Strong/Strong Titration Graph

Let's look at how the pH changes during a strong/strong titration from a graphical perspective. (A weak titration graph is found here) 

We will be looking at this for the problem here:

50.00 mL of 2.00 M HA (an imaginary acid) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

We did all of the calculations here, so if you haven't looked through those yet, it would be worth checking out.

If we graph the points from that problem, we get this:

Adding in a few more points to "smooth" things out a bit, we get this:

In simplest terms, at the beginning, when we have excess acid, the solution is acidic (with a low pH). When we to equivalence, the pH becomes 7. Once we pass the equivalence point, the solution contains excess base and the pH is high.

The equivalence point is the middle of the steep vertical part of the graph. Stated differently, the equivalence point is the point in the graph where the slope is the most vertical.

We can find that point graphically, by making a graph of the slope vs. volume of base added. For those of you with some level of higher math, this is simply the derivative of the previous graph.

If it helps to visualize what this graph is showing, here are the two graphs on the same axes:

Determining the pH During a Strong/Strong Titration

A strong/strong titration, that is a titration using a strong acid and a strong base, is really an exercise in stoichiometry and limiting reagents as described here. (A weak/strong titration problem is done here.)

It is important to remember that stoich is done with moles, and the problem we are trying to solve (shown below) gives us volume and molarity. We will need to deal with that by remembering that:

\(moles = Molarity (\frac{moles}{liter}) \cdot Volume (L)\)

Here's the problem we are going to look at:

50.00 mL of 2.00 M HA (an imaginary acid) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

The first point is easy. We have 2.00 M strong acid and we know that the \([H_3O^{+1}]\) of a strong acid is just the concentration of the acid. Remembering that pH is the -log of \([H_3O^{+1}]\) means we can do the following math:

\(-log[H_3O^{+1}] = -log(2.00) = -0.301\)

To solve the rest of the problem, we are going to need to keep track of the following:

• volume of base added
• total volume of the solution
• moles of acid (\(M_a \cdot V_a\))
• moles of base (\(M_b \cdot V_b\))

We can do all this by putting the relevant information into a table, where each column will be one of the bits of information listed above and each row of the table will be a step (a different volume of base) in the problem.

Let's tackle the first addition of base. We can start by filling out what we know:

Then we know that strong acids and bases react ~100%, so we can simply subtract to determine the amount of acid that remains:

Lastly, we can find the concentration of the acid (\(\frac{moles}{liters}\)) and then pH.

Each step before the equivalence point works the same way

and

and

Suddenly, things change dramatically -- at the equivalence point the initial reaction leaves neither acid or base in the solution.

After the equivalence point, things change again. In this case, we have used up all of the acid, and have extra base in solution. 

This means that we can calculate the \([XOH]\). Since this is a strong base, the \([OH^{-1}]\) will be the same as \([XOH]\). We can then take the -log of \([OH^{-1}]\), which will be the pOH, and subtract that from 14 to get the pH.

Now that you've been through the mathematics of this process it is worth looking at how the pH changes graphically.

How pH changes during a weak titration - the big ideas

Let's look at how the pH changes during the titration of a weak acid with a strong base. For this, we'll assume that you've already mastered strong/strong titration theory, math and graphs.

So, for the discussion on weak acids, we’ll focus our thoughts on this problem, which is a slightly altered version of the strong/strong titration problem we did before.: 

50.00 mL of 2.00 M HA (an imaginary weak acid with a \(K_a=4.23\cdot10^{-6}\)) are titrated with 1.00M XOH (an imaginary strong base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

First, let’s consider what is happening in the flask. At the beginning, there is only weak acid in the flask, so the pH is relatively low. Determining that actual pH will require us to do an equilibrium problem (with an ICE table) using \(K_a\) and the reaction below:

\(HA + H_2O \rightleftharpoons H_3O^{+1} + A^{-1}\)

When we start to add the base, a reaction will occur between the acid and base. Since the base is strong, that reaction will (effectively) go 100%. This makes for a simple limiting reagent problem.

