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# Ionization Constants

A simple way to understand an ionization constant is to think of it in a clear-cut way: To what degree will a substance produce ions in water? In other words, to what extent will ions be formed?

### Introduction

Water has a very low concentration of ions that are detectable. Water undergoes self-ionization, where two water molecules interact to from a hydronium ion and a hydroxide ion.

$H_2O + H_2O \rightarrow H_3O^+ + OH^- \tag{1}$

To understand this process in further detail, we can take a closer look.

Even though water does not form a lot of ions, the existence of them is evident as is proved by the electrical conductivity of pure water. Water undergoes ionization because the powerfully electronegative oxygen atom takes the electron from a hydrogen atom, making the proton dissociate.

$\ce{H-O-H} \rightarrow H^+ + OH^- \tag{2}$

with two ions are formed:

• Hydrogen ions $$H^+$$
• Hydroxyl ions $$OH^-$$

The hydrogen ions then react with water to form hydronium ions:

$H^+ + H_2O \rightarrow H_3O^+ \tag{3}$

Now, we since we have seen the break down of the process, this brings us back to our first analysis of self-ionization. Typically, hydrogen atoms are bonded with another water molecule resulting in a hydration that forms a hydronium and hydroxide ion.

$H_2O + H_2O \rightarrow H_3O^+ + OH^- \tag{4}$

In 1L of pure water at 25 degrees Celsius, there is 10-7 moles of hyrodronium ions or hydroxide ions at equilibrium. Let’s come back to the question of to what degree will a substance form ions in water? For water, we have learned that it will occur until 10-7 moles of either ion will ionize at the previous given conditions. Since this is during equilibrium, a constant can be formed.

$H_2O \rightleftharpoons H^+ + OH^- \tag{5}$

$K_{eq}= \dfrac{[H^+] [OH^-]}{[H_2O]} \tag{6}$

Each molecule is expressed in concentrations of moles per Liter. You might be thinking… Hold on, what is this equation talking about? I will explain further on, once we understand the self-ionization of water.

Each H+, OH-, H2O are expressed in concentrations. The concentration of a substance is usually defined in moles per Liter. To find the concentration for water, we must calculate the moles per Liter then. Let’s think about it, this is relatively simple. Since we have 1 L of water, we have 1000 grams of water. To convert to moles, we simply divide by the molecular mass. Concentration therefore is calculated by: (1000/18) The answer is 55M of water. Now, we have (10-7)2/55.5 = 1.8 * 10-16 M

However, since there is not a lot of ionization, the change in water concentration is very small. It does not decrease very much. All of these concentrations are generally reliable and dependable. Experimentation has proved their accuracy as scientists use these standard values. The concentration of water is considered a constant. Kw = 55.5M But because water is a liquid and its molarity is invariable, we multiply the concentrations of the ions and use that answer for other empirical calculations. Keq = 10-14 M = [OH-] [H+] In chem. 2B, this standard equation is used

$K_w = [H_3O^+] [OH^-] = 10^{-14} \tag{7}$

### What about acids and bases?

Now that we have an idea of what an ionization constant is, let’s take a look at how acids and bases play in this scenario. Strong acids and bases are those that completely dissociate into ions once placed in solution. For example: KOH + H2O → K+ + OH- So, if we had 2M of KOH, then we would also have two moles of OH- ions.

### How about if the acid or base is weak?

Weak acids and bases, however do not behave the same way. Their amounts cannot be calculated as easily. This is because the ions do not fully dissociate in the solution. Weak acids have a higher pH than strong acids, because weak acids do not release all of its hydrogens. The acid dissociation constant tell us the extent to which an acid dissociates

$HCN_{(aq)} + H_2O \rightleftharpoons H_3O^+ + CN^- \tag{8}$

$K_a= \dfrac{[H_3O^+] [CN^-]}{[HCN]} \tag{9}$

This equation is used fairly often when looking at equilibrium reactions. During equilibrium, the rates of the forward and backward reaction are the same. However, the concentrations tend to be varied. Since concentration is what gives us an idea of how much substance has dissociated, we can relate concentration ratios to give us a constant. K is found by first finding out the molarity of each substance. Then, just as shown in the equation, we divide the products by the reactants, excluding solids and liquids. Also when there is more than one product or reactant, their concentrations must be multiplied together. Even though you will not see a multiplication sign, if there are two molecules associated, remember to multiply them. If there is a coefficient in front of a molecule, the concentration must be raised to that power in the calculations.

