Imagine you're cleaning the kitchen counters. You'll probably use cleaning products to keep your house clean! These everyday cleaning products contain weak acids and bases. We are going to explore how weak acids and bases react in equilibrium.
Weak acid and base equilibrium refer to the equilibrium in weak acid and base reactions because of their partial ionization in a solution.
- This article is about weak acid and base equilibria.
- First, we are going to broadly discuss the ionization of weak acids and bases
- Then, we will consider the weak acid and weak base equilibrium equations
- After, we will study and work through some acid-base equilibrium problems
- Finally, we will explore some common examples of weak acids and base
Ionization of Weak Acids and Bases
Before we can get into the specifics of weak acid-base equilibrium, let's consider what makes an acid or base weak and the equilibrium in weak acid-base reactions.
Weak acids are acids that only partially ionize in an aqueous solution.
Weak bases are bases that only partially ionize in an aqueous solution.
This differs from strong acids and bases, which fully ionize in aqueous solutions!
A relatable metaphor to visualize this complete vs. partial ionization can be found in the context of friendships!
- If you have a strong friendship with someone, you may give them most of your trust and accept many gifts from them. But if you have a weaker relationship with someone, you may withhold most of your trust and be slow to accept gifts.
- In the same way, a strong acid completely donates all of its H+ protons, and a strong base readily accepts all H+ protons. Whereas a weak acid only donates some of its H+ protons, and a weak base only accepts some H+ protons.
While weak acids and bases aren't in a friendship, they are definitely in chemical "relationships" partially giving H+ or accepting H+ protons, respectively!
How are weak acid-base equilibria formed?
Since weak acids and bases don't fully ionize when they react with water, the resulting solution is a mixture of ions. Let's look at an example of a weak base to conceptualize this. When ammonia reacts with water, we get the following chemical equation:
$$NH_{3\ (aq)}+H_{2}O_{(l)}\rightleftharpoons NH_{4\ (aq)}^{+}+OH_{(aq)}$$
In the aqueous solution, we will have:
- The base: NH3.
- The hydroxide ion: OH-.
- The conjugate acid: NH4+ .
Because the reaction doesn't go to 100% completion, we are left with ions and the unionized base species, which create a dynamic equilibrium! This means that the ammonia molecules accept hydrogen ions from the water molecules and form hydroxide ions and ammonium ions at the same rate that hydroxide ions donate hydrogen ions to ammonium ions to reform ammonia molecules and water molecules!
Try to say that sentence five times fast... it's a tongue twister for sure! If you have difficulty understanding equilibrium, check out Dynamic Equilibrium and Reversible Reactions for a refresher!
If weak acids and bases aren't 100% ionizing, how do we know to what percent they are ionizing? Well, that's where equilibrium equations and constants come into play! We will spend time on acids and bases separately as they are each essential concepts to grasp!
Weak Acid Equilibrium Equation
Let's examine the general weak acid equilibrium equation and constant! In a weak acid reaction, the aqueous acid molecules, HA (aq), react with liquid water by donating H+ ions to the water. This reaction forms aqueous hydronium ions and aqueous anions, A-(aq).
The following generic equation can represent the weak acid reaction:
$$HA_{(aq)}+H_{2}O_{(l)}\rightleftharpoons H_{3}O^{+}_{(aq)}+A^{-}_{(aq)}$$
Where:
- HA = acid species.
- A- = conjugate base of the acid.
Based on this generic equation, we can construct the equilibrium constant expression, referred to as the acid ionization constant, Ka.
The acid ionization constant, Ka, represents the relative strength of an acid. The strength of an acid is determined by the extent to which an acid dissociates in an aqueous solution.
$$$K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}$
This acid ionization constant expression reveals two complementary things to us about the relationship between Ka and the strength of an acid.
- the more HA acid molecules dissociate → the higher the Ka value → the stronger the acid → lower the pH
- the less HA acid molecules dissociate → the lower the Ka value → the weaker the acid → higher the pH
Note, in our case "ionization" and "dissociation" are often used interchangeably. But to be specific:
- Ionization is the specific process involving the formation of ions
- Dissociation is a more general term involving separating a substance into smaller molecules, ions, or atoms.
Let's see how well you understand these equations so far!
Write out the equation and equilibrium expression for the reaction of nitrous acid (HNO2).
The answer is worked out below, but try and do it by yourself before referencing it!
Well, we know that the generic equation is: HA (aq) + H2O (l) ⇌ H3O+(aq) + A–(aq) .
So, plugging nitrous acid into this, we should get: HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq)
Note: we will use both of the following symbols to represent the hydronium ions: H+ and H3O+.
Once we have the correct equation, we can plug them into the Ka equilibrium expression.
$$K_{a}=\frac{[H^{+}][NO_{2}^{-}]}{[HNO_{2}]}$$
Weak Base Equilibrium Equation
Now let's look at the weak base equilibrium equation and ionization constant. In a weak base reaction, the basic aqueous molecules, B (aq), react with liquid water by accepting the H+ ions from the water. This forms aqueous hydroxide ions, OH-, and conjugate acid, BH+(aq).