Say we have 10 moles of acid in the flask and add 2 moles of base. The following reaction will occur: 

\(HA + XOH \rightarrow HOH + X^{+1} + A^{-1}\)

Since the acid and base react in a 1:1 ratio, we will use up 2 moles of the acid, leaving 8. 

\(2~moles~XOH \cdot \frac{1~HA}{1~XOH} = 2~moles~HA~used\) 

and

\(10~moles~HA~originally~present~-2~moles~HA~used = 8~moles~HA~remaining\)

Since unreacted acid remains in the flask, the pH will still be relatively low. We can again do an equilibrium problem using \(K_a\) to find the \([H_3O^{+1}]\) and then the pH. However, it has now become more complicated. Since the acid (HA) is weak, it's conjugate base (\(A^{-1}\)) cannot be ignored. It is easy to find that amount, since it is produced in a 1:1 ratio with the amount of acid that was used up in the reaction. So, 

 \(2~moles~HA \cdot \frac{1~A^{-1}}{1~HA} = 2~moles~A^{-1}~produced\)

This amount of \(A^{-1}\) will also appear in the "I" row of the ICE table.

For all points up to (but not including) the equivalence point of the titration this will be our situation: we will react the strong base completely. That reaction will leave some of the weak acid and will create some of the conjugate base. Solving for pH will require us to do an equilibrium problem using an ICE table that initially contains both the remaining acid and the created conjugate base. 

At the equivalence point, things will change. Again, let's imagine that we started with 10 moles of the acid, but now we have added 10 moles of the base. The same reaction occurs, again ~100%. This time, however, we use up all of the acid and all of the base. That leaves us with only the conjugate base (in this case \(A^{-1}\)). The pH will not be 7 because the solution contains a base. 

To determine the pH we will need to rethink what can happen in the solution. Our previous reaction can't occur any more since we have no HA present. Instead, the base will react with water like this

\(A^{-1} + H_2O \rightleftharpoons HA + OH^{-1}\)

We know the amount of base present (from our simple stoichiometry) so we can use that in the "I" row of an ICE table but, of course, we need an appropriate K value. Since this is a basic reaction, we need a \(K_b\). Remembering that 

\(K_w=K_a \cdot K_b\) 

we can rearrange to solve for \(K_b\)

\(K_b=\frac{K_w}{K_a}\)

All of that will allow us to find the \([OH^{-1}]\). From that we can calculate the pOH and then the pH. 

Once we pass the equivalence point, things change again. Now the acid becomes the limiting reagent. Again, let's imagine that we started with 10 moles of the acid, but now we have added 12 moles of the base. Since the acid and base react 1:1, we will use all of the acid and 10 moles of the base, leaving 2 moles of base unreacted. 

\(10~moles~HA \cdot \frac{1~XOH}{1~HA} = 10~moles~XOH~used\) 

and

\(12~moles~XOH~added~-10~moles~XOH~used = 2~moles~XOH~remaining\)

Since there is extra base in the flask, the pH will be high. And, more importantly, since the base is strong, we will be able to find \([OH^{-1}]\) easily since, for strong bases \([OH^{-1}] = [XOH]\)

How pH Changes During a Titration

Titration, as first discussed here, involves adding one solution to another. Most commonly, these are acid and base solutions and we generally add a basic solution to an acidic one. We are going to work through these ideas using that as our model, but it is important to know that acids can be added to bases and that titration can be done with solutions that are not acidic or basic. 

For this discussion, we’ll focus our thoughts on this problem: 

50.00 mL of 2.00 M HA (an imaginary acid) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

How this plays out, and as a result, how we deal with it depends on whether the acid and base are strong or weak. On this page, we will assume that both the acid and the base are strong. On this page we will look at the situation if the acid is weak (and the base is strong).

First, let’s consider what is happening in the flask. At the beginning, there is only strong acid in the flask, so the pH is low.

When we start to add the base, a reaction will occur between the acid and base. Since both are strong, that reaction will (effectively) go 100%. This makes for a simple limiting reagent problem.

Say we have 10 moles of acid in the flask and add 2 moles of base. The following reaction will occur: 

\(HA + XOH \rightarrow HOH + XA\)

Since the acid and base react in a 1:1 ratio, we will use up 2 moles of the acid, leaving 8. 