The amount of Ka can help us identity how weak an acid is. A weaker acid tends to have a smaller Ka. This is because the concentration on the bottom of the equation is larger. There is more of the acid, and less of the ions. You can think of Ka as a way of relating concentration in order to find out other calculations, typically the pH of a substance. A pH tells you how basic or acidic something is, and as we have learned that depends on how much ions become dissociated.

Example 1: Calculate Ionization Constant

Calculate the ionization constant of a weak acid. Solve for Ka given 0.8M of hydrogen cyanide and 0.39 M for hyrodronium and cyanide ions:

$HCN_{(aq)} + H_2O → H_3O^+ + CN^-$

Solution

(HCN) is 0.8M (H3O+) is 0.0039M (CN-) is 0.0039M.

We can assume here that [H3O+] = [CN-]

The equilibrium constant would be

$K_a= \dfrac{0.0039^2}{0.8} = 1.9 \times -^{5}$

We can use the similar method to find the Kb constant for weak bases. Again, an ionization constant is a measure of the extent a base will dissociate. Kb relates these molarity quantities. A smaller Kb corresponds to a weaker base, as a higher one a stronger base. Some common weak bases: NH3, NH4OH, HS-, C5H5N

$C_5H_5N_{(aq)} + H_2O_{(l)} → C_5H_5N^+_{(aq)} + OH^-_{ (aq)} \tag{10}$

We would find Kb the same way we did Ka. However, most problems are not as simple and obvious. Let’s do an example that is a little more challenging to help you understand this better.

Example 2: Calculate Concentration

Calculate the concentration of [OH-] in a 3M Pyridine solution with Kb= 1.5 * 10-9.

Solution

Since we already have our equation, let’s write the expression for the constant as discussed earlier (concentration of products divided by concentration of reactants)

$K_b= \dfrac{[OH^-] [C_5H_5NH^+]}{[C_5H_5N]} = 1.5 \times 10^{-9}$

Because pyridine is a weak base, we can assume that not much of it will disassociate. And since we have 3M, let’s make 3M-X our calculation. Subtracting X is going to be the change in molarity due to dissociation. Since our ions are unknown and they are both one mole, we can solve for them as being X. THUS: (X2/3-X) = 1.5 * 10-9

To make the calculation simple, we can estimate 3-X to be approximately 3 because of the weak base. Now our equation becomes: (X2/3) = 1.5 * 10-9. When we solve for X our answer is $$4.7 \times 10^{-5}\; M$$

This approximation was effective because x is small, which must be less than 5% of the initial concentration for that estimation to be justified.

### Summary

We have learned that an ionization constant quantifies the degree a substance will ionize in a solution (typically water). Ka, Kb, and Kw are constants for acids, bases, and finally water, respectively and are related by

$K_a \times K_b = K_w$

However, these constants are also used to find concentrations as well as pH.

### References

1. Grisham, Charles. Biochemistry: Updated Third Edition. Belmont, Ca; Thomas Learning Inc. 2007. pgs. 37-40

### Contributors

• Tatiana Litvin (UCD)

Viewing 3 of 3 comments: view all
Your explanation of the mathematics behind Kw is incorrect. The concentration of water does not play any role in this calculation. Equilibrium constants are actually ratios of activities, not concentrations. The activity of a pure solvent in a dilute solution is defined as unity (1). Thus, the activity of the H2O is 1. Because we are assuming these are dilute solutions, the activity coefficients of the solutes (H+ and OH-) are nearly unity, thus the activity of these ions are almost exactly the same value as their concentrations. With those assumptions, you can express the equilibrium constant in terms of the concentrations of H+ and OH-. But the water concentration is NOT involved in any way. Details come from Raoult's Law and Henry's law for solutions, and from the fundamental description of equilibrium in a physical chemistry text. This erroneous inclusion of the concentration of water seems to have first appeared (and is now ubiquitous) in organic textbooks, so if you want to get confirmation of my comments, ask a physical chemist, not an organic chemist.
Posted 08:26, 23 Oct 2014
Yes.
Posted 09:14, 9 Dec 2014
Looks like there is an error in the first Example box.

"0.39 M for hyrodronium and cyanide ions" should probably be 0.0039 M.
Posted 05:50, 25 Jan 2016
Viewing 3 of 3 comments: view all
10:00, 29 Dec 2015

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This material is based upon work supported by the National Science Foundation under Grant Numbers 1246120, 1525057, and 1413739.