- This article is about
**weak acids and bases**in chemistry. - First of all, we'll
**define weak acids and bases.** - We'll then consider the
**pH of weak acids and bases**. - To consolidate our learning, we'll
**compare strong and weak acids and bases**, before looking at a**weak acid and base chart**. - After that, we'll explore
**K**and_{a}**K**_{b}. - Finally, we'll turn our attention to
**weak acid and base titrations**.

## Defining weak acids and bases

What are **weak acids and bases**? Let's look at a few definitions to make this clear.

### Weak acids

To understand weak acids, we first need to define **strong acid**, so that you can accurately compare the two.

A **strong acid** is an acid that dissociates fully in solution.

All acids are **proton donors**. Every molecule of a strong acid donates a proton when it reacts. We can represent this using the following equation. Note how the reaction isn’t reversible:

$\mathrm{HA}\left(\mathrm{aq}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

On the other hand, **weak acids** behave a little differently.

A **weak acid** is an acid that only partially dissociates in solution.

Weak acids form an **equilibrium**, in which the majority of the molecules present are acid molecules and only a tiny fraction donate their protons and dissociate into ions. The stronger an acid is, the more the equilibrium shifts to the right and the greater the concentration of hydrogen ions in solution.

$\mathrm{HA}\left(\mathrm{aq}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

Remember that acids in solution actually dissociate and react with water to form the hydronium ion, H_{3}O^{+}. In order to simplify the equation, we omit the water and replace the hydronium ion with the hydrogen ion. Here's the original version of the equation:

$\mathrm{HA}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{H}}_{3}{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

In contrast, here is the simplified version:

$\mathrm{HA}\left(\mathrm{aq}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

In fact, whenever you see the hydrogen ion in acid-base reactions, you should know that it actually represents the hydronium ion.

### Weak bases

We can also find **strong **and **weak bases**. Remember that a **base** is a proton acceptor.

A **strong base **is a base that dissociates fully in solution.

An example of a strong base is sodium hydroxide, NaOH. Every molecule of NaOH accepts a proton from water, dissociating in solution to form sodium ions and hydroxide ions:

$\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\to {\mathrm{NaH}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

We also have a simplified way of representing this reaction. Once again, we omit the water molecule:

$\mathrm{NaOH}\left(\mathrm{aq}\right)\hspace{0.17em}\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

In contrast to sodium hydroxide, ammonia is only a** weak** base.

A weak base is a base that only partially dissociates in solution.

Like with weak acids, weak bases form an **equilibrium** in which the backward reaction is strongly favoured, and only a small proportion of the molecules ionise. This means not every molecule of the weak base accepts a proton. We represent the dissociation of ammonia with the following equation:

${\mathrm{NH}}_{3}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {{\mathrm{NH}}_{4}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

The general equations for the dissociation of a strong base and weak base are shown below, with the dissociation of a strong base first:

$\mathrm{B}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{BH}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{B}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{BH}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Don’t get strong and concentrated mixed up - they mean totally different things. **Concentration** is a measure of how many acid or base molecules are dissolved in solution, whilst **strength** is a measure of the proportion of these molecules that dissociate into ions. You can concentrated weak acids, as well as dilute strong acids!

## pH of weak acids and bases

Weak acids have higher pH values than strong acids. Likewise, weak bases have lower pH values than strong bases. This is because weak acids and bases only partially ionise in solution.

Remember that pH is a measure of hydrogen ion concentration in solution. Furthermore, a lower pH means there is a *higher* concentration of hydrogen ions.

## Strong and weak acids and bases

Let's now take some time to compare strong and weak acids and bases. This table should recap the differences we've already mentioned, as well as introduce you to their reactions with metals and conductivities.

Strong acid | Weak acid | Strong base | Weak base | |

Behaviour in solution | Fully ionises | Partially ionises | Fully ionises | Partially ionises |

Equation | $\mathrm{HA}\to {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$ | $\mathrm{HA}\leftrightharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$ | $\mathrm{B}+{\mathrm{H}}_{2}\mathrm{O}\to {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}$ | $\mathrm{B}+{\mathrm{H}}_{2}\mathrm{O}\leftrightharpoons {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}$ |

pH (colour range) | 1-3 (red-orange) | 4-6 (orange-yellow) | 8-10 (aqua-blue) | 11-13 (blue-purple) |

Reaction with reactive metal | Vigorous reaction | Slow reaction | Vigorous reaction | Slow reaction |

Conductivity | Very good | Good-poor | Very good | Good-poor |

Example | HCl | CH_{3}COOH | NaOH | NH_{3} |

## Weak acids and bases chart

You’ll encounter both strong and weak acids and bases in everyday life. For example, concentrated hydrochloric acid, a strong acid, is used to remove algae from the bottoms of boats, whilst a more dilute solution is used in toilet cleaners. Hydrochloric acid is also the acid found in our stomachs that helps digest our food. Citric acid and ethanoic acid, the acids found in lemons and malt vinegar respectively, are both weak acids. Basic sodium bicarbonate is also known as baking soda, a useful leavening agent in baking, whilst the base calcium hydroxide is used to help neutralise acidic soil.