\(2~moles~XOH \cdot \frac{1~HA}{1~XOH} = 2~moles~HA~used\) 

and

\(10~moles~HA~originally~present~-2~moles~HA~used = 8~moles~HA~remaining\)

Since unreacted acid remains in the flask, the pH will still be low. This will be the situation up to (but not including) the equivalence point of the titration. 

At the equivalence point, things will change. Again, let's imagine that we started with 10 moles of the acid, but now we have added 10 moles of the base. The same reaction occurs, again ~100%. This time, however, we use up all of the acid and all of the base. That leaves us with only the ionic product of the reaction (in this case XA). the pH of this solution will be 7. (The reason is explained here.)

Once we pass the equivalence point, things change again. Now the acid becomes the limiting reagent. Again, let's imagine that we started with 10 moles of the acid, but now we have added 12 moles of the base. Since the acid and base react 1:1, we will use all of the acid and 10 moles of the base, leaving 2 moles of base unreacted. 

\(10~moles~HA \cdot \frac{1~XOH}{1~HA} = 10~moles~XOH~used\) 

and

\(12~moles~XOH~added~-10~moles~XOH~used = 2~moles~XOH~remaining\)

Since there is extra base in the flask, the pH will be high.

Let's look at how this plays out in the context of our problem above.

Hydrolysis

When a solid acid or a compound containing hydroxides is dissolved, the pH of the solution is affectedin obvious ways. However, many ionic compounds which are not obviously acidic or basic also affect the pH when they are dissolved. 

Understanding why ionic compounds can influence the pH of a solution when they dissolve requires us to think about the ions produced during the solvation of the compound. For instance when sodium acetate is dissolved in water it produces sodium ions and acetate ions:

\(NaC_2H_3O_2 \rightleftharpoons Na^{+1} + C_2H_3O_2^{-1}\)

This, by the way is why we call this hydrolysis. "Hydro" means water, and "lysis" means splitting or breaking. So, water breaks apart ionic compounds - hydro-lysis.

Then what?

Acetate, as a negative ion, is able to "attract" and \(H^{+1}\) from a water molecule creating acetic acid and hydroxide ions (\(OH^{-1}\)):

\(C_2H_3O_2^{-1} + H_2O \rightleftharpoons HC_2H_3O_2 + OH^{-1}\)

The creation of the hydroxide ions, makes the pH of the solution go up.

Of course, you may be wondering about the sodium ions. Since \(Na^{+1}\) doesn't have an \(H^{+1}\) to give to water and the positive charge won't attract any \(H^{+1}\) ions from water, the \(Na^{+1}\) will not have any effect on the pH.

The complication of strong acids

Of course, every negative ion can act as a base (attracting \(H^{+1}\) from water), but not all solutions are basic. The reason has to do with how strong a base the negative ion is. For example, chloride ion (\(Cl^{-1}\)) can act as a base in water according to the reaction here:

\(Cl^{-1} + H_2O \rightleftharpoons HCl + OH^{-1}\)

However, solutions made of chloride are not basic. To understand that we need to think about HCl (the acid on the right side of the reaction above. We know that HCl is a strong acid. As such, reactions involving HCl run (nearly) to completion. That means that in the reaction above, the backward reaction occurs ~100%. Stated differently, the forward reaction happens ~0%. In fact a better way to write that reaction might be:

\(Cl^{-1} + H_2O \leftarrow HCl + OH^{-1}\)

So, because the conjugate of chloride is a strong acid, \(Cl^{-1}\) is a horrible base and can be ignored. This is the reason that a solution of table salt (sodium chloride, NaCl) is neutral -- the \(Na^{+1}\) is not an acid or a base and the \(Cl^{-1}\) is such a bad base that it has no measurable effect.

One more example

Let's ponder what happens when ammonium nitrate (\(NH_4NO_3\))is dissolved in water. The compound breaks up into ions as it dissolves.

