Finding Ka

How many** weak acids** can you think of in your everyday life? We can name a few. **Citric acid** is an obvious choice - as the name suggests, it is mainly found in citrus fruits like lemons, grapefruits, kumquats, and satsumas. **Lactic acid **is produced in your cells when there isn't enough oxygen present to fully oxidise a molecule called **pyruvate** into carbon dioxide; it causes the burning ache you associate with a hard workout. However, there is also** chlorogenic acid **(found in coffee and green tea),** salicylic acid** (an acid derived from willow trees, used in acne treatments and anti-dandruff shampoos), and **glutamic acid** (the amino acid responsible for the 'umami' taste of certain foods, such as soy sauce and Parmesan cheese). To easily compare the strength of weak acids, we use their **Ka value**. It tells us the extent of their dissociation in water. In this article, we'll explore different methods of **finding Ka**.

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Jetzt kostenlos anmeldenHow many** weak acids** can you think of in your everyday life? We can name a few. **Citric acid** is an obvious choice - as the name suggests, it is mainly found in citrus fruits like lemons, grapefruits, kumquats, and satsumas. **Lactic acid **is produced in your cells when there isn't enough oxygen present to fully oxidise a molecule called **pyruvate** into carbon dioxide; it causes the burning ache you associate with a hard workout. However, there is also** chlorogenic acid **(found in coffee and green tea),** salicylic acid** (an acid derived from willow trees, used in acne treatments and anti-dandruff shampoos), and **glutamic acid** (the amino acid responsible for the 'umami' taste of certain foods, such as soy sauce and Parmesan cheese). To easily compare the strength of weak acids, we use their **Ka value**. It tells us the extent of their dissociation in water. In this article, we'll explore different methods of **finding Ka**.

- This article is about
**finding Ka**in physical chemistry. - We'll
**define Ka**before looking at the**expression for Ka**, and its**relationship to****pKa**. - We'll then explore
**different ways of finding Ka**. - This will include finding Ka from
**pH**,**titration curves and equivalence point**, and**Kb**. - We'll also take a deep dive into
**Ka and percentage ionisation**.

If you've read a little about acids and bases before, you might know what a **weak acid** is.

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

We can represent the** dissociation **(also known as **ionisation**) of the weak acid HA with the following equilibrium reaction:

$$HA\rightleftharpoons H^++A^-$$

Here, HA dissociates into H^{+} and A^{-} ions in a** reversible reaction**. This reversible reaction forms a **dynamic ****equilibrium**, and like all chemical equilibria, it can be represented by an **equilibrium constant**. The equilibrium constant for the dissociation of a weak acid is known as **Ka**.

**Ka** is an equilibrium constant that measures the extent of dissociation of a weak acid in solution at equilibrium. It has the units mol dm^{-3}.

For the reaction that we saw above, the Ka expression looks like this:

$$Ka=\frac{[H^+]\space [A^-]}{HA}$$

You might also come across **pKa**. **pKa** is simply an alternative way of measuring the dissociation of a weak acid. You see, Ka is often very small and this makes it tricky to work with. For example, it is easy to accidently omit a decimal place in chemical calculations. pKa, on the other hand, is a typically larger number. Here's how Ka and pKa are related:

$$pKa=-\log_{10}Ka$$ $$Ka=10^{-pKa}$$

Be aware that the *higher* the value of Ka, the* lower* the value of pKa. This often catches students out!

Ka is useful because it tells us about the strength of the acid:

- The
**higher**the value of Ka, the**stronger**the acid, and the**greater**the extent of its dissociation. - The
**lower**the value of Ka, the**weaker**the acid, and the**lesser**the extent of its dissociation.

Therefore, with Ka, it is easy to compare different acids and find out which is the strongest, and which is the weakest. But how do we actually calculate Ka? Well, there are a few different ways, which we'll spend the rest of the article exploring. These include finding Ka from:

**pH**. This involves working with the**Ka expression**.**Titration curves**. This involves looking at**equivalence point**,**half-neutralisation**, and**pKa**.- From Kb. This involves
**conjugate acid-base pairs**. - We'll also briefly consider finding Ka from
**percentage ionisation**.