Here's a handy chart with common chemical examples of strong and weak acids and bases.

## Weak acids and K_{a}

Now let's move on to calculations involving weak acids and bases. We'll start by focusing on **K**_{a}.

In **The Ionic Product of Water**, we discussed **K**_{w}. K_{w} is a modified equilibrium constant for the dissociation of water. We can also get K_{a}, a modified equilibrium constant for the dissociation of weak acids.

The general equation of the equilibrium constant for the reaction $\mathrm{aA}+\mathrm{bB}\leftrightharpoons \mathrm{cC}+\mathrm{dD}$ is shown below:

${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[\mathrm{C}\right]}^{\mathrm{c}}{\left[\mathrm{D}\right]}^{\mathrm{d}}}{{\left[\mathrm{A}\right]}^{\mathrm{a}}{\left[\mathrm{B}\right]}^{\mathrm{b}}}$

Square brackets represent concentration, and the small letter represents the number of moles of each species in the chemical equation. For example, the equilibrium reaction, ${\mathrm{H}}_{2}\left(\mathrm{g}\right)+{\mathrm{I}}_{2}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{HI}\left(\mathrm{g}\right)$, has the following equilibrium constant:

${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[\mathrm{HI}\right]}^{2}}{\left[{\mathrm{H}}_{2}\right]\left[{\mathrm{I}}_{2}\right]}$

Let’s look at this from the point of view of a weak acid. It dissociates in solution with the equation $\mathrm{HA}\left(\mathrm{aq}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$. The reactant is the acid and the products are the hydrogen ions and the negative ions. This gives us the following equilibrium constant, known as K_{a}:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right(\mathrm{aq}\left)\right]\left[{\mathrm{A}}^{-}\right(\mathrm{aq}\left)\right]}{\left[\mathrm{HA}\right(\mathrm{aq}\left)\right]}$

You can remove the state symbols to simplify the equation if you want.

“But hang on!” We hear you cry. “How about the equation of a weak acid involving water and hydronium ions! Where have they gone?”

If we write out that equation, we get $\mathrm{HA}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{H}}_{3}{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$. This produces the following equation for the equilibrium constant:

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{H}}_{3}{\mathrm{O}}^{+}\right(\mathrm{aq}\left)\right]\left[{\mathrm{A}}^{-}\right(\mathrm{aq}\left)\right]}{\left[\mathrm{HA}\right(\mathrm{aq}\left)\right]\left[\mathrm{H}2\mathrm{O}\right(\mathrm{l}\left)\right]}$

However, the concentration of water is so large that it completely dominates all the other values in the equation. To form K_{a}, we simply omit it.

You should also remember that we use the hydrogen ion to represent the hydronium ion in acid-base reactions. This gives us the familiar equation for K_{a}:

${\mathrm{K}}_{\mathrm{a}}:\frac{\left[{\mathrm{H}}^{+}\right(\mathrm{aq}\left)\right]\left[{\mathrm{A}}^{-}\right(\mathrm{aq}\left)\right]}{\left[\mathrm{HA}\right(\mathrm{aq}\left)\right]}$

### The units of K_{a}

To find the units of K_{a}, we multiply and cancel down the units of all the species involved in the equation. All three species, [H^{+}], [A^{-}], and [HA], have the units mol dm^{-3}. The equation now looks like this:

$\frac{\mathrm{mol}{\mathrm{dm}}^{-3}\mathrm{x}\mathrm{mol}{\mathrm{dm}}^{-3}}{\mathrm{mol}{\mathrm{dm}}^{-3}}$

One of the mol dm^{-3} from the top of the fraction cancels out with the one on the bottom, leaving just one mol dm^{-3}:

$\frac{\mathrm{mol}{\mathrm{dm}}^{-3}\mathrm{x}\overline{)\mathrm{mol}{\mathrm{dm}}^{-3}}}{\overline{)\mathrm{mol}{\mathrm{dm}}^{-3}}}$

### K_{a} and pK_{a}

Just as pK_{w} is the negative log of K_{w}, pK_{a} is the negative log of K_{a}:

${\mathrm{pK}}_{\mathrm{a}}=-\mathrm{log}\left({\mathrm{K}}_{\mathrm{a}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{a}}={10}^{-{\mathrm{pK}}_{\mathrm{a}}}$

You should note the following relationships between K_{a}, pK_{a}, acid strength and pH:

- As K
_{a}increases, pK_{a}and pH both decrease. - As K
_{a}increases, acid strength increases.