\(NH_4NO_3 \rightleftharpoons NH_4^{+1} + NO_3^{-1}\)

The nitrate ion (\(NO_3^{-1}\)) is the conjugate of the strong acid \(HNO_3\) and, as such, has no effect on the pH of the solution.

The ammonium ion (\(NH_4^{+1}\)), however, is the conjugate of the base ammonia (\(NH_3\)). It can react with water:

\(NH_4^{+1} + H_2O \rightleftharpoons NH_3 + H_3O^{+1}\)

Since the reaction makes hydronium ions, the pH will go down and the solution will be acidic.

OK, but what about...

Using the logic above, you should be able to determine whether an ionic compound will make a acidic, neutral or basic solution when dissolved. But, there is one situation where the situation gets complicated: when the solid is composed of a positive ion that is a weak acid and the negative ion is a weak base.

An example of this type of compound might be ammonium nitrite (\(NH_4NO_2\)). The positive ion (\(NH_4^{+1}\)) is a weak acid, and the negative ion (\(NO_2^{-1}|)) is a weak base. That means that both of the following reactions will occur:

\(NH_4NO_3 \rightleftharpoons NH_4^{+1} + NO_3^{-1}\)

and 

\(NO_2^{-1} + H_2O \rightleftharpoons HNO_2 + OH^{-1}\)

So, we are creating both hydronium (\(H_3O^{+1}\)) and hydroxide (\(OH^{-1}\)). So the pH is pushed down and up. It would be nice if the answer were just "It's Neutral!" but life isn't that kind. In a situation like this, the answer lies in the relative strengths of the \(K_a\) and \(K_b\). Whichever is larger, "wins" and will determine whether the pH of the solution is below or above 7.

Oxidation Numbers

 

Oxidation number is, simply the charge on each atom, whether it is alone, found as an ion or within a compound. That means that in many (most?) cases, it is relatively easy to determine the oxidation number of an atom. For instance, we know that pure elements are neutral, so the oxidation number on atoms of chlorine in a cloud of chlorine gas (\(Cl_2\)) is 0. Ions with the “ide” ending have the negative charge that makes the atom more stable, so chloride is always -1. That is true whether the chloride is part of a compound, like sodium chloride, or found independently in an aqueous solution. However, chlorine also has some other possible charges like those in the compounds \(HClO, HClO_2, HClO_3\), and \(HClO_4\).

For these more complicated cases, we need a set of rules to help us. The most common set of rules has some serious weaknesses. For instance, I was taught this rule:

Oxygen is always -2 except in peroxides

The problem, of course is that you must then be able to recognize peroxides when you see them. In addition, the rule is still not really correct. It should be that Oxygen is always -2 except pure oxygen, ozone, peroxides and hypoflorous acid.

The rule I was taught for hydrogen was not better:

Hydrogen is always +1, except for metal hydrides

What I will present here is a set of rules of my own invention. There are no exceptions and no other labels or particular types of compounds you need to watch for. It always works and it is always correct.

The McAfoos Method of Assigning Oxidation Numbers 

What follows are a series of rules, NOT steps. In other words, Rule #1 is the most important rule and beats all other rules. Rule #2 beats all others except rule #1, etc.

Rule #1 – The sum of the oxidation numbers = the total charge. This rule assigns oxidation numbers to pure elements, single element ions and the “last” element left without an oxidation number. Examples are shown below.

Rule #2 – Single charge elements get their charge. This rule assigns oxidation numbers to those elements that only have one non-zero charge. The list is short: alkali metals are always +1 (not H), alkaline earth metals are always +2, F is always -1 (not all the halogens), Al (+3), Zn (+2), and Ag (+1). Fair notice: if you forget the last three you’ll still probably be fine.

Rule #3 – Hydrogen is always +1. This doesn’t contradict the rule above if you remember that these rules are in descending importance. In other words, the only times that H is NOT +1 will have already been worked out based on the two rules above.

Rule #4 – Oxygen is always -2. As with the hydrogen rule, any “contradictions” to this rule will already have been found and dealt with.

Rule #5 – The most electronegative element gets its logical negative charges. This rule will only be used on rare occasions, since the earlier rules will handle almost everything.

