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## Chemistry library

### Course: Chemistry library>Unit 13

Lesson 2: Acid-base equilibria

# 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 start text, H, A, end text with conjugate base start text, A, end text, start superscript, minus, end superscript, the equilibrium constant has the form:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction
• The acid dissociation constant K, start subscript, start text, a, end text, end subscript quantifies the extent of dissociation of a weak acid. The larger the value of K, start subscript, start text, a, end text, end subscript, the stronger the acid, and vice versa.
• For a generic weak base start text, B, end text with conjugate acid start text, B, H, end text, start superscript, plus, end superscript, the equilibrium constant has the form:
K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, B, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, B, end text, close bracket, end fraction
• The base dissociation constant (or base ionization constant) K, start subscript, start text, b, end text, end subscript quantifies the extent of ionization of a weak base. The larger the value of K, start subscript, start text, b, end text, end subscript, 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, start subscript, start text, a, end text, end subscript, the acid dissociation constant, and K, start subscript, start text, b, end text, end subscript, the base dissociation constant.

### Warm-up: Comparing acid strength and $\text{pH}$start text, p, H, end text

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

We have two aqueous solutions: a 2, point, 0, start text, M, end text solution of hydrofluoric acid, start text, H, F, end text, left parenthesis, a, q, right parenthesis, and a 2, point, 0, start text, M, end text solution of hydrobromic acid, start text, H, B, r, end text, left parenthesis, a, q, right parenthesis. Which solution has the lower start text, p, H, end text?

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

This time we have a 2, point, 0, start text, M, end text solution of hydrofluoric acid, start text, H, F, end text, left parenthesis, a, q, right parenthesis, and a 1, point, 0, start text, M, end text solution of hydrobromic acid, start text, H, B, r, end text, left parenthesis, a, q, right parenthesis. Which solution has the lower start text, p, H, end text?
Assume we don't know the equilibrium constant for the dissociation of hydrofluoric acid.

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

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, start subscript, start text, a, end text, end subscript, the equilibrium constant for the acid dissociation reaction.
For a generic monoprotic weak acid start text, H, A, end text, the dissociation reaction in water can be written as follows:
start text, H, A, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, A, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
Based on this reaction, we can write our expression for equilibrium constant K, start subscript, start text, a, end text, end subscript:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction
The equilibrium expression is a ratio of products to reactants. The more start text, H, A, end text dissociates into start text, H, end text, start superscript, plus, end superscript and the conjugate base start text, A, end text, start superscript, minus, end superscript, the stronger the acid, and the larger the value of K, start subscript, start text, a, end text, end subscript. Since start text, p, H, end text is related to open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, the start text, p, H, end text of the solution will be a function of K, start subscript, start text, a, end text, end subscript as well as the concentration of the acid: the start text, p, H, end text decreases as the concentration of the acid and/or K, start subscript, start text, a, end text, end subscript increase.

## Common weak acids

Malic acid, start text, C, end text, start subscript, 4, end subscript, start text, H, end text, start subscript, 6, end subscript, start text, O, end text, start subscript, 5, end subscript, 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 minus, start text, C, O, O, H, end text. Malic acid left parenthesis, start text, C, end text, start subscript, 4, end subscript, start text, H, end text, start subscript, 6, end subscript, start text, O, end text, start subscript, 5, end subscript, right parenthesis, 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, start subscript, start text, a, end text, end subscript values.
NameFormulaK, start subscript, start text, a, end text, end subscript, left parenthesis, 25, degrees, start text, C, end text, right parenthesis
Ammoniumstart text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript5, point, 6, times, 10, start superscript, minus, 10, end superscript
Chlorous acidstart text, H, C, l, O, end text, start subscript, 2, end subscript1, point, 2, times, 10, start superscript, minus, 2, end superscript
Hydrofluoric acidstart text, H, F, end text7, point, 2, times, 10, start superscript, minus, 4, end superscript
Acetic acidstart text, C, H, end text, start subscript, 3, end subscript, start text, C, O, O, H, end text1, point, 8, times, 10, start superscript, minus, 5, end superscript

Concept check: Based on the table above, which is a stronger acidminusacetic acid or hydrofluoric 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 start text, H, A, end text can be calculated as follows:
$\text{% dissociation} =\dfrac{[\text A^-(aq)]}{[\text{HA}(aq)]} \times 100\%$
If nitrous acid left parenthesis, start text, H, N, O, end text, start subscript, 2, end subscript, right parenthesis has a K, start subscript, start text, a, end text, end subscript of 4, point, 0, times, 10, start superscript, minus, 4, end superscript at 25, degrees, start text, C, end text, what is the percent dissociation of nitrous acid in a 0, point, 400, start text, space, M, end text 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 start text, H, N, O, end text, start subscript, 2, end subscript in water. Nitrous acid can donate a proton to water to form start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis:
start text, H, N, O, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis

### Step 2: Write the expression for $K_\text{a}$K, start subscript, start text, a, end text, end subscript

From the equation in Step 1, we can write the K, start subscript, start text, a, end text, end subscript expression for nitrous acid:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, N, O, end text, start subscript, 2, end subscript, close bracket, end fraction, equals, 4, point, 0, times, 10, start superscript, minus, 4, end superscript

### Step 3: Find $[\text H^+]$open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and $[\text{NO}_2^-]$open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket at equilibrium

Next, we can use an start text, I, C, E, end text table to determine algebraic expressions for the equilibrium concentrations in our K, start subscript, start text, a, end text, end subscript expression:
start text, H, N, O, end text, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis\rightleftharpoonsstart text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscriptstart text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript
Initial0, point, 400, start text, M, end text00
Changeminus, xplus, xplus, x
Equilibrium0, point, 400, start text, M, end text, minus, xxx
Plugging the equilibrium concentrations into our K, start subscript, start text, a, end text, end subscript expression, we get:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, left parenthesis, x, right parenthesis, left parenthesis, x, right parenthesis, divided by, left parenthesis, 0, point, 400, start text, M, end text, minus, x, right parenthesis, end fraction, equals, 4, point, 0, times, 10, start superscript, minus, 4, end superscript
Simplifying this expression, we get the following:
start fraction, x, squared, divided by, 0, point, 400, start text, M, end text, minus, x, end fraction, equals, 4, point, 0, times, 10, start superscript, minus, 4, end superscript
This is a quadratic equation that can be solved for x either by using the quadratic formula or an approximation method.
Either method will give x, equals, 0, point, 0126, start text, space, M, end text. Therefore, open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket, equals, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, 0, point, 0126, start text, space, M, end text.

### Step 4: Calculate percent dissociation

To calculate percent dissociation, we can use the equilibrium concentrations we found in Step 3:
\begin{aligned}{\text{% dissociation}}&=\dfrac{[\text{NO}_2^-]}{[\text{HNO}_2]}\\ \\ &=\dfrac{0.0126\text{ M}}{0.400\text{ M}}\times100\%\\ \\ &=3.2\%\end{aligned}
Therefore, 3, point, 2, percent of the start text, H, N, O, end text, start subscript, 2, end subscript in solution has dissociated into start text, H, end text, start superscript, plus, end superscript and start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript ions.

## Weak bases and $K_\text{b}$K, start subscript, start text, b, end text, end subscript

Let's now examine the base dissociation constant (also called the base ionization constant) K, start subscript, start text, b, end text, end subscript. We can start by writing the ionization reaction for a generic weak base start text, B, end text in water. In this reaction, the base accepts a proton from water to form hydroxide and the conjugate acid, start text, B, H, end text, start superscript, plus, end superscript:
start text, B, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, B, H, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
We can write the expression for equilibrium constant K, start subscript, start text, b, end text, end subscript as follows:
K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, B, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, B, end text, close bracket, end fraction
From this ratio, we can see that the more the base ionizes to form start text, B, H, end text, start superscript, plus, end superscript, the stronger the base, and the larger the value of K, start subscript, start text, b, end text, end subscript. As such, the start text, p, H, end text of the solution will be a function of both the value of K, start subscript, start text, b, end text, end subscript as well as the concentration of the base.

## Example 2: Calculating the $\text{pH}$start text, p, H, end text of a weak base solution

What is the start text, p, H, end text of a 1, point, 50, start text, space, M, end text solution of ammonia, start text, N, H, end text, start subscript, 3, end subscript? left parenthesis, K, start subscript, start text, b, end text, end subscript, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript, right parenthesis
This example is an equilibrium problem with one extra step: finding start text, p, H, end text from open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket. 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, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript:
start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis

### Step 2: Write the expression for $K_\text{b}$K, start subscript, start text, b, end text, end subscript

From this balanced equation, we can write an expression for K, start subscript, start text, b, end text, end subscript:
K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, N, H, end text, start subscript, 3, end subscript, close bracket, end fraction, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript

### Step 3: Find $[\text{NH}_4^+]$open bracket, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, close bracket and $[\text{OH}^-]$open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket at equilibrium

To determine the equilibrium concentrations, we use an start text, I, C, E, end text table:
start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis\rightleftharpoonsstart text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscriptstart text, O, H, end text, start superscript, minus, end superscript
Initial1, point, 50, start text, M, end text00
Changeminus, xplus, xplus, x
Equilibrium1, point, 50, start text, M, end text, minus, xxx
Plugging the equilibrium values into our K, start subscript, start text, b, end text, end subscript expression, we get the following:
K, start subscript, start text, b, end text, end subscript, equals, start fraction, left parenthesis, x, right parenthesis, left parenthesis, x, right parenthesis, divided by, 1, point, 50, start text, M, end text, minus, x, end fraction, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript
Simplifying, we get:
start fraction, x, squared, divided by, 1, point, 50, start text, M, end text, minus, x, end fraction, equals, 1, point, 8, times, 10, start superscript, minus, 5, end superscript
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, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, 5, point, 2, times, 10, start superscript, minus, 3, end superscript, start text, space, M, end text

