# Weak acid-base equilibria

Weak acid and base ionization reactions and the related equilibrium constants,  Ka and Kb. Relating Ka and Kb to pH, and calculating percent dissociation.

## Key points:

• For a generic monoprotic weak acid $\text{HA}$ with conjugate base $\text{A}^-$, the equilibrium constant has the form:
$K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$
• The acid dissociation constant $K_\text{a}$ quantifies the extent of dissociation of a weak acid. The larger the value of $K_\text{a}$, the stronger the acid, and vice versa.
• For a generic weak base $\text{B}$ with conjugate acid $\text{BH}^+$, the equilibrium constant has the form:
$K_\text{b}=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$
• The base dissociation constant (or base ionization constant) $K_\text{b}$ quantifies the extent of ionization of a weak base. The larger the value of $K_\text{b}$, the stronger the base, and vice versa.

## Strong vs. weak acids and bases

Strong acids and strong bases refer to species that completely dissociate to form ions in solution. By contrast, weak acids and bases ionize only partially, and the ionization reaction is reversible. Thus, weak acid and base solutions contain multiple charged and uncharged species in dynamic equilibrium.
In this article, we will discuss acid and base dissociation reactions and the related equilibrium constants: $K_\text{a}$, the acid dissociation constant, and $K_\text{b}$, the base dissociation constant.

### Warm-up: Comparing acid strength and $\text{pH}$

#### Problem 1: Weak vs. strong acids at the same concentration

We have two aqueous solutions: a $2.0\,\text M$ solution of hydrofluoric acid, $\text{HF}(aq)$, and a $2.0\,\text{M}$ solution of hydrobromic acid, $\text{HBr}(aq)$. Which solution has the lower $\text{pH}$?

Answer: $\text{HBr}(aq)$ has the lower $\text{pH}$
We want to compare the relative amounts of $\text{H}^+$ in solution to compare the $\text{pH}$ values. Since the two acids have the same concentration, the acid that dissociates more in solution will have the lower $\text{pH}$.
$\text{HBr}(aq)$ is a strong acid that dissociates fully in solution:
$\text{HBr}(aq) \rightarrow \text{Br}^-(aq)+\text{H}^+(aq)$
In comparison, $\text{HF}(aq)$ is a weak acid that only partially dissociates in solution:
$\text{HF}(aq) \leftrightharpoons \text{F}^-(aq)+\text{H}^+(aq)$
Since $\text{HBr}(aq)$ is a stronger acid compared to $\text{HF}(aq)$, it will have a higher $[\text{H}^+]$ and lower $\text{pH}$.

#### Problem 2: Weak vs. strong acids at different concentrations

This time we have a $2.0\,\text M$ solution of hydrofluoric acid, $\text{HF}(aq)$, and a $1.0\,\text{M}$ solution of hydrobromic acid, $\text{HBr}(aq)$. Which solution has the lower $\text{pH}$?
Assume we don't know the equilibrium constant for the dissociation of hydrofluoric acid.

We want to compare the relative amounts of $\text{H}^+$ (or $\text{H}_3 \text O^+$) in solution to compare the $\text{pH}$ values. However, this time the two acids do not have the same concentration: $\text{HBr}(aq)$ is a strong acid that dissociates fully in solution, but we have a lower concentration of $\text{HBr}(aq)$ compared to the weak acid $\text{HF}(aq)$.
In order to answer this question, we will need to calculate $[\text H^+]$ for both of our solutions. This will require knowing the equilibrium constant for the dissociation of $\text{HF}(aq)$.
$\text{HF}(aq) \leftrightharpoons \text{F}^-(aq)+\text{H}^+(aq)$
$K_\text a=\dfrac{[\text F^-(aq)][\text H^+(aq)]}{[\text{HF}(aq)]}$