A word of warning  – Compounds containing multiple instances of the same element should be broken into ions before assigning oxidation numbers. Although this is not generally an issue, when it is an issue, it matters immensely.

The only way to understand how these rules work is to see them in action, so let's take a look. 

Environmental Impacts of Base Anhydrides

As discussed here, metal oxides such as \(CaO\) form basic solutions when mixed into water. Since most metal oxides have relatively low solubility, this is not as important as the environmental impacts of acid anhydrides, but it is worth understanding. 

The most common base anhydride (at least in normal life) is lime, CaO. This compound is called lime since it is derived from limestone (\(CaCO_3\)) through decomposition:

\(CaCO_3 \rightarrrow CO_2 + CaO\)

Lime has a number of industrial uses, but the use you are most likely to come in contact with is the production of cement and concrete. Lime is a major component of these. (Don't get confused. Cement is not lime just like cookies are not flour. Lime is an ingredient in cement.)

The inclusion of lime in cement is one of the reasons that working with cement can cause skin trouble for workers and why the recommendation is to wear gloves when working with wet cement. 

Additionally, soil in contact with cement will be slightly more basic than soil further away. That means that acid-loving plants, such as azaleas will do better planted away from the foundation of your house than right up against it. 

On the other side of the situation, some soils are naturally acidic which can, in some cases, inhibit grass from growing well. In those cases, adding lime to the soil can bring the pH up to a healthier level.

Environmental Impacts of Acid Anhydrides

 As discussed here, non-metal oxides react with water to create acidic solutions. 

The reason that this is so important environmentally is that the combustion of fossil fuels, and especially coal, release non-metal oxides into the air. All fossil fuels are composed primarily of carbon and carbon chains and thus their combustion releases \(CO_2\) into the air. Here is the combustion of octane, the primary compound found in gasoline:

\(2 ~ C_8H_18 + 25 ~ O_2 \rightarrow 16 ~ CO_2 + 18~ H_2O\)

Coal is primarily carbon, but it contains both sulfur and nitrogen as impurities (as well as other elements such as mercury). When the sulfur and nitrogen in coal burn, they release \(SO_2, SO_3, and ~ NO_x\) (a mixture of a number of different nitrogen oxygen compounds). That means that any factory that is burning coal, and everything (factories, cars, etc.) that is burning fossil fuels is releasing non-metal oxides into the air. When those non-metal oxides meet clouds in the atmosphere, they form acids which then fall as rain.

That acid, falling as rain can cause damage to buildings and statues as seen here.

There is good news and bad news associated with this. 

The good news is that we are MUCH better at "scrubbing" what comes out of our smoke stacks than we used to be. At one point in the last century, rain was measured with a pH as low as 3.5, that 100x's as acidic as normal rain (with a pH of 5.5). That means that we are doing much less damage to statues and buildings than we used to.

The bad news is that we are adding more and more \(CO_2\) to the atmosphere. Although carbonic acid is a relatively weak acid and we are no longer seeing acid-rain associated damage in the way we did before, there is another, more worrisome impact.

The majority of our planet is covered with water. As we increase the amount of \(CO_2\) in the atmosphere, we increase the amount that dissolves in the ocean. Although the ocean is buffered and huge, we have already, in the past century, seen a change in the pH of the ocean. Before the industrial revolution, the pH of the ocean was about 8.2. It is now about 8.1. Although this seems like a tiny change, due to the logarithmic nature of the pH scale, it actually represents about a 30% increase in the acidity of the ocean (source).

This change in the pH of the ocean is especially worrying because the bottom of the ocean food web is comprised primarily of tiny crustaceans often lumped together as "krill". These creatures form their protective shells from carbonate found in the ocean. As the acidity of the ocean increases (as the pH goes down) the concentration of carbonate decreases according to the reaction:

\(CO_3^{-2} + H^{+1} \rightleftharpoons HCO_3^{-1}\)

Not only will this make it harder and harder for these creatures to form shells, but the increased acidity can even dissolve shells that have formed.

