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 \(\ce{N2O4}\) that you added must be converted to NO 2 . 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 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, \(\ce{BH}\), that dissociates in water to form \(\ce{B^{–}}\) and hydronium ion:
\[\ce{BH(aq) + H2O(l) <=> B^{–}(aq) + H3O^{+}(aq)} \nonumber\]
we can write a simple equilibrium expression, as follows:
\[K_{C}=\frac{[H_{3}O^{+}][B^{-}]}{[BH]}=''K_{a}'' \nonumber \]
You should note two things in this equation. Because the activity of water, as the solvent, is defined to have a value of 1, the activity for water does not affect the value of the equilibrium constant (remember, solids and liquids and solvents all have an activity of 1, and so do not affect the value of K) and the equilibrium constant for K C is written as K a 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 K a 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 K a for a weak acid is “small”. For most common weak acids, the values for K a will be in the range of 10 -3 to 10 -6 .
Consider acetic acid (the acidic component of vinegar) where K a = 1.8 × 10 -5 .
\[\ce{CH3COOH(aq) + H2O(l) <=> CH3COO^{–}(aq) + H3O^{+}(aq)} \nonumber\]
Solution
\[K_{a}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=1.8\times 10^{-5} \nonumber \]
-
A series of acids have the following
K
a 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 × 10 2
-
At 25.0
o
C, the concentrations of H
3
O
+
and OH
–
in pure water are both 1.00 × 10
-7
M, making
K
c
= 1.00 × 10
-14
(recall that this equilibrium constant is generally referred
to as K w ). At 60.0 o C, K w increases to 1.00 × 10 -13 . What is the p H of a sample of pure water at 60.0 o C?