Not sure about the ideas we've mentioned? **Acids and Bases **will give you an introduction to the whole topic of acid-base chemistry, whilst **Weak Acids and Bases **looks more specifically at - you guessed it - weak acids. The latter article also shows you how to calculate pH from Ka. On the other hand, check out **Equilibrium Constants **and **Chemical Equilibrium **for a guide to reversible reactions, equilibria, and writing equilibrium constant expressions.

The first way of finding Ka that you could be tested upon in your exams is **finding Ka from ****pH**. You'll be given information about a weak acid dissolved in solution and told the resulting pH. You must then use your knowledge of how weak acids behave in solution, alongside a few basic assumptions and the expression for Ka, to calculate Ka.

To find Ka from pH:

- Write an expression for Ka.
- Find [H
^{+}] using the pH given in the question. Hence, find [A^{-}]. - Find [HA] using other information given in the question.
- Substitute values for [H
^{+}], [A^{-}], and [HA] into the expression for Ka and solve to get your final answer.

Here are the assumptions that we make. They make our calculations a lot simpler:

- Firstly, we know that weak acids (HA) only
**partially ionise**in solution at equilibrium. This means that only a minimal proportion of the acid dissociates into H^{+}and A^{-}ions - the vast majority remains as HA molecules. Therefore, for our first assumption, we assume that**the concentration of HA molecules at equilibrium is the same as the original amount of acid added to the solution**. - Secondly, we know that
**water molecules also****partially ionise**in solution, producing H^{+}and OH^{-}ions. But once again, this dissociation only affects a tiny proportion of water molecules, and so the amount of H^{+}ions in solution from water molecules is minimal. Therefore, we assume that**the entire equilibrium concentration of H**.^{+}is due to the dissociation of the weak acid, HA

Still confused about finding Ka from pH? Read through our worked example to get a feel for the method, then try the problem by yourself. This is a useful way of practising chemical calculations.

**0.020 moles of the weak acid HF dissolve in 1.00 dm ^{3} of pure water. The resulting solution has a pH of 2.42. Find the value of Ka for HF.**

First, let's write an expression for Ka:

$$Ka=\frac{[H^+]_{eqm}\space [F^-]_{eqm}}{[HF]_{eqm}}$$

Now we can use the pH given in the question to find [H^{+}]:

$$10^{-pH}=[H^+]$$ $$10^{-2.42}=3.80\times 10^{-3}\space mol\space dm^{-3}$$

This helps us find [F^{-}]. HF is a monoproctic acid, meaning that one molecule of HF dissociates into one H^{+} ion and one F^{-} ion. Therefore, the concentration of H^{+} ions equals the concentration of F^{-} ions:

$$[H^+]=[F^-]$$ $$[F^-]=3.80\times 10^{-3}\space mol\space dm^{-3}$$

We can also use other information from the question to find [HF]. Because weak acids only partially dissociate in solution, we assume that [HF] is equal to the amount of HF initially added. Here, 0.020 moles of HF is added to 1.00 dm^{3} of water, and so we get the following concentration of HF molecules:

$$[HF]=\frac{0.020}{1.00}$$ $$[HF]=0.020\space mol\space dm^{-3}$$

We're now ready to substitute all of these values into our Ka expression. If we rearrange and solve the equation, we'll get our final answer:

$$Ka=\frac{(3.80\times 10^{-3})\space (3.80\times 10^{-3})}{0.020}$$ $$Ka=7.22\times 10^{-4}\space mol\space dm^{-3}$$

Hence, the weak acid HF has a Ka value of **7.22 ****×**** 10 ^{-4} mol dm**

Do you want to calculate Ka in a more hands-on way? You could carry out a **titration reaction** and **find Ka using ****the equivalence point and the ****half-neutralisation point**.