Of course, if you damage or remove the bottom of a food web, there is the potential for a complete collapse of the rest of the web. Given that 10% of the world's population depends on fisheries for their livelihood and 4.3 billion people are reliant on fish for 15% of their animal protein intake (source), this problem looms over everything we do. 

Formal Charge

There are several rules that help us decide, when we are drawing Lewis Dot Structures, which element is at the center of a molecule. Most of these are not really rules as much as they are suggestions.

• Carbon goes in the middle
• Hydrogen is only the outside
• If you have only one of an element, it goes on the inside
• The element that makes the most bonds goes on the inside
• The least electronegative element goes in the inside

Of course, sometimes these “rules” conflict. For instance in the compound N2O, there is only one oxygen (which should put it in the middle) but nitrogen makes more bonds and is less electronegative. So, how do we decide? We use formal charge.

Formal Charge

Formal charge is a way of looking at a Lewis Dot Structure and determining how “normal” it is. 

Let’s try to make sense of that idea. If we look at a simple Lewis Dot Structure and “break” the bonds in half (so that each element gets ONE of the two electrons in the bond, we can then compare that to the “expected” number of electrons.

Let's look at the Lewis Dot Structures of methane and cyanide.

If we break all of the bonds in methane, as shown below, each hydrogen has one electron. That’s what we expect, so the formal charge is 0. The carbon has 4 electrons, which is also what we expect (counting only valence) electrons, the formal charge is also 0.

In cyanide, however, things are a little different. If we break all of the bonds, as shown below, the carbon has 5 electrons, rather than the normal 4, so the formal charge on the C atom is -1. The nitrogen also has 5 electrons, but that’s normal for nitrogen, so the formal charge is 0.

Note that the sum of the formal charges is the overall charge. That is ALWAYS true.

Now let's look at the two possible arrangements for the atoms in N₂O.

When we break the bonds in the structure on the left, both of the nitrogen atoms have 6 electrons - that’s one more than normal, so the formal charge on each of them is -1. At the same time, the oxygen has only 4 electrons - two less than normal, giving a formal charge of +2.

In the structure on the right, breaking the bonds leaves a different situation. The nitrogen on the left has 6 electrons (formal charge of -1), the nitrogen in the middle has 4 electrons (formal charge of +1) and the oxygen has 6 electrons (formal charge of 0)

We compare those two results by adding the absolute value of the formal charges. For the structure on the left this adds up to 4 (1+2+1), while for the structure on the right, the sum is 2 (1+1+0). Since formal charge is a measure of “how far from normal” something is, this sum tells us that the structure on the right is “more normal.” We interpret that to mean that it will be favored by nature.

Acid and Base Anhydrides

Acid Anhydrides

In our discussion of double displacement reactions, we said that these reactions occur when a gas is produced.  There were four gases that we look for: \(H_2S, H_2CO_3, H_2SO_3, and ~ NH_4OH\). However, as discussed here, three of those compounds are actually not gases. They are compounds that immediately decompose to release gases.

For example, \(H_2CO_3\) breaks down to produce carbon dioxide and water according to the reaction here:

\(H_2CO_3 \rightarrow CO_2 + H_2O\)

But, now that we've learned a bit about reversible reactions, we know that this should actually be written as the following:

\(H_2CO_3 \rightleftharpoons CO_2 + H_2O\)

Of course that means that we could also write the reaction this way:

\(CO_2 + H_2O \rightleftharpoons  H_2CO_3\)

This is tremendously important because it shows that adding \(CO_2\) to water will produce carbonic acid and will therefore lower the pH of the solution. This same process can be seen for sulfur dioxide and sulfur trioxide:

\(SO_2 + H_2O \rightleftharpoons H_2SO_3\)

and

\(SO_3 + H_2O \rightleftharpoons H_2SO_4\)

Nitrogen oxides can do something similar, although nitrogen chemsitry is more varied and more complex. Nitrogen forms a series of oxygen compounds with various ratios. As sch, chemists often write the collection of compounds as \(NO_x\). When \(NO_x\) reacts with water it can create both nitrous and nitric acid:

\(NO_x + H_2O \rightleftharpoons HNO_2 ~ and ~ HNO_3\)

Of course we can't balance that reaction as written, but for our purposes here, what's important is the following:

• ALL of these cases (\(CO_2, SO_2, SO_3, and ~ NO_x\) are non-metal oxides
• ALL of these compounds form acidic solutions when mixed with water

These compounds (non-metal oxides) are, therefore, called acid anhydrides. A term which simply means that if the compound is put into water it will be an acid. 