## Weak acids and the acid dissociation constant, $K_\text{a}$

Weak acids are acids that don't completely dissociate in solution. In other words, a weak acid is any acid that is not a strong acid.
The strength of a weak acid depends on how much it dissociates: the more it dissociates, the stronger the acid. In order to quantify the relative strengths of weak acids, we can look at the acid dissociation constant $K_\text{a}$, the equilibrium constant for the acid dissociation reaction.
For a generic monoprotic weak acid $\text{HA}$, the dissociation reaction in water can be written as follows:
A monoprotic acid is an acid that can lose one acidic proton per molecule of acid, such as $\text{HCl}$.
Some acids can lose more than one proton per acid molecule. Sulfuric acid, $\text H_2 \text{SO}_4$, is an example of a diprotic acid, which can lose up to two protons per molecule of $\text H_2 \text{SO}_4$. Triprotic acids have three acidic protons, such as phosphoric acid $(\text H_3 \text{PO}_4)$.
$\text{HA}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{H}_3\text{O}^+(aq)+\text{A}^-(aq)$
Based on this reaction, we can write our expression for equilibrium constant $K_\text{a}$:
$K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$
Here is a quick review of the rules for writing equilibrium constants:
• Concentrations of products are multiplied together and placed in the numerator.
• Concentrations of reactants are multiplied together and placed in the denominator.
• All concentrations are raised to a power equal to the molar coefficient of each species from the balanced equation. (Therefore, before writing the equilibrium expression, we'll need to have a balanced equation!)
• Only species in the aqueous and gaseous phases are included in the equilibrium expression. Pure solids and liquids are omitted.
The equilibrium expression is a ratio of products to reactants. The more $\text{HA}$ dissociates into $\text{H}^+$ and the conjugate base $\text{A}^-$, the stronger the acid, and the larger the value of $K_\text{a}$. Since $\text{pH}$ is related to $[\text H_3 \text O^+]$, the $\text{pH}$ of the solution will be a function of $K_\text{a}$ as well as the concentration of the acid: the $\text{pH}$ decreases as the concentration of the acid and/or $K_\text{a}$ increase.

## Common weak acids

Malic acid, $\text{C}_4\text{H}_6\text{O}_5$, is an organic acid found in apples. Image from Wikimedia Commons, CC BY-SA 3.0.
Carboxylic acids are a common functional group in organic weak acids, and they have the formula $-\text{COOH}$. Malic acid $(\text{C}_4\text{H}_6\text{O}_5)$, an organic acid that contains two carboxylic acid groups, contributes to the tart flavor of apples and some other fruits. Since there are two carboxylic acid groups in the molecule, malic acid can potentially donate up to two protons.
The table below lists some more examples of weak acids and their $K_\text{a}$ values.
NameFormula$K_\text{a}(25\,^\circ\text{C})$
Ammonium$\text{NH}_4^+$$5.6\times10^{-10}$
Chlorous acid$\text{HClO}_2$$1.2\times10^{-2}$
Hydrofluoric acid$\text{HF}$$7.2\times10^{-4}$
Acetic acid$\text{CH}_3\text{COOH}$$1.8\times10^{-5}$

Concept check: Based on the table above, which is a stronger acid$-$acetic acid or hydrofluoric acid?
The $K_\text{a}$ value for $\text{HF}$ is greater than the $K_\text{a}$ value for $\text{CH}_3\text{COOH}$ $(7.2\times10^{-4}>1.8\times 10^{-5})$. The larger $K_\text{a}$ value means that hydrofluoric acid will dissociate in solution more than acetic acid. Therefore, hydrofluoric acid is a stronger acid than acetic acid.

## Example 1: Calculating % dissociation of a weak acid

One way to quantify how much a weak acid has dissociated in solution is to calculate the percent dissociation. The percent dissociation for weak acid $\text {HA}$ can be calculated as follows:
If nitrous acid $(\text{HNO}_2)$ has a $K_\text{a}$ of $4.0\times10^{-4}$ at $25\,^\circ\text{C}$, what is the percent dissociation of nitrous acid in a $0.400\text{ M}$ solution?
Let's go through this example step-by-step!

### Step 1: Write the balanced acid dissociation reaction

First, let's write the balanced dissociation reaction of $\text{HNO}_2$ in water. Nitrous acid can donate a proton to water to form $\text{NO}_2^-(aq)$:
$\text{HNO}_2(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{H}_3\text{O}^+(aq)+\text{NO}_2^-(aq)$

### Step 2: Write the expression for $K_\text{a}$

From the equation in Step 1, we can write the $K_\text{a}$ expression for nitrous acid:
$K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{NO}_2^-]}{[\text{HNO}_2]}=4.0\times10^{-4}$