The following generic equation can represent this weak base reaction:
$$B_{(aq)}+H_{2}O_{(l)}\rightleftharpoons BH^{+}_{(aq)}+OH^{-}_{(aq)}$$
Where:
- B = basic species.
- BH+ = conjugate acid of the base.
The equilibrium constant for weak bases is known as the base dissociation constant, Kb.
The base dissociation constant, Kb refers to the relative strength of bases, defined as the extent to which the base dissociates into an aqueous solution.
The Kb of a weak base is expressed by the equation:
$$K_{b}=\frac{[BH^{+}][OH^{-}]}{[B]}$$
Similar to Ka, this shows us that the more the base ionizes (BH+) → the higher the Kb → the stronger the base!
You will be provided the Ka and Kb equilibrium constants in problems. You should know when to use it, how to write the equilibrium constant expression from a chemical equation, and what the equilibrium constants tell you!
Acid-Base Equilibrium Problems
So, how does all this information about Ka, Kb, and equilibrium reactions help us? You will usually be asked to do one of three things:
Calculate the pH (or pOH) given an initial concentration of an acid (or base) and Ka (or Kb)
Calculate Ka (or Kb) given an initial concentration of an acid (or base) and pH (or pOH)
Calculate the concentration of H+ (or OH-) ions given the initial concentration of acid (or base) and Ka (or Kb)
Consider, this weak acid problem example!
What are the hydronium ion concentration and pH of a 0.010 M solution of acetic acid, CH3COOH?
Ka for acetic acid is 1.8 x 10–5. In other words, calculate the pH ( logarithm base 10 of the hydronium ion concentration) given the initial concentration of an acid and Ka.
Let's first identify what we have been given and what we are trying to find.
We have been given:
- Ka of acetic acid as 1.8 x 10-5
- Initial concentration of acetic acid is 0.010 M
We want to find:
- H3O+ concentration at equilibrium
- pH of acetic acid
1. The first step is to write out the balanced chemical equation.
In this case, we know that the weak acid is acetic acid, CH3COOH, reacting with water to produce the conjugate base, CH3COO-, and hydronium ions. Then:
$$CH_{3}COOH_{(aq)}+H_{2}O_{(l)}\rightleftharpoons CH3COO^{-}_{(aq)}+H_{3}O^{+}$$
2. Next, we write out the Ka constant expression,
$$1.8\cdot 10^{-5}=\frac{[H_{3}O^{+}][CH_{3COO^{-}}]}{[CH_{3}COOH]}$$
3. We next create an ICE table! (Notice that we are dropping the dimension of molarity, M, and only working with the numerical values in our ICE table):
| Reaction | CH3COOH (HA) | CH3COO- (A-) | H3O+ |
| Initial Concentration (I) | 0.010 | 0* | 0* |
| Change in Concentration (C) | -x | +x | +x |
| Equilibrium (E) | 0.010 - x | x | x |
Let's spend some time here unpacking how to use and fill in this ICE table. We can use this to figure out the concentration of H3O+.
- For Initial reaction, (I): We know the initial concentration of acetic acid is 0.010 M, and we assume that the initial concentrations of CH3COO- and H3O+ are zero.
- For the change in concentration, (C): we will say x is the amount of acid dissociating. Since acetic acid is dissociating (experiencing a reduction in concentration) we use, -x, to represent this change. Since the molar ratio of CH3COO- and H3O+ is 1:1 we can say [CH3COO-] = [H3O+], and because the concentration of these species is increasing we use the symbol, +x, to represent the change in both.
- At equilibrium (E), we add the two lines "Initial Concentration (I)" and "Change in Concentration (C)" to get the line "Equilibrium (E)". Then we place these values into our equilibrium expression.
4) Next, we put our equilibrium concentration values into our equilibrium constant expression and solve for x! This will tell us the concentration of H3O+.
$$1.8\cdot 10^{-5}=\frac{[x][x]}{[0.010-x]}=\frac{[x^{2}]}{[0.010-x]}$$
From here, we can either use the quadratic equation to solve for x, OR we can assume that because the weak acid will only slightly ionize, the initial concentration of [HA] is approximately the same as its equilibrium concentration!
If we make this assumption then we can say [0.010-x] ≈ [0.01]:
$$1.8\cdot 10^{-5}=\frac{x^{2}}{[0.010]}=\sqrt{1.8\cdot 10^{-5}(0.010)}$$
$$x=4.26\cdot 10^{-4}$$
Now, reinserting the original dimension of molarity, M, that we were originally working with, we find that the concentration change is,
$$x=4.26\cdot 10^{-4}M$$
Generic forms of these equations that are helpful to memorize are:
$$K_{a}=\frac{[H_{3}O^{+}]^{2}}{[Concentration\ of\ HA]}$$
$$[H_{3}O^{+}]=\sqrt{K_{a}\cdot (Concentration\ of\ HA)}$$
So we now know that the concentration of hydronium ions for the ionization of 0.01 M acetic acid is 4.26 x 10-4 M.