This property has significant environmental importance, which is discussed here.

Base Anyhydrides

A related chemistry occurs when a metal oxide is placed in water. For instance, \(CaO\) will react with water to produce calcium hydroxide according to the reaction here:

\(CaO + H_2O \rightleftharpoons Ca(OH)_2\)

The same reaction will occur with other metal oxides. For example, we can look at the reactions of sodium oxide and aluminum oxide with water.

\(Na_2O + H_2O \rightleftharpoons 2 NaOH\)

and

\(Al_2O_3 + 3 H_2O \rightleftharpoons 2 Al(OH)_3\)

As a result of these reactions, metal oxides are called base anhydrides.

This property also has an environmental importance, which is discussed here.

Redox Reactions

 

Electrochemistry is one of the units that is neglected in most first year high school chemistry courses. This is a shame because electrochemistry is all around us. Anytime you carry around an electronic device that does not need to be plugged in to work, you are using electrochemistry. 

In addition, most reactions are electrochemical in nature. For instance, the single displacement reaction here

\(Mg + Cu(NO_3)_2 \rightarrow Mg(NO_3)_2 + Cu\) 

can be thought of as a simple trading of places between the copper and the magnesium. However a closer look shows us that there is more going on. 

Any pure metal (actually any pure element) is neutral. That means that the atoms of magnesium on the left side of the reaction have no charge. But, on the right side the magnesium has a charge of +2. (We know it must be a positive ion because it is hanging out with nitrate – a negative ion.) That means that in the course of this reaction, each Mg atom lost 2 electrons. 

If we start to wonder where those electrons went, we don't need to look any further than the copper. On the left side of the reaction, the copper is composed of +2 ions and on the right it is pure, uncharged copper. So, this reaction is not simply a “trading of places”. This reaction involves copper ions taking electrons away from magnesium atoms.

Be sure to avoid the idea that the magnesium “gave” the electrons to the copper. NO positive nucleus has EVER given away negative electrons. This reaction is, pure and simple, electron theft.

Let’s look at another single displacement reaction, this time between copper and silver.

\(AgNO_3 + Cu \rightarrow Cu(NO_3)_2 + Ag\)

Again we can see the electron theft occurring. On the left side of the reaction, the copper has no charge and on the right it has a charge of +2. Something has stolen 2 electrons. The culprit, of course is silver. On the left the silver has a charge of +1 and on the right it is neutral. 

There is a problem however. The silver has stolen 1 electron, but the copper lost 2. The solution of course is that TWO silver ions “gang up” on the copper, each taking one electron, and we could have seen that if the reaction had been properly balanced. Here is it, written correctly.

\(2~ AgNO_3 + Cu \rightarrow Cu(NO_3)_2 + 2~ Ag\)

Let’s look at one more reaction – the combustion of methane.

\(CH_4 + 2~ O_2 \rightarrow CO_2 + 2 ~H_2O\)

Given what we know about elements in general and oxygen in particular, we can find the electron theft occurring here. On the left side, oxygen is a pure element, and therefore has no charge. If we look at the water on the right side we can guess that oxygen has a -2 charge. That guess makes sense for 2 reasons. This first reason is that -2 is the preferred charge on oxygen (since 2 extra electrons fill the energy level). The second reason is that we know that hydrogen is generally +1 (think acid chemistry).

There is a problem with that logic however – water is not ionic, and therefore, doesn’t actually have any charges. That means that electron theft is not as clear here as it was in the other reactions. Rather than outright theft, what has happened here is a transfer of control. The bonds in water are polar, due to the high electronegativity of oxygen. So, in a sense, the oxygen has “taken control” of those electrons, making it “feel” -2ish. This “ish” type of charge is called oxidation number and being able to determine it for separate atoms within compounds is an important skill discussed here.