**Half-neutralisation** (also known as **h****alf-equivalence**) is the point at which exactly half of the acid has been neutralised by a base.

Half-neutralisation is significant for one particular reason. Here, the value of Ka equals the concentration of H^{+} ions in solution, and so pKa is identical to the value of pH. **At half-neutralisation, pKa = pH**. Therefore, if you know the pH of a solution at half-neutralisation, you know the acid's pKa. Once you know pKa, you can easily find Ka, simply by applying the formula we learned earlier that relates the two values: \(pKa=-\log_{10}Ka\)

From looking at the definition of the word, we can say that at half-neutralisation, exactly half of the acid molecules have dissociated into H+ ions and A- ions - no more, no fewer. Some H+ ions react with the titrant to form water, but all the A- ions remain in the solution. Therefore, the concentration of A- ions is the same as the concentration of HA molecules. At half-neutralisation, [HA] ≡ [A^{-}].

Why is that helpful? Well, let's look at the expression for Ka. You'll see that [A^{-}] in the fraction's numerator cancels out with [HA] in the fraction's denominator. Therefore, Ka equals just [H^{+}]:

$$Ka=\frac{[H^+]_{eqm}\space [A^-]_{eqm}}{[HA]_{eqm}}$$

$$Ka=[H^+]$$

If we take negative logs of both sides, we find that the left-hand side becomes pKa, whilst the right-hand side becomes pH:

$$-\log_{10}Ka=-\log_{10}[H^+]$$

$$pKa=pH$$

Here's how you find Ka from titration curves, equivalence point, and half-neutralisation:

- Find the pH of the solution at half-neutralisation using either a titration experiment or titration curve.
- Hence, infer pKa.
- Use pKa to calculate pKa.

We gave you two ways of finding the pH at half-neutralisation. This first way is more practical - it involves a titration reaction. Let's now look at how you carry out the experiment and use the results to find the pH at half-neutralisation.

- Set up a titration experiment in the standard way by suspending a burette above a conical flask using a stand and clamp. In addition, prepare and calibrate a pH probe.
- Use a volumetric flask to measure 25 cm
^{3}of a weak acid into the conical flask. Add a few drops of a suitable indicator. - Fill the burette with a strong base. This is your titrant.
- Add the base titrant to the acid in 1 cm
^{3}intervals. As you approach the equivalence point, add the base drop-wise. - Stop adding the base when the indicator
*just*changes colour. Then, immediately another 25 cm^{3}of the weak acid to the conical flask. - Measure the pH of the solution.

See what we've done here? At the equivalence point, *all* of the acid has been neutralised. By adding the same volume of unreacted acid to the solution, we end up at half-neutralisation - the concentration of A^{-} ions equals the concentration of HA molecules. Notice that this means the volume of titrant at half-neutralisation is half the volume of titrant at the equivalence point. Once at half-neutralisation, we can then use a pH probe to accurately find the solution's pH.

If you are working with unfamiliar acids and bases, you may have to carry out the experiment a couple of times to find out roughly how much titrant to add before you reach the equivalence point. Then, you can repeat the experiment again, adding the titrant dropwise as you near the end of the titration so that you can reach that crucial point exactly. Check out **pH Curves and Titrations **for more examples of this type of practical.

Don't have the time or facilities to carry out a full-blown titration? Not to worry! You can also use existing data to find the pH of a weak acid at half-neutralisation, in the form of a titration curve.

A **pH curve**, also known as a **titration curve**, is a graph showing how the pH of a solution changes when an acid or alkali is added to it. It measures pH on the y-axis against the volume of titrant added on the x-axis

Here's how you identify the equivalence point and the pH at half-neutralisation on a weak acid-strong base titration curve:

- When you first add the base to the acid, the pH of the solution rises slowly. This creates a curve with an initial shallow gradient.
- However, once you've added a certain amount of base, the pH increases rapidly. This results in a steep vertical section.
- The curve then levels off again, creating another section with a shallow gradient.
- We first locate the equivalence point. This point lies in the middle of the steep vertical section of the curve.
- We use the equivalence point to find the equivalence point volume - the volume of titrant needed to fully neutralise the acid.
- We then find the half-neutralisation point volume. This is precisely half of the equivalence point volume.
- Finally, we use our graph to identify the half-neutralisation pH. This is simply the pH of the solution when we've added the half-neutralisation point volume of titrant.