This assumption can only be made when x (change in concentration) is less than 5% of the initial concentration that is
( x / initial concentration of HA) x 100 < 5%. In this case, x is around 4% so we can use the simplified equation. Most questions will fall under this assumption, but it is important to know how to justify the use of it with this equation!
5) Finally, we find the pH by using the equation:
$$pH=-log(H^{+})=-log(4.26\cdot 10^{-4})=3.37$$
Here again, we inserted into the pH formula the concentration of hydronium ions for the ionization of 0.01 M acetic acid, 4.26 · 10-4 M, using only the numerical value by dropping the dimension of molarity, M.
You can easily apply these four steps to any weak acid or base equilibrium problem asking you to find pH. If you were working with a weak base, you would simply be solving for Kb and then need to find the pH from the pOH value, as will be explained in greater detail below.
Note: Given that, 14 = pH + pOH, we find the value for the pH by rearranging the equation to: pH = 14 - pOH.
The other type of question you will be asked deals with calculating Ka (or Kb) given an initial concentration of an acid (or base) and pH (or pOH).
A 0.73 M solution of ammonia has a pH value of 12.87. Determine the Kb for ammonia. In other words, we are asked to calculate the base dissociation constant, Kb, given the initial concentration of a base and the pH.
We have been given:
- pH of ammonia as 12.87
- Initial concentration of ammonia as 0.73 M
We want to find:
1) Write out chemical equation of ammonia
$$NH_{3\ (aq)} + H_{2}O_{(l)} \rightleftharpoons NH_{4\ (aq)}^{+} + OH^{-}_{(aq)}$$
2) We can use the pH of ammonia to find the [OH-] concentration:
$$pOH=14-12.37=1.13$$
$$[OH^{-}]=10^{-1.13}=7.4\cdot 10^{-2}$$
Notice, in this last step we have used the following formula, [OH- ] = 10-pOH, to find the concentration of hydroxide, OH-, ions.
3) We create an ICE table, plugging in the numerical value, 7.4 x 10-2, the concentration of hydroxide ions, for the change, x, and the numerical value for the initial concentration of ammonia, 0.73 M.
| Reaction | NH3 (B) | NH4+ (BH+) | OH- |
| Initial Concentration (I) | 0.73 | 0 | 0 |
| Change in Concentration (C) | -7.4 x 10-2 | +7.4 x 10-2 | +7.4 x 10-2 |
| Equilibrium (E) | 0.656 | 7.4 x 10-2 | 7.4 x 10-2 |
4) Plugging in our equilibrium values:
$$K_{b}=\frac{[NH_{4}^{+}][OH]}{[NH_{3}]}=\frac{[7.4\cdot 10^{-2}][7.4\cdot 10^{-2}]}{[0.656]}$$
$$K_{b}=8.3\cdot 10^{-3}$$
Here, is a summary of important equations:
| Trying to Find | Weak Acid | Weak Base |
| [H3O+]/[OH-] | $$K_{a}=\frac{[H_{3}O^{+}]^{2}}{[Concentration\ of\ HA]}$$ $$H_{3}O^{+}=\sqrt{K_{a}\cdot (Concentration\ oF\ HA)}$$ | $$K_{b}=\frac{[OH^{-}]^{2}}{[Concentration\ of\ base]}$$ $$OH^{-}=\sqrt{K_{b}\cdot (Concentration\ of\ base)}$$ |
| pH/pOH | $$pH=-log(H_{3}O^{+})$$ | $$pOH=-log(OH^{-})$$ $$14-pOH=pH$$ |
| Ka/Kb | $$[H_{3}O^{+}]=10^{-pH}$$ $$K_{a}=\frac{[H_{3}O^{+}]^{2}}{[Concentration\ of\ HA]}$$ | $$[OH^{-}]=10^{-pOH}$$ $$K_{b}=\frac{[OH^{-}]^{2}}{[Concentration\ of\ base]}$$ |
Weak Acid and Weak Base Examples
Common examples:
Weak Acids
- Acetic acid (CH3COOH).
- Formic acid (HCOOH).
- Hydrocyanic acid (HCN).
Weak Bases
- Ammonium Hydroxide (NH4OH).
- Aniline (C6H5NH2).
- Ammonia (NH3).
Weak acid and base equilibria - Key takeaways
- Weak Acids and Bases only partially ionize
- the extent to which they ionize is quantified by dissociation constants
- Weak Acids can be represented by the generic equation HA (aq) + H2O (l) ⇌ H3O+(aq) + A–(aq), and the equilibrium constant
- Weak Bases can be represented by the generic equation B (aq) + H2O l) ⇌ BH+(aq) + OH-(aq) and equilibrium constant
- To find the concentration of H3O+ or OH-, you can use the equation
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