### Step 3: Find $[\text H^+]$ and $[\text{NO}_2^-]$ at equilibrium

Next, we can use an $\text{ICE}$ table to determine algebraic expressions for the equilibrium concentrations in our $K_\text{a}$ expression:
$\text{HNO}_2(aq)$$\rightleftharpoons$$\text{H}_3\text{O}^+$$\text{NO}_2^-$
Initial$0.400\,\text M$$0$$0$
Change$-x$$+x$$+x$
Equilibrium$0.400\,\text M-x$$x$$x$
Plugging the equilibrium concentrations into our $K_\text{a}$ expression, we get:
$K_\text{a}=\dfrac{(x)(x)}{(0.400\,\text M-x)}=4.0\times10^{-4}$
Simplifying this expression, we get the following:
$\dfrac{x^2}{0.400\,\text M-x}=4.0\times10^{-4}$
This is a quadratic equation that can be solved for $x$ either by using the quadratic formula or an approximation method.
To learn about using approximation methods, check out the video on the small x approximation for small Kc values.
Either method will give $x=0.0126\text{ M}$. Therefore, $[\text{NO}_2^-]=[\text{H}_3\text{O}^+]=0.0126\text{ M}$.

### Step 4: Calculate percent dissociation

To calculate percent dissociation, we can use the equilibrium concentrations we found in Step 3:
Therefore, $3.2\%$ of the $\text{HNO}_2$ in solution has dissociated into $\text H^+$ and $\text{NO}_2^-$ ions.

## Weak bases and $K_\text{b}$

Let's now examine the base dissociation constant (also called the base ionization constant) $K_\text{b}$. We can start by writing the ionization reaction for a generic weak base $\text{B}$ in water. In this reaction, the base accepts a proton from water to form hydroxide and the conjugate acid, $\text{BH}^+$:
$\text{B}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{BH}^+(aq)+\text{OH}^-(aq)$
Good question! Perhaps a better name would be the "base ionization constant." But for whatever reason (perhaps to be consistent with the naming of $K_\text{a}$), $K_\text{b}$ is called the "base dissociation constant" in most chemistry textbooks, so we will stick with that convention here.
We can write the expression for equilibrium constant $K_\text{b}$ as follows:
$K_\text{b}=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$
From this ratio, we can see that the more the base ionizes to form $\text{BH}^+$, the stronger the base, and the larger the value of $K_\text{b}$. As such, the $\text{pH}$ of the solution will be a function of both the value of $K_\text{b}$ as well as the concentration of the base.

## Example 2: Calculating the $\text{pH}$ of a weak base solution

What is the $\text{pH}$ of a $1.50\text{ M}$ solution of ammonia, $\text{NH}_3$? $(K_\text{b}=1.8\times10^{-5})$
This example is an equilibrium problem with one extra step: finding $\text{pH}$ from $[\text {OH}^-]$. Let's go through the calculation step-by-step.

### Step 1: Write the balanced ionization reaction

First, let's write out the base ionization reaction for ammonia. Ammonia will accept a proton from water to form ammonium, $\text{NH}_4^+$:
$\text{NH}_3(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{NH}_4^+(aq)+\text{OH}^-(aq)$

### Step 2: Write the expression for $K_\text{b}$

From this balanced equation, we can write an expression for $K_\text{b}$:
$K_\text{b}=\dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}=1.8\times10^{-5}$

### Step 3: Find $[\text{NH}_4^+]$ and $[\text{OH}^-]$ at equilibrium

To determine the equilibrium concentrations, we use an $\text{ICE}$ table:
$\text{NH}_3(aq)$$\rightleftharpoons$$\text{NH}_4^+$$\text{OH}^-$
Initial$1.50\,\text M$$0$$0$
Change$-x$$+x$$+x$
Equilibrium$1.50\,\text M-x$$x$$x$
Plugging the equilibrium values into our $K_\text{b}$ expression, we get the following:
$K_\text{b}=\dfrac{(x)(x)}{1.50\,\text M-x}=1.8\times10^{-5}$
Simplifying, we get:
$\dfrac{x^2}{1.50\,\text M-x}=1.8\times10^{-5}$
This is a quadratic equation that can be solved by using the quadratic formula or an approximation method. Either method will yield the solution
$x=[\text{OH}^-]=5.2\times10^{-3}\text{ M}$

### Step 4: Find $\text{pH}$ from $[\text{OH}^-]$

Now that we know the concentration of hydroxide, we can calculate $\text{pOH}$:
\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(5.2\times10^{-3})\\ \\ &=2.28\end{aligned}
Recall that at $25\,^\circ\text{C}$, $\text{pH}+\text{pOH}=14$. Rearranging this equation, we have:
$\text{pH}=14-\text{pOH}$
Plugging in our value for $\text{pOH}$, we get:
$\text{pH}=14.00-(2.28)=11.72$
Therefore, the $\text{pH}$ of the solution is 11.72.

