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### Course: Chemistry archive > Unit 8

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
with conjugate base$\text{HA}$ , the equilibrium constant has the form:${\text{A}}^{-}$

- The
**acid dissociation constant** quantifies the extent of dissociation of a weak acid. The larger the value of${K}_{\text{a}}$ , the stronger the acid, and vice versa.${K}_{\text{a}}$ - For a generic weak base
with conjugate acid$\text{B}$ , the equilibrium constant has the form:${\text{BH}}^{+}$

- The
**base dissociation constant**(or**base ionization constant**) quantifies the extent of ionization of a weak base. The larger the value of${K}_{\text{b}}$ , the stronger the base, and vice versa.${K}_{\text{b}}$

## 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 $2.0{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ solution of hydrofluoric acid, $\text{HF}(aq)$ , and a $2.0{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ solution of hydrobromic acid, $\text{HBr}(aq)$ . Which solution has the lower $\text{pH}$ ?

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

This time we have a $2.0{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ solution of hydrofluoric acid, $\text{HF}(aq)$ , and a $1.0{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{M}$ 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.*

## 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:

Based on this reaction, we can write our expression for equilibrium constant ${K}_{\text{a}}$ :

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 ${K}_{\text{a}}$ . 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.

*stronger*the acid, and the larger the value of## Common weak acids

*Carboxylic acids*are a common functional group in organic weak acids, and they have the formula

The table below lists some more examples of weak acids and their ${K}_{\text{a}}$ values.

Name | Formula | |
---|---|---|

Ammonium | ||

Chlorous acid | ||

Hydrofluoric acid | ||

Acetic acid |

**Concept check: Based on the table above, which is a stronger acid**$-$ acetic 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 $\text{HA}$ can be calculated as follows:

**If nitrous acid**$({\text{HNO}}_{2})$ has a ${K}_{\text{a}}$ of $4.0\times {10}^{-4}$ at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ , 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 ${\text{NO}}_{2}^{-}(aq)$ :

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

From the equation in ${K}_{\text{a}}$ expression for nitrous acid:

**Step 1**, we can write the### 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:

Initial | ||||

Change | ||||

Equilibrium |

Plugging the equilibrium concentrations into our ${K}_{\text{a}}$ expression, we get:

Simplifying this expression, we get the following:

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 $x=0.0126\text{M}$ . 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,**$3.2\mathrm{\%}$ 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, ${\text{BH}}^{+}$ :

We can write the expression for equilibrium constant ${K}_{\text{b}}$ as follows:

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, ${\text{NH}}_{3}$ ? $({K}_{\text{b}}=1.8\times {10}^{-5})$

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, ${\text{NH}}_{4}^{+}$ :

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

From this balanced equation, we can write an expression for ${K}_{\text{b}}$ :

### Step 3: Find $[{\text{NH}}_{4}^{+}]$ and $[{\text{OH}}^{-}]$ at equilibrium

To determine the equilibrium concentrations, we use an $\text{ICE}$ table:

Initial | ||||

Change | ||||

Equilibrium |

Plugging the equilibrium values into our ${K}_{\text{b}}$ expression, we get the following:

Simplifying, we get:

This is a quadratic equation that can be solved by using the quadratic formula or an approximation method. Either method will yield the solution

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

Now that we know the concentration of hydroxide, we can calculate $\text{pOH}$ :

Recall that at $25{{\textstyle \phantom{\rule{0.167em}{0ex}}}}^{\circ}\text{C}$ , $\text{pH}+\text{pOH}=14$ . Rearranging this equation, we have:

Plugging in our value for $\text{pOH}$ , we get:

**Therefore, the**$\text{pH}$ of the solution is 11.72.

## Common 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
with conjugate base$\text{HA}$ , the equilibrium constant has the form:${\text{A}}^{-}$

- The acid dissociation constant
quantifies the extent of dissociation of a weak acid. The larger the value of${K}_{\text{a}}$ , the stronger the acid, and vice versa.${K}_{\text{a}}$ - For a generic weak base
with conjugate acid$\text{B}$ , the equilibrium constant has the form:${\text{BH}}^{+}$

- The base dissociation constant (or base ionization constant)
quantifies the extent of ionization of a weak base. The larger the value of${K}_{\text{b}}$ , the stronger the base, and vice versa.${K}_{\text{b}}$

## Try it!

## 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?(9 votes)
- 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,(19 votes)

- 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.(6 votes)
- 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.(8 votes)

- Is it possible to find the percent dissociation of a weak base, or is it only applicable to weak acids?(4 votes)
- 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.(8 votes)

- 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?(5 votes)
- 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:

https://www.youtube.com/watch?v=60vtFe42sGs

(NB: the demonstrator shows a deplorable lack of concern for modeling safe laboratory practices!)(5 votes)

- After reading the article I understood that ICE Table applies only to the weak acid and bases and not to the strong acid and bases. In case of the strong acid and base we can directly use the concentration of the compound given because it dissociates totally. I am correct right?(4 votes)
- I guess you are correct, because, as strong acids and bases dissociate completely in an aqueous solution, it is safe to say that their concentrations can be used in calculations. ICE literally stands for Initial, Change and Equilibrium, so, while it IS true that we have an equilibrium in even strong acids and bases, I think the reaction is favored so strong in the direction of the forward reaction of dissociation, so, the effect of the reverse reaction is negligible. Hence, there is no need for ICE tables.(1 vote)

- why are we making ICE tables here?(3 votes)
- 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.(4 votes)

- Well i'm a 3rd grader and I want to learn this and isn't OH weak?(2 votes)
- OH- is actually considered to be a strong base, as its conjugate acid, water (H2O), is a weak acid.(5 votes)

- In the percent dissocation example above, and in the last step (step 4), why did we use the [HNO3] as 0.400 M rather than (0.400-x) which should be the more accurate concentration (after we found x=0.0126)?(2 votes)
- Because that’s how percent ionisation is defined. It’s dissociated / initial.(3 votes)

- Which is more dangerous: a dilute strong acid or a concentrated weak base.(2 votes)
- In the ICE tables, is the change always -x? If not, under what conditions would be higher (e.g. -2x or some other number)?

Thank you(2 votes)