### Step 4: Find $\text{pH}$start text, p, H, end text from $[\text{OH}^-]$open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket

Now that we know the concentration of hydroxide, we can calculate start text, p, O, H, end text:
\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(5.2\times10^{-3})\\ \\ &=2.28\end{aligned}
Recall that at 25, degrees, start text, C, end text, start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14. Rearranging this equation, we have:
start text, p, H, end text, equals, 14, minus, start text, p, O, H, end text
Plugging in our value for start text, p, O, H, end text, we get:
start text, p, H, end text, equals, 14, point, 00, minus, left parenthesis, 2, point, 28, right parenthesis, equals, 11, point, 72
Therefore, the start text, p, H, end text 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 start text, H, end text, start superscript, plus, end superscript. Ammonia, start text, N, H, end text, start subscript, 3, end subscript is an example of an amine base. Pyridine, start text, C, end text, start subscript, 5, end subscript, start text, H, end text, start subscript, 5, end subscript, start text, N, end text, is another example of a nitrogen-containing base.

## Summary

• For a generic monoprotic weak acid start text, H, A, end text with conjugate base start text, A, end text, start superscript, minus, end superscript, the equilibrium constant has the form:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction
• The acid dissociation constant K, start subscript, start text, a, end text, end subscript quantifies the extent of dissociation of a weak acid. The larger the value of K, start subscript, start text, a, end text, end subscript, the stronger the acid, and vice versa.
• For a generic weak base start text, B, end text with conjugate acid start text, B, H, end text, start superscript, plus, end superscript, the equilibrium constant has the form:
K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, B, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, B, end text, close bracket, end fraction
• The base dissociation constant (or base ionization constant) K, start subscript, start text, b, end text, end subscript quantifies the extent of ionization of a weak base. The larger the value of K, start subscript, start text, b, end text, end subscript, the stronger the base, and vice versa.

## Try it!

### Problem 1: Finding $K_\text{b}$K, start subscript, start text, b, end text, end subscript from $\text{pH}$start text, p, H, end text

A 1, point, 50, start text, space, M, end text solution of pyridine, start text, C, end text, start subscript, 5, end subscript, start text, H, end text, start subscript, 5, end subscript, start text, N, end text, has a start text, p, H, end text of 9, point, 70 at 25, degrees, start text, C, end text. What is the K, start subscript, start text, b, end text, end subscript of pyridine?

## Want to join the conversation?

• for the example 1: calculating the % dissociation, the part where the ICE table is used and you can use the quadratic formula to find concentration "x", the two answers I got for x was x= -0.01285M and x=0.01245M. You guys said the concentration I should have found is 0.0126M. I was trying to figure out which of the "x" is the correct one ( I assume since a negative concentration can not exist, the concentration has to be 0.01245M) and I have gone through my calculations a few times, and I don't know where I went wrong. Is it a rounding error?
• What you've calculated using the quadratic formula is correct. As you rightly say, you can't have a negative concentration, so the viable answer is 0.01245 M.

If you use the small x assumption, then it comes out as 0.012649 M. You would expect a small difference in the result depending on which method is used. So it's strange that the article says that either method will give x=0.0126 M.

My guess is that the author of the article probably used the small x method and perhaps didn't check the quadratic formula result.

However, the difference is so small that it's negligible,
• In example 1, why is the formula for % dissociation [A-]/[HA]*100% and not [H3O+]/[HA]*100% or [H3O+][A-]/[HA]*100%? Is this a stupid question? Sorry, if it is.
• It's not a stupid question.
You can use either [A⁻]/[HA]₀× 100 % or [H₃O⁺]/[HA]₀× 100 % to calculate % dissociation of a weak acid.
Your last formula is just Kₐ×100 %, so that one is wrong.
• Is it possible to find the percent dissociation of a weak base, or is it only applicable to weak acids?
• You can find the percent ionization of a weak base. This is analogous to finding the percent dissociation of an acid, except you are interested in what percentage of the base became ionized by bonding to an H+ ion.
• So all of these are happening in water. What if these reactions aren't happening in water? Is there a situation like that? Like in gas? or something? Not something necessary to think about?
• This is an interesting area, but I don't know much about this – I certainly don't remember learning about this in 1st or 2nd year undergraduate chemistry. However, acid-base reactions definitely take place in solvents other than water and even in the gas phase.

For non-aqueous solvents, the pKa values for an acid may be higher or lower than they are in water depending on the acid and the solvent. The wikipedia article on this is somewhat readable:
https://en.wikipedia.org/wiki/Acid_dissociation_constant#Acidity_in_nonaqueous_solutions

An example I found for a gas phase acid-base reaction:
(NB: the demonstrator shows a deplorable lack of concern for modeling safe laboratory practices!)
• why are we making ICE tables here?
• ICE tables are just a way of organizing data. You don't have to use them, but it often is one of the best ways to keep track of lots of different numbers.
• Well i'm a 3rd grader and I want to learn this and isn't OH weak?
• OH- is actually considered to be a strong base, as its conjugate acid, water (H2O), is a weak acid.