It's your turn: use the graph above to find Ka, with the help of the following worked example.

**25 cm ^{3 }NaOH is required to fully neutralise 25 cm^{3} benzoic acid, resulting in the titration curve shown above. Use the half-neutralisation point to find the value of Ka for this acid.**

We know that at half-neutralisation, pKa equals pH. The equivalence point volume of NaOH is 25 cm^{3}, and so the half-neutralisation point volume is 12.5 cm^{3}. The titration curve tells us that when we add 12.5 cm^{3} of NaOH, the solution has a pH of about 4.19. Therefore:

$$pKa=4.19$$

We can now use this value to find Ka:

$$Ka=10^{-pKa}$$ $$Ka=6.46\space mol\space dm^{-3}$$

For our last way of calculating Ka, let's explore how we can **find Ka from Kb**. This involves looking at **conjugate acid-base pairs**.

A **conjugate acid** is a base that has accepted a proton from an acid. It can act just like a normal acid by giving up its proton. On the other hand, a **conjugate base** is an acid that has donated a proton to a base. It can act just like a normal base by accepting a proton.

When a weak acid (HA) dissociates in solution, it splits up into a proton (H^{+}) and a negative ion (A^{-}). However, as we learned at the start of the article, weak acids only **partially dissociate** in solution, and so the reaction is **reversible**. This means that the negative A^{-} ion can recombine with a positive H^{+} ion to form HA once again. In other words, it gains a proton, and so acts as a **base**. Therefore, **A ^{-} is HA's**

Head over to **Brønsted-Lowry Acids and Bases **to read more about the basic definition of acids and bases, as well as further examples of conjugate acids and bases.

Just like **Ka** measures the extent of dissociation of a weak acid, the equilibrium constant **Kb** measures the extent of dissociation of a weak base. Kb applies to conjugate bases, too. In fact, the value of Ka for a weak acid and the value of Kb for its conjugate base partner have a special relationship:

$$Ka\times Kb=Kw$$

Note that Kw is a **fixed value at a certain temperature**. This means that Ka and Kb are **inversely proportional**:

- The
**higher**the value of Ka, the**stronger**the acid. This results in a**weaker**conjugate base with a**lower**Kb value. - The lower the value of Kb, the weaker the acid. This results in a
**stronger**conjugate base with a**higher**Kb value.

Remember **Kw**? It is the equilibrium constant for the dissociation of water, and you can learn about it in the **Ionic Product of Water. **At 25 °C, Kw equals 1.00 × 10^{-14} mol^{2} dm^{-6}. The other value that we mentioned, **Kb**, is covered in more depth in **Weak Acids and Bases**.

Want to know how the formula linking Ka, Kb, and Kw came about? It's not too tricky, but it involves looking at weak acids and bases in terms of their interaction with water molecules. Let's take a look.

Consider our general weak acid (HA) again. It partially dissociates into H^{+} and A^{-} ions. The H^{+} ion is picked up by a water molecule to form H_{3}O^{+}:

$$HA+H_2O\rightleftharpoons H_3O^++A^-$$

We can write a Ka expression for this equilibrium:

$$Ka=\frac{[H_3O^+]\space [A^-]}{[HA]}$$

Now consider A^{-}, the conjugate base of HA. It can take H^{+} ions from water to form HA, leaving behind OH^{-} ions:

$$A^-+H_2O\rightleftharpoons HA+OH^-$$

We can write a Kb expression for this equilibrium:

$$Kb=\frac{[HA] [OH^-]} {[A^-]}$$

If we multiply Ka and Kb together, we find that [HA] and [A^{-}] cancel out from the numerator and denominator, leaving us with just [H_{3}O^{+}] multiplied by [OH^{-}]. This equals Kw:

$$Ka\times Kb=\frac{[H_3O^+]\space [A^-]}{[HA]} \times \frac{[HA][OH^-]} {[A^-]}$$

$$Ka\times Kb=[H_3O^+]\space [OH^-]$$

$$Ka\times Kb=Kw$$

Finding Ka from Kb is probably the simplest type of Ka calculation. Here's the method, if you can even call it a method - it is just one step!

- Divide Kw by Kb to find Ka.

We told you it was easy!

Now, give the following question a go. If you get stuck, our worked example will guide you through the problem.

**The conjugate base of the weak acid CH _{3}COOH is CH_{3}COO^{-}. At 25 **

Well, we know that for a conjugate acid-base pair, Ka × Kb = Kw. At this temperature, Kw = 1.00 × 10^{-14 }mol^{2} dm^{-6}. Dividing Kw by the value of Kb given in the question will give us Ka:

$$Ka=\frac{1.00\times 10^{-14}}{5.70\times 10^{-10}}$$ $$Ka=1.75\times 10^{-5}\space mol\space dm^{-3}$$

This is our final answer.

Ka, above all, is an equilibrium constant that tells us about the extent of a weak acid's dissociation. Remember that *dissociation* is simply just another term for **ionisation**. Therefore, it makes sense that we can calculate Ka using **percentage ionisation**: the percentage of acid molecules that ionise (dissociate) in solution. The concept is relatively straightforward:

- The
**higher**the value of Ka, the**stronger**the acid, and so the**greater**the percentage ionisation in solution. - The
**lower**the value of Ka, the**weaker**the acid, and so the**lower**the percentage ionisation in solution.

However, the calculations themselves go beyond anything you are expected to know at A level, and so we won't cover them today - we don't want to overwhelm you with too much information! You are welcome to do some further reading yourself if you are interested in learning about this type of question.

**Ka**is the equilibrium constant for the dissociation of a weak acid in solution.- For the dissociation reaction \(HA\rightleftharpoons H^++A^-\), Ka has the expression \(Ka=\frac{[H^+]\space [A^-]}{[HA]}\)
- The
**higher**the value of Ka, the**stronger**the acid, and the**greater**its dissociation in solution. - We can find Ka from:
**pH**. This involves working with the**Ka expression**.**Titration curves**. This involves looking at the**equivalence point**,**half-neutralisation**, and**pKa**.**Kb**. This involves working with**conjugate acid-base pairs**.**Percentage ionisation**.

To find percentage ionisation from pH and Ka:

- Use pH to find [H
^{+}]. - Use the expression for Ka to find [HA].
- Hence, find the original concentration of acid added to solution.
- Divide [H
^{+}] by the original concentration of acid, then multiply this value by 100. This gives you percentage ionisation.

Note that you don't have to worry about percentage ionisation if you are studying chemistry A level - it isn't covered in the specification and won't be tested in your exam!

What is Ka?

An equilibrium constant that measures the extent of dissociation of a weak acid in solution at equilibrium.

What are the units of Ka?

mol dm^{-3}

The higher the value of Ka, the ____.

Stronger the acid.

The higher the value of pKa, the ____.

Weaker the acid.

Outline the steps used to find Ka from pH.

- Write an expression for Ka.
- Find [H
^{+}] using pH. Hence, find [A^{-}]. - Find [HA].
- Substitute values for [H
^{+}], [A^{-}], and [HA] into the expression for Ka and solve.

What two assumptions do we make when calculating Ka from pH?

- The concentration of HA molecules at equilibrium is the same as the original amount of acid added to the solution.
- The entire equilibrium concentration of H
^{+}is due to the dissociation of the weak acid, HA; we ignore any H^{+ }ions from the dissolution of water.

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