### Finding Ka from pH

To calculate the pH of weak acids, you use the relationships between K_{a,} PK_{a}, and the concentration of the acid in solution. You’ll be given information about the acid’s PK_{a} or K_{a}. There are a few more steps compared to working out the pH of a strong acid, but it isn’t too tricky. Let’s have a look at an example together.

**Ethanoic acid, CH _{3}COOH, has **

**K**

_{a}= 1.74 x 10^{-5}. Calculate the pH of a 0.100 mol dm^{-3}solution of this weak acid.First of all, let’s look at the equation for the dissociation of ethanoic acid:

${\mathrm{CH}}_{3}\mathrm{COOH}\left(\mathrm{aq}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)$

To find pH, we need to know [H^{+}], the concentration of hydrogen ions in solution. Well, what do we know about K_{a}? It is a modified equilibrium constant for the dissociation of a weak acid such as ethanoic acid and involves [H^{+}]. For ethanoic acid, it looks like this:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{CH}3\mathrm{COOH}\right]}$

Let’s think about those values. The concentration of ethanoic acid was originally 0.100 mol dm^{-3}. At equilibrium, it will be a little less than that because some of the molecules will dissociate into ions. However, ethanoic is a weak acid and hardly any of the molecules dissociate - the equilibrium lies far to the left. We can therefore say that the concentration of ethanoic acid at equilibrium is still roughly 0.100 mol dm^{-3}. Let’s put that value into our equation:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{CH}3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{0.100}$

Look back at the equation. When one mole of ethanoic acid dissociates, it forms one mole of positive hydrogen ions, H^{+}, and one mole of acetate ions, CH_{3}COO^{-}. This means that the number of hydrogen ions in solution equals the number of acetate ions in solution, and thus they have the same concentrations:

${\mathrm{K}}_{\mathrm{a}}=\left[{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right]=\left[{\mathrm{H}}^{+}\right]$

We can replace [CH_{3}COO^{-}] with [H^{+}] in our equation for K_{a}:

${\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\mathrm{H}}^{+}\right]}^{2}}{0.100}$

The question gives us K_{a} , so we can substitute that in. Now we have an equation where the only unknown is [H^{+}]. We can solve it normally, as shown:

$1.74\mathrm{x}{10}^{-5}=\frac{{\left[{\mathrm{H}}^{+}\right]}^{2}}{0.100}\phantom{\rule{0ex}{0ex}}1.74\mathrm{x}{10}^{-5}\mathrm{x}0.100={\left[{\mathrm{H}}^{+}\right]}^{2}\phantom{\rule{0ex}{0ex}}\sqrt{1.74\mathrm{x}{10}^{-5}\mathrm{x}0.100}=\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}1.319\mathrm{x}{10}^{-3}=\left[{\mathrm{H}}^{+}\right]$

You should remember the equation for pH. Substituting our value for [H^{+}], we get our final answer:

$\mathrm{pH}=-\mathrm{log}\left(\right[{\mathrm{H}}^{+}\left]\right)=-\mathrm{log}(1.319\mathrm{x}{10}^{-3})\phantom{\rule{0ex}{0ex}}\mathrm{pH}=2.88$

## Weak bases and K_{b}

We know what a **weak base** is - a base that only partially dissociates in solution. It forms an **equilibrium reaction**. Just like for acids, we can find an equilibrium constant, this time known as K_{b}. The equation for K_{b} is given below using B to represent the base:

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{BH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\mathrm{B}\right]}$

### Units of K_{b}

Like K_{a} , K_{b} has the units mol dm^{-3}.

### Kb and pKb

You can probably guess how we calculate pK_{b}. It is simply the negative log of K_{b}:

${\mathrm{pK}}_{\mathrm{b}}=-\mathrm{log}\left({\mathrm{K}}_{\mathrm{b}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{b}}={10}^{-{\mathrm{pK}}_{\mathrm{b}}}$

There’s a special relationship between pK_{w} , pK_{a} and pK_{b}. It is very similar to the relationship we saw between pK_{w}, pH and pOH in **The Ionic Product of Water**:

${\mathrm{pK}}_{\mathrm{w}}={\mathrm{pK}}_{\mathrm{a}}+{\mathrm{pK}}_{\mathrm{b}}$

### Finding the pH of weak bases

Finding the pH of a weak base is similar to finding the pH of a weak acid. However, there are a few differences. Let’s look at an example together.

