Acids and bases

Calculations of pH and pOH can be boiled down to one thing: what is the actual concentration of H^{+}, [H^{+}], or OH^{−}, [OH^{−}], in the solution?

Obviously the calculations also work the other way around, so if you know the pH or pOH value from a measurement on the solution, you can calculate their respective [H^{+}] and [OH^{−}].

The degree of difficulty and complexity in isolating one of these concetrations varies a lot, which si why pH and pOH are not always easy.

An extremely important detail to be aware of, is the autoprotolysis of water. For the concentrations of acids and bases used in chemistry in practice, you can disregard this, but for highly diluted solutions, the buffer capacity of water start making an impact. This will be adressed in the section about pitfalls.

At highly concentrated solutions you also have to be alert. For an acid to give off H^{+}, there must be a water molecule present to receive it and form an oxonium ion. An acid like concentrated sulfuric acid is approximately 97 %. This means that there is 3 % water for the 97 % acid to dissociate in. That is not possible, and the way to deal with this is adressed in the section about pitfalls.

Let us take a look at some examples.

Obviously the calculations also work the other way around, so if you know the pH or pOH value from a measurement on the solution, you can calculate their respective [H

The degree of difficulty and complexity in isolating one of these concetrations varies a lot, which si why pH and pOH are not always easy.

An extremely important detail to be aware of, is the autoprotolysis of water. For the concentrations of acids and bases used in chemistry in practice, you can disregard this, but for highly diluted solutions, the buffer capacity of water start making an impact. This will be adressed in the section about pitfalls.

At highly concentrated solutions you also have to be alert. For an acid to give off H

Let us take a look at some examples.

Strong acids and bases in aqueous solution are the easiest to work with. Since they are strong, we can assume that the dissociation is close to 100 %. Looking at a 2.0 M HCl, dissociation looks like this:

We see that*c*(HCl) = [H_{3}O^{+}] i.e. [H^{+}] = 2.0 M, resulting in:

pH = -log([H^{+}]) = -log(2.0) = -0.3

For divalent and trivalent acids the procedure is the same, as long as the acid dissociates completely, like sulfuric acid. For a 0.5 M sulfuric acid, it looks like this:

We can see that*c*(H_{2}SO_{4}) = 0.5 · [H_{3}O^{+}] i.e. [H^{+}] = 1.0 M, which gives us:

pH = -log([H^{+}]) = -log(1.0) = 0.0

Strong bases are handled exactly like the acids, with the notable difference that we determine pOH first, and then determine pH. The calculation of pH for a 1.0 M strong base like potassium hydroxide looks like this:

We can see that*c*(KOH) = [OH^{−}] i.e. [OH^{−}] = 1.0 M, which gives us:

pOH = -log([OH^{−}]) = -log(1.0) = 0.0

H_{2}O(l) | + | HCl(aq) | H_{3}O^{+}(aq) | + | Cl^{−}(aq) | ||

Start | - | 2.0 M | 0.0 M | 0.0 M | |||

End | - | 0.0 M | 2.0 M | 2.0 M |

We see that

pH = -log([H

For divalent and trivalent acids the procedure is the same, as long as the acid dissociates completely, like sulfuric acid. For a 0.5 M sulfuric acid, it looks like this:

2 H_{2}O(l) | + | H_{2}SO_{4}(aq) | 2 H_{3}O^{+}(aq) | + | SO_{4}^{2−}(aq) | ||

Start | - | 0.5 M | 0.0 M | 0.0 M | |||

End | - | 0.0 M | 1.0 M | 0.5 M |

We can see that

pH = -log([H

Strong bases are handled exactly like the acids, with the notable difference that we determine pOH first, and then determine pH. The calculation of pH for a 1.0 M strong base like potassium hydroxide looks like this:

KOH(aq) | K^{+}(aq) | + | OH^{−}(aq) | ||

Start | 1.0 M | 0.0 M | 0.0 M | ||

End | 0.0 M | 1.0 M | 1.0 M |

We can see that

pOH = -log([OH

⇕

pH = 14 - pOH = 14 - 0 = 14When we get to the acids and bases that do not dissociate 100 % things become a bit more complicated. Using acetic acid, HAc, as an example, this is how we do it:

The start concentration, before the acid start dissociating, is the formal concentration*c*. When the acid has dissociated, we have a concentration of H^{+}, which we do not know. We will set [H^{+}] to x. Then start and end for the dissociation looks like this:

If you look at some of the older educational material, a couple of formulas were used for calculating pH for weak and medium strength acids, that were simpler and easier to use for the calculations. Today they make no sense using. They are demonstrated in the next section for those who feel like doing old school calculations. Here we take the direct approach:

Looking at the available tables for p*K*_{a} values, we can see that for HAc p*K*_{a} = 4.76 i.e. *K*_{a} = 1.74·10^{-5} M. The equilibrium constant, i.e. *K*_{a} for HAc is:

Inserting the concentrations from the reaction equation, it now looks like this:

Mathematically what we have here is a second degree equation that can be solved. If we are informed that*c*(HAc) = 1.0 M, we can solve the equation. Here x = -4.2·10^{-3} og 4.2·10^{-3}, but since you cannot have negative concentrations, the correct answer must be [H^{+}] = 4.2·10^{-3} M and thus pH = 2.4.

The weak and medium strength bases are handled just like the acids. A good exampel could be the weak base ammonia, NH_{3}:

Notice that you do not write the concentration of water, when it acts as a solvent. You do not do this because in these cases, you do not get a measurable change in concentration, and because concentrations of solvents in equilibrium equations are defined as 1, i.e. they vanish from the equation.

Because it is a base, we use*K*_{b} for the calculations:

*K*_{b} for ammonia is 1.8 · 10^{-5} M, and just like the acids, except using *K*_{b} we get:

If we are informed that*c*(NH_{3}) = 1.0 M, we can solve the second degree equation. Here x = -4.3·10^{-3} and 4.2·10^{-3}, but since you cannot have negative concentrations, the result must be [OH^{−}] = 4.2·10^{-3} M and thus pOH = 2.4 ⇔ pH = 11.6.

The start concentration, before the acid start dissociating, is the formal concentration

HAc(aq) | H^{+}(aq) | + | Ac^{−}(aq) | ||

Start | c | 0 M | 0 M | ||

End | c-x | x | x |

If you look at some of the older educational material, a couple of formulas were used for calculating pH for weak and medium strength acids, that were simpler and easier to use for the calculations. Today they make no sense using. They are demonstrated in the next section for those who feel like doing old school calculations. Here we take the direct approach:

Looking at the available tables for p

K_{a} = | [H^{+}] · [Ac^{−}] |

[HAc] |

Inserting the concentrations from the reaction equation, it now looks like this:

1.74·10^{-5} M = | x · x |

c-x |

Mathematically what we have here is a second degree equation that can be solved. If we are informed that

The weak and medium strength bases are handled just like the acids. A good exampel could be the weak base ammonia, NH

NH_{3}(aq) | + | H_{2}O(l) | NH_{4}^{+}(aq) | + | OH^{−}(aq) | ||

Start | c | - | 0 M | 0 M | |||

End | c-x | - | x | x |

Notice that you do not write the concentration of water, when it acts as a solvent. You do not do this because in these cases, you do not get a measurable change in concentration, and because concentrations of solvents in equilibrium equations are defined as 1, i.e. they vanish from the equation.

Because it is a base, we use

K_{b} = | [NH_{4}^{+}] · [OH^{−}] |

[NH_{3}] |

1.8·10^{-5} M = | x · x |

c-x |

If we are informed that

The divalent and trivalen acids and bases may appear a bit messy at first, when you start working on them. Fortunately there are some circumstances that makes them easy to do calculations on.

Looking at the fictional acid H_{2}B, the reaction equations looks like this:

Starting with the equilibrium equations, they look like this:

For the acid we now get an overall*K*_{a} by multiplying the *K*_{a} values for the two reaction steps:

*K*_{a} are values you can look up, and it may be tempting to just go ahead and do calculation on the entire equilibrium, just like you would do for the monovalent acids, but this is a waste of effort, that will often give you the wrong result. For the medium strength and weak acids, the first dissociation will usually be dominant, and the subsequent ones so small that they are neglegible.

If we take a look at a divalent acid like maleic acid (1,4-butane dicarboxylic acid), we can see that for this molecule*K*_{a1} = 1.5·10^{-2} M and *K*_{a2} = 2.6·10^{-7} M. With a factor 10^{5} difference between the two *K*_{a} values, the change in concentrations due to the dissociation of HB^{−} will be so small that they are neglegible. So, **what you do with the polyvalent acids is to calculate as if the dominant equilibrium is a monovalent acid**, as shown in the previous section.

