# 10.5: Equilibria involving Acids and Bases

Consider a simple chemical system that is at equilibrium, such as dinitrogen tetroxide: nitrogen dioxide. The Law of Mass Action states that when this system reaches equilibrium, the ratio of the products and reactants (at a given temperature) will be defined by the equilibrium constant, K. Now imagine that, after equilibrium has been reached, more dinitrogen tetroxide is introduced into the container. In order for the ratio to remain constant (as defined by K) some of the N2O4 that you added must be converted to NO2. The addition of reactants or products to a system at equilibrium is commonly referred to as a “stress”. The response of the system to this stress is dictated by Le Chatelier's Principle.

Le Chatelier's Principle

Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. Again, stress refers to a change in concentration, a change in pressure or a change in temperature, depending on the system being examined. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change.

We will consider temperature and pressure effects in General Chemistry, but for now, remember; in a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.

In Chapter 8, we learned that a “weak acid” was only partially dissociated in solution, while a “strong acid” was fully dissociated. Now that we better understand the concept of equilibrium, these two classes of Brønsted acids can simply be differentiated based on their equilibrium constants. For an acid, BH, that dissociates in water to form B and hydronium ion, we can write a simple equilibrium expression, as follows:

BH(aq) + H2O(l) ⇄ B(aq) + H3O+(aq)

$K_{C}=\frac{[H_{3}O^{+}][B^{-}]}{[BH]}=''K_{a}''$

You should note two things in this equation; the term for water does not appear (remember, solids and liquids are never written in equilibrium expressions) and the equilibrium constant for KC is written as Ka to denote that this is an acid dissociation equilibrium. Now, as we learned in  Chapter 8, a strong acid is “fully dissociated”, which simply means that [BH] is very, very small, thus Ka for a strong acid is very, very large. A weak acid is only “partially dissociated” which means that there are significant concentrations of both BH and B in solution, thus Ka for a weak acid is “small”. For most common weak acids, the values for Ka will be in the range of 10-3 to 10-6.

Example $$\PageIndex{1}$$:

Consider acetic acid (the acidic component of vinegar) where Ka = 1.8 × 10-5.

CH3COOH(aq) + H2O(l) ⇄ CH3COO(aq) + H3O+(aq)

Solution

$K_{a}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=1.8\times 10^{-5}$

Exercise $$\PageIndex{1}$$

1. A series of acids have the following Ka values: rank these in descending order from the strongest acid to the weakest acid.

A. 6.6 × 10–4 B. 4.6 × 10–4 C. 9.1 × 10–8 D. 3.0 × 102

2. At 25.0 oC, the concentrations of H3O+ and OH in pure water are both 1.00 × 10-7 M, making Kc = 1.00 × 10-14 (recall that this equilibrium constant is generally referred
to as Kw). At 60.0o C, Kw increases to 1.00 × 10-13. What is the pH of a sample of pure water at 60.0o C?

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