve in 100 mL, rather than in 1 L.

Solving this is easy...IF you think. If you try to set up a conversion process, you will get it wrong. I promise.

Here’s the simple logic. 100 mL is \(\frac{1}{10}\) of a liter. Therefore you can just divide the value we had before by 10.

So, the solubility of \(PbCl_2\) is 0.445 \(\frac{g}{100~mL}\).

Finding Molar Solublity

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the molar solubility of the compound?

This problem is the SAME as finding the concentrations in a saturated solution, with the exception that the answer is ALWAYS “x”. Be sure to read all the way to the bottom of this page to make sure that you understand the implications of that statement!

Our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

Simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x=0.016 M

Molar solubility is “x” therefore the answer is 0.016M.

What's interesting here is that it does NOT matter what the formula for the ionic compound is, the answer is always "x". 

A simple ionic compound like \(AgCl\) would simplify to:

\(K_{sp} = x^2\)

while a compound like \(Pb_3(PO_4)_2\) would work out to be:

\(K_{sp} = 108x^5\)

However, in both cases, the left column of the ICE table would be "some - x" and therefore the molar solubility is just "x".

Finding the Concentrations - Common Ion

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\). What is the concentration of chloride ion 

in a saturated solution that already has \([Pb^{+2}]\ = 0.25~M\)?

There are several things that are implied, but not stated, that we need to recognize:

The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).

The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [0.25 + x] \cdot [2x]^2\)

This will be a “little x” problem, so we can assume that 0.25 + x = 0.25. That simplifies the math to:

\(1.7 \cdot 10^{-5} = (0.25)\cdot 4x^2\)

which gives us x=0.0041 M

Since \([Cl^{-1} = 2x\) we know that \([Cl^{-1}]=0.0082~M\).

This is an example of the common ion effect. Simply put, the presence of lead ions in the solution did not allow as much of the lead II chloride to dissolve as would have in pure water.

Another, perhaps more useful, way to think about the common ion effect is to relate it to Le Chatelier’s Principle. For the reaction:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

additional amounts of \(Pb^{+2}\) ions will drive the reaction to the left, reducing the amount of chloride ion in solution.

Finding Concentrations in a Saturated Solution

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the concentration of chloride ion in a saturated solution?

There are several things that are implied, but not stated, that we need to recognize:

• The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).
• The solution is made by adding solid \(PbCl_2\) to pure water. This is implied by the term saturated solution. If the situation was anything OTHER than solid added to water, we would be told.
• The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

Although writing “some” in the midst of an ICE table seems odd, we can get away with it, since the solid is not part of the equilibrium expression.

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~or~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x = 0.016 M

Since \([Cl^{-1}] = 2x\) we know that \([Cl^{-1}]=0.032~M\).

Solubility

Understanding solubility requires us to understand how things (especially ionic compounds) dissolve, reversible reactions, and the mathematics of equilibrium - especially “little x” problems.

Let’s look at solubility with respect to one particular ionic compound - lead II chloride (\(PbCl_2\)). All of the ideas that are developed here, will then be applicable to other compounds. We can write the process of lead II chloride dissolving as a simple reaction:

\(PbCl_{2~(s)} \rightleftharpoons PbCl_{2~(aq)}\)

Although that reaction is correct, it isn’t very useful. It is much more interesting to write it as an ionic equation:

\(PbCl_{2~(s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

Remembering that solids are not included in the equilibrium expression, we can write the following:

\(K=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

Since this equilibrium expression has an “odd” look (nothing on the bottom), we give it a special name and add a subscript to the K:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

where \(K_{sp}\) stands for the Solubility Product.

There are several important ideas embedded in this:

1. \(K_{sp}\) is always for an ionic solid dissolved in water breaking up into its constituent ions
2. Any problem that mentions \(K_{sp}\) does not need to give the reaction, because it is implied.

There is also some vocabulary that flows from this:

• Saturated solutions are solutions for which \(Q_{sp} = K_{sp}\). In other words, a saturated solution is at equilibrium. Remember that this does NOT mean that the process of dissolving has stopped. It means that the rate at which ions are entering solution is matched by the rate at which ions are coming together and dropping out of solution. Therefore the concentrations of the ions will not be changing, nor will the amount of solid present. In addition, no solid must be visibly present. Saturation is based on the concentrations of dissolved ions only.
• Unsaturated solutions are those that can hold more solute than is currently dissolved. That is \(K_{sp} > Q_{sp}\). If solid is present, then the rate of solvation is faster than the rate of precipitation and the concentrations will be rising. If no solid is present, then the solution will remain as it is. 
• Supersaturated solutions are those where “too much” solute is dissolved. That is, solutions where \(K_{sp} < Q_{sp}\). Generally, we would expect such a situation to result in ions in solution coming together and precipitating, thus decreasing the concentrations of the ions and increasing the amount of solid present in the container. In rare occasions, when the crystalline structure of the solid is complex, it is possible for a solution to temporarily remain in a state of supersaturation until random chance allows the beginning of precipitation or until crystalization is seeded.
• Molar Solublity is the amount of a solid that will dissolve in water in units of \(\frac{moles~solid}{liters}\)

The only way to really understand all of this is, of course, to look at some problems. We’ll look at five different problems: 

1. Finding the concentrations of ions in a saturated solution 
2. Finding the concentrations of ions in a saturated solution in which one ion was already present (this is called the common ion effect) 
3. Finding the molar solubility of an ionic compound 
4. Finding the solubility of an ionic compound in \(\frac{g}{100~mL}\) 
5. Determining whether a precipitate will form when solutions are mixed