**A 0.15 mol dm ^{-3} solution of NH_{3 }has K_{b} = 1.77 x 10^{-5}. What is its pH at 25℃?**

First of all, let’s take our equation for the dissociation of ammonia and our equation for K_{b}:

${\mathrm{NH}}_{3}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {{\mathrm{NH}}_{4}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{b}}=\frac{\left[{{\mathrm{NH}}_{4}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_{3}\right]}$

We know from the first equation that the amounts of ammonium ions, NH_{4}^{+}, and hydroxide ions, OH^{-}, are equal. They therefore have equal concentrations. This simplifies the equation for K_{b}:

${\mathrm{K}}_{\mathrm{b}}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^{2}}{\left[{\mathrm{NH}}_{3}\right]}$

We know the value for K_{b} and we know the concentration of ammonia, NH_{3}. Our original solution had a concentration of 0.15 mol dm^{-3}. Though the equilibrium solution will have a slightly lower concentration, the proportion of molecules that have dissociated into ions is so small that we can largely ignore it. Therefore, the equilibrium concentration of NH_{3} molecules still roughly equals 0.15 mol dm^{-3}. We can substitute these values in to the equation and rearrange to find [OH^{-}]:

$1.77\mathrm{x}{10}^{-5}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^{2}}{0.15}\phantom{\rule{0ex}{0ex}}\sqrt{1.77\mathrm{x}{10}^{-5}\mathrm{x}0.15}=\left[{\mathrm{OH}}^{-}\right]=1.63\mathrm{x}{10}^{-3}\mathrm{mol}{\mathrm{dm}}^{-3}$

We can then use the relationships between [OH^{-}], pOH, pK_{w} and pH to work out the pH, just like we did to find the pH of a strong base. (If you aren’t too sure about this, check out **Brønsted-Lowry Acids and Bases** for a detailed explanation.)

$\mathrm{pOH}=-\mathrm{log}\left(\right[{\mathrm{OH}}^{-}\left]\right)\phantom{\rule{0ex}{0ex}}\mathrm{pOH}=-\mathrm{log}(1.63\mathrm{x}{10}^{-3})=2.79\phantom{\rule{0ex}{0ex}}\mathrm{pH}={\mathrm{pK}}_{\mathrm{w}}-\mathrm{pOH}\phantom{\rule{0ex}{0ex}}\mathrm{pH}=14-2.79=11.21$

## Steps for finding the pH of weak acids and bases

Congratulations! You made it through some tricky calculations. You should now be able to work out the pH values for all sorts of acids, bases, and mixtures.

The following flowchart provides a summary of the steps taken to work out the pH of weak acids and bases. Remember to check out all of the other articles we mentioned here for more information about various acid-base calculations.

## Weak acid and base titrations

We've learned that weak acids have higher pH values than strong acids, whilst weak bases have lower pH values than storng bases. This means that they produce slightly different pH curves in titration experiments. You can explore this more in **pH Curves and Titrations**.

## Weak Acids and Bases - Key takeaways

- A
**weak acid**is an acid that only partially dissociates in solution. These form an equilibrium represented by the equation $\mathrm{HA}\left(\mathrm{aq}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$ - A
**weak base**is a base that only partially dissociates in solution. These form an equilibrium represented by the equation $\mathrm{B}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{BH}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$ **K**_{a}is a modified equilibrium constant for the dissociation of weak acids. It is represented by the equation ${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$**K**is a modified equilibrium constant for the dissociation of weak bases. It is represented by the equation ${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{BH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\mathrm{B}\right]}$_{b}- Both K
_{a}and K_{b}take the units**mol dm**^{-3}. - ${\mathrm{pK}}_{\mathrm{a}}=-\mathrm{log}\left({\mathrm{K}}_{\mathrm{a}}\right)$ and ${\mathrm{pK}}_{\mathrm{b}}=-\mathrm{log}\left({\mathrm{K}}_{\mathrm{b}}\right)$
- We can use K
_{a}and K_{b}to find the pH of solutions containing weak acids and bases.

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##### Frequently Asked Questions about Weak Acids and Bases

How do you identify weak acids and bases?

You can identify weak acids and bases based on the fact that they only partially dissociate in solution. They tend to have pH values closer to 7 than strong acids and bases.

What are weak acids and bases?

Weak acids and bases are, as the name suggests, types of acids and bases. They differ from strong acids and bases in that they only partially dissociate in solution.

How do you find the pH of weak acids and weak bases?

To work out the pH of weak acids and bases, you use the equilibrium constants K_{a} and K_{b} to work out the concentrations of hydrogen ions or hydroxide ions in solution. You can then calculate pH.

What indicator is used for a weak acid and strong base?

Phenolphthalein would be a suitable indicator for a titration reaction between a weak acid and a strong base.

What is the difference between strong and weak acids?

Strong acids fully dissociate in solution whereas weak acids only partially dissociate.

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