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:

\(CaCO_3 \to CO_2 + CaO \)

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.