## Common weak bases

At left, structure of pyridine. On right, structure of a generic amine: a neutral nitrogen atom with single bonds to R1, R2, and R3.
Pyridine (left) is cyclic nitrogen-containing compound. Amines (right) are organic compounds containing a neutral nitrogen atom with three single bonds to hydrogen or carbon. Both molecules act as weak bases.
From soaps to household cleaners, weak bases are all around us. Amines, a neutral nitrogen with three bonds to other atoms (usually a carbon or hydrogen), are common functional groups in organic weak bases.
Amines act as bases because nitrogen's lone pair of electrons can accept an $\text H^+$. Ammonia, $\text{NH}_3$ is an example of an amine base. Pyridine, $\text C_5 \text H_5 \text N$, is another example of a nitrogen-containing base.

## Summary

• For a generic monoprotic weak acid $\text{HA}$ with conjugate base $\text{A}^-$, the equilibrium constant has the form:
$K_\text{a}=\dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$
• The acid dissociation constant $K_\text{a}$ quantifies the extent of dissociation of a weak acid. The larger the value of $K_\text{a}$, the stronger the acid, and vice versa.
• For a generic weak base $\text{B}$ with conjugate acid $\text{BH}^+$, the equilibrium constant has the form:
$K_\text{b}=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$
• The base dissociation constant (or base ionization constant) $K_\text{b}$ quantifies the extent of ionization of a weak base. The larger the value of $K_\text{b}$, the stronger the base, and vice versa.

1. Weak Acids and Bases” from UC Davis ChemWiki, CC BY-NC-SA 3.0

Zumdahl, S.S., and Zumdahl S.A. (2003). Atomic Structure and Periodicity. In Chemistry (6th ed., pp. 290-94), Boston, MA: Houghton Mifflin Company.

## Try it!

### Problem 1: Finding $K_\text{b}$ from $\text{pH}$

A $1.50\text{ M}$ solution of pyridine, $\text{C}_5\text{H}_5\text{N}$, has a $\text{pH}$ of $9.70$ at $25\,^\circ\text{C}$. What is the $K_\text{b}$ of pyridine?

Let's first write the base ionization reaction for pyridine:
$\text{C}_5\text{H}_5\text{N}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{C}_5\text{H}_5\text{NH}^+(aq)+\text{OH}^-(aq)$
The corresponding $K_\text{b}$ expression is:
$K_\text{b}=\dfrac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]}$
To calculate $K_\text{b}$, we'll need to determine the concentrations of the species in the equilibrium expression.
We can make an $\text{ICE}$ table for the ionization reaction to find the concentrations at equilibrium:
$\text{C}_5\text{H}_5\text{N}(aq)$$\rightleftharpoons$$\text{C}_5\text{H}_5\text{NH}^+$$\text{OH}^-(aq)$
Initial$1.50\text{ M}$$0\text{ M}$$0\text{ M}$
Change$-x$$+x$$+x$
Equilibrium$1.50\text{ M}-x$$x$$x$
Plugging our expressions for the equilibrium concentrations of each species into the $K_\text{b}$ equation, we have:
$K_\text{b}=\dfrac{(x)(x)}{(1.50\text{ M}-x)}=\dfrac{x^2}{1.50\text{ M}-x}$
We can use $\text{pH}$ to solve for $[\text{OH}^-]$, which will give us $x$. First we can use $\text{pH}$ to calculate the $\text{pOH}$ of the solution:
$\text{pOH}=14-9.70=4.30$
We can then use $\text{pOH}$ to solve for $[\text{OH}^-]$, which will give us $x$.
\begin{aligned}[\text{OH}^-]&=10^{-\text{pOH}}\\ \\ &=10^{-4.30}\\ \\ &=5.05\times10^{-5}\text{ M}\end{aligned}
Therefore, $x=5.05\times10^{-5}\text{ M}$.
Finally, we can plug this value for $x$ into our $K_\text{b}$ expression to solve for $K_\text{b}$:
\begin{aligned}K_\text{b}&=\dfrac{x^2}{1.50-x}\\ \\ &=\dfrac{(5.05\times10^{-5})^2}{(1.50-5.05\times10^{-5})}\\ \\ &=1.7\times10^{-9}\end{aligned}
Therefore, the $K_\text{b}$ of pyridine is $1.7\times10^{-9}$.