 Titration - how the math works.

As explained here, titration allows us to find the concentration of a solution by reacting it with another solution, about which we have more information. 

Most commonly, titration is done with acids and bases, so we’ll discuss those first and then discuss other possibilities.

An acid base reaction is “done” when the number of hydroxide ions matches the number of hydronium ions. In other words, the reaction is done when moles of \(H^{+1}\) = moles of \(OH^{-1}\). Assuming that the acid and base are both monoprotic (that is the acid only has one H+ ion to donate and the base can only take one H+ ion) then we can say that the reaction is “done” when

\(mole _{acid} = mole _{base}\)

If we are working with solutions, then \(moles = concentration ( \frac{mol}{L})\cdot Volume(L)\). So,

\(mole_a = mole_b ~~becomes~~ M_aV_a = M_bV_b\)

In all cases, titration is done so that three of these values are known and the fourth can, therefore be found. Here’s an example:

25.00 mL of an acid with an unknown concentration are titrated with 32.64 mL of 0.445 M NaOH. What is the concentration of the acid?

In this case we know everything except the \(M_a\) so we can solve for that algebraically, getting

\(M_a = \frac{M_bV_b}{V_a}\)

substituting in our values, we get 

\(\frac{(0.445M)\cdot(32.64 mL)}{(25.00mL)}\) = 0.581 M

Note that although this math should be done with liters, we can “get away with” using mL since the two units will cancel each other out.

Dealing with Solids 

It is important to remember that the equation that matters is NOT \(M_aV_a= M_bV_b\), but rather, \(moles_a = moles_b\), That means that we can use other ways to get moles. If, for instance, the acid is a solid, then \(moles_a = \frac{mass_a}{Molar~Mass_a}\). If the base was still a solution, that would give us the following: 

\(\frac{mass_a}{MM_a} = M_bV_b\)

Of course the base could be the solid. The point is not to "learn" the equation above, but rather to understand that the "moles" on either side of the equation can be ANYTHING that translates into moles.

Dealing with diprotic and triprotic acids and bases 

To understand the impact that di- and triprotic acids and bases have on the math of titrations, you need to remember that the equation is NOT moles acid = moles base, but rather moles \(H^{+1}\) ions = moles \(OH^{-1}\) ions.

Imagine that you had 30.0 mL of 0.500 M \(H_3PO_4\). If you needed the moles of phosphoric acid, you could calculate \(M_aV_a = (0.500\frac{mol}{L}) \cdot (0.0300 L) = 0.015 mols H_3PO_4\). However, in a titration, the phosphoric acid has 3 hydrogens that can react with the base, so the moles of \(H^{+1}\) ions is 3 times the moles of \(H_3PO_4\). 

The same idea plays out with a polyprotic base. For instance, 0.12 moles of \(Mg(OH)_2\) can produce 0.240 moles of hydroxide ions. We can compensate for this by adding in a factor on both sides of the equation, so \(M_aV_a= M_bV_b\) is more correctly written as \((\#_a)M_aV_a = (\#_b)M_bV_b\) where \(\#_a\) is the number of H’s in the acid and \(\#_b\) is the number of OH’s in the base.

Here's a sample problem involving polyprotic acids and bases:

How many mL of a 0.115 M solution of barium hydroxide are needed to fully titrate 18.5 mL of 0.331 M phosphoric acid?

Barium hydroxide is \(Ba(OH)_2\) meaning that each mole will produce 2 moles of \(OH^{-1}\) ions. Phosphoric acid is \(H_3PO_4\), so each mole of phosphoric acid will result in 3 moles of \(H^{+1}\) ions. Our math, therefore, looks like this:

\((\#_a)M_aV_a = (\#_b)M_bV_b\)

We are trying to solve for the volume of base needed, so we can rearrange algebraically which will give us:

\(V_b=\frac{(\#_a)M_aV_a}{(\#_b)M_b}\)

Substituting in our values from the problem, gives us this:

\(V_b=\frac{(3)(0.331M)(18.5 mL)}{(2)(0.115M)}=79.8 mL\)