Polyvalent bases are handles in exactly the same manner, doing the calculations on the dominant step as if it was a monovalent base.

Since it is only "usually", that calculations on polyvalent acids and bases can be reduced to calculations on the domainant step as if it was a monovalent acid/base, this also implies that there must be exceptions. There is. If the*K*_{a} values for the reaction steps are so low that the overall *K*_{a} is a strong acid, you will calculate as if it is a strong divalenet or trivalent strong acid. This is actually how it is done with H_{2}SO_{4}. Here *K*_{a1} = 10^{3} M and *K*_{a2} = 1.2·10^{-2} M, so eventhough the second H^{+} is "only" a medium strength acid, the combined acidic strength is so high, that that in practice both H^{+} are released. At the other end of the scale, we have bases like the divalent base urea. Here the difference between *K*_{b1} and *K*_{b2} is so small, that it cannot be determined, and you only have one *K*_{b} = 1.3·10^{-14} M^{2}. **If the two reaction steps have approximately the same ***K*_{a} or *K*_{b} values, there is no dominant reaction step. In that case, you have to solve the third degree equation that comes out of the total equilibrium equation (fourth degree equation at trivalent acids/bases).

Looking at the fictional acid H

Reaction step 1: | H_{2}B(aq) H^{+}(aq) + HB^{−}(aq) |

Reaction step 2: | HB^{−}(aq) H^{+}(aq) + B^{2−}(aq) |

Total: | H_{2}B(aq) 2 H^{+}(aq) + B^{2−}(aq) |

Starting with the equilibrium equations, they look like this:

K_{a1} = | [H^{+}] · [HB^{−}] |

[H_{2}B] |

K_{a2} = | [H^{+}] · [B^{2−}] |

[HB^{−}] |

For the acid we now get an overall

K_{a} = K_{a1} · K_{a2} = | [H^{+}] · [HB^{−}] | · | [H^{+}] · [B^{2−}] | = | [H^{+}]^{2} · [B^{2−}] |

[H_{2}B] | [HB^{−}] | [H_{2}B] |

If we take a look at a divalent acid like maleic acid (1,4-butane dicarboxylic acid), we can see that for this molecule

Polyvalent bases are handles in exactly the same manner, doing the calculations on the dominant step as if it was a monovalent base.

Since it is only "usually", that calculations on polyvalent acids and bases can be reduced to calculations on the domainant step as if it was a monovalent acid/base, this also implies that there must be exceptions. There is. If the

One of the classic pitfalls is at very low concentrations. Let's say that we dilute an acid to [H

Since H

In the oldfashioned calculations of pH and pOH two formulas are used, one for weak acids/bases and one for medium strength acids/bases. For the weak and medium strength acids the respective formulas are

For the weak and medium strength bases the respective formulas are

For a 1.0 M acetic acid, which is a weak acid with*pK*_{a} = 4.76, the calculations looks like this:

For a 1.0 M hydrofluoric acid (HF), which is a medium strength acid where*K*_{a} = 6.76·10^{-4}, the calculations looks like this:

Both equations are approximations of the equilibrium equations, where it has been possible to simplify the math, because some of the factors becomes negligible. The equations works just fine, as long as the concentrations are not too low, and can easily be used, if you want to.

pH = | pK_{a} - log c |

2 |

pH = - log | -K_{a} + √ K_{a}² + 4 · K_{a} · c |

2 |

For the weak and medium strength bases the respective formulas are

pOH = | pK_{b} - log c |

2 |

pOH = - log | -K_{b} + √ K_{b}² + 4 · K_{b} · c |

2 |

For a 1.0 M acetic acid, which is a weak acid with

pH = | 4.76 - log (1.0) |

2 |

⇕

pH = 2.4For a 1.0 M hydrofluoric acid (HF), which is a medium strength acid where

pH = - log | -6.76·10^{-4} - √ (6.76·10^{-4})² + 4 · 6.76·10^{-4} · 1 |

2 |

⇕

pH = 1.6Both equations are approximations of the equilibrium equations, where it has been possible to simplify the math, because some of the factors becomes negligible. The equations works just fine, as long as the concentrations are not too low, and can easily be used, if you want to.