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# Buffer Solutions

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Your body is clever. Your cells continuously produce substances that cause changes in pH – from lactic acid to carbon dioxide to ketones. If the pH of your blood or any other of your bodily fluids gets too high, you enter a state of alkalosis. Early symptoms include numbness and muscle spasms. On the other hand, if it gets too low, you enter acidosis. But don’t worry – most people’s bodies are able to keep their inner pH levels more or less stable using systems called buffer solutions.

A buffer solution is a solution that maintains a constant pH when small amounts of acid or alkali are added to it.

• We’ll look at both acidic and basic buffer solutions and what happens to them when you add in extra acids or alkalis.
• You’ll be able to see buffer solutions in action with a real-life example.
• After that, we’ll learn how to carry out buffer solution calculations, and you’ll be able to put your newly-learned knowledge into practice.

## Types of buffer solutions

Buffer solutions can be acidic or basic. They are designed to keep the concentrations of hydrogen ions and hydroxide ions roughly the same by reacting with whatever substances are added to them, be it another acid or another base.

## Acidic buffer solutions

Acidic buffer solutions are formed by mixing a weak acid with one of its salts in solution. The salt is also the acid’s conjugate base.

A conjugate base is a base formed when an acid loses a proton.

Let’s take a look at how acidic buffer solutions work.

Suppose we have the weak acid HA and its soluble salt MA. You should know from the article ‘Weak acids and bases’ that weak acids partially dissociate in solution. This is important for keeping a constant pH if we add an alkali. For our acid HA, that dissociation looks something like this:

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

Because the acid only partially dissociates, the concentrations of H+ ions and OH- ions are very small.

On the other hand, the soluble salt MA fully ionises in solution:

$\mathrm{MA}\left(\mathrm{aq}\right)\to {\mathrm{M}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

This is important for keeping a constant pH if we add an acid.

### Adding an alkali to acidic buffer solutions

If we add an alkali to the buffer solution, we are essentially adding extra hydroxide ions, OH-. We’d expect the pH to increase. However, the OH- ions instead react with the weak acid in the buffer solution to form water and A- ions:

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

This means that overall, the numbers of hydrogen ions and hydroxide ions haven’t changed, meaning [H+] and [OH-] are the same. The buffer has resisted change in pH.

OH- ions can also be removed by a second process – they react with the hydrogen ions produced when a tiny proportion of the weak acid HA dissociates, producing water. As soon as they react, the equilibrium shifts to the right and so more of the weak acid dissociates until all of the alkali is used up.

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

Although we’ve written these equations using single-headed forward arrows, you should note that they are actually reversible reactions. However, the equilibrium lies far over to the right, and so barely any of the reverse reaction occurs.

### Adding an acid to acidic buffer solutions

If we add an acid to the buffer solution, we’d expect the pH to decrease as the acid donates protons, which are just hydrogen ions, to the solution. However, this is where the soluble salt MA comes in. Any protons react with the A- ions produced when the salt ionises to form the weak acid HA:

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

Overall, the concentrations of hydrogen ions and hydroxide ions have largely remained the same. The buffer has again resisted a change in pH.

You’ll notice that adding an acid to an acidic buffer solution produces more of the weak acid, HA. This is a reversible reaction – the acid could then dissociate back into H+ and A-, increasing [H+]. But remember that weak acids only partially dissociate in solution, so the actual increase in [H+] is very small.

### An alternative acidic buffer solution

You can also create an acidic buffer by half-neutralising a weak acid with a strong base.

Half-neutralisation is the point where half of the acid in a solution has been neutralised by a base. This means that half of the protons have been used up.

This forms the same solution we explored above – a solution containing a weak acid and its salt. However, because the acid is half-neutralised, the buffer solution has a unique property: its pKa equals its pH.

Not sure what pKa is? Take a look at ‘Weak Acids and Bases’ to find out more.

Let’s look at this more closely. Take one mole of the weak acid HA and half a mole of the base MOH in a solution with a volume of 1000 cm3. At half-neutralisation, exactly half of the weak acid reacts with the base to form a salt, MA, and water, H2O. The equation below shows the number of moles before the reaction, the change in the number of moles, and the number of moles after the reaction for each of the species:

The change in moles in a reaction between a weak acid and a strong base. Anna Brewer, StudySmarter Original

We have half a mole for each of the weak acid HA and the salt MA. MA ionises in solution into M+ and A-. This means that we also have half a mole of A- ions. There are the same amount of HA molecules as A- ions – their concentrations are equal. In other words, [HA] = [A-].

Let’s now look at the equation for Ka:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{A}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{HA}\right]}\phantom{\rule{0ex}{0ex}}$

Because [HA] and [A-] are equal, we find that the [A-] at the top of the equation cancels out with the [HA] at the bottom of the equation:

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

We are left with just Ka equalling H+. This means that the pH of this buffer solution equals the pKa of the acid:

${\mathrm{K}}_{\mathrm{a}}=\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}{\mathrm{pK}}_{\mathrm{a}}=\mathrm{pH}$

## Basic buffer solutions

Similarly to acidic buffer solutions, basic buffer solutions are made from a weak base and one of its salts. This salt is the base’s conjugate acid.

A conjugate acid is an acid formed when a base gains a proton.

Let’s look at an example, such as the buffer solution formed when ammonia is mixed with ammonium chloride. Ammonia is a weak base. This means that it partially dissociates in solution, as shown by the following equation:

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

The ammonium chloride is a salt and so fully ionises in solution:

${\mathrm{NH}}_{4}\mathrm{Cl}\left(\mathrm{aq}\right)\to {{\mathrm{NH}}_{4}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

Let’s now explore how the buffer solution reacts to extra acids and bases.

### Adding an acid to basic buffer solutions

If we add an acid, the H+ ions it releases react with aqueous ammonia to form ammonium ions:

${\mathrm{NH}}_{3}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {{\mathrm{NH}}_{4}}^{+}\left(\mathrm{aq}\right)$

There may also be a second reaction. Remember, a small proportion of the ammonia in solution dissociates into ammonium ions, NH4+, and hydroxide ions, OH-. The acid’s hydrogen ions can also react with the hydroxide ions to form water:

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

Because the hydrogen ions have reacted and been used up, the overall concentrations of hydrogen ions and hydroxide ions remain constant. The buffer solution has resisted change in pH.

### Adding an alkali to basic buffer solutions

If we add an alkali to our buffer solution, it reacts with the ammonium ions in solution to form ammonia and water:

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

Overall, the hydroxide ion concentration remains largely unchanged. The buffer solution has resisted change in pH.

Once again, these reactions are actually reversible reactions. However, the presiding reaction is the forward reaction, and so it is easier to think of them as one-way.

## Examples of buffer solutions

Let’s now look at an example of a buffer solution in action. A typical buffer solution can be made from ethanoic acid and sodium ethanoate. The ethanoic acid partially dissociates in solution:

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

The sodium ethanoate fully ionises in solution:

${\mathrm{CH}}_{3}\mathrm{COONa}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)+{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)$

If we add hydrogen ions, H+, they react with the ethanoate ions, CH3COO-, that came from the sodium ethanoate:

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

If we add hydroxide ions, OH-, they either react with the ethanoic acid, CH3COOH, to form ethanoate ions and water, or they react with the hydrogen ions produced when a small proportion of ethanoic acid dissociates:

${\mathrm{CH}}_{3}\mathrm{COOH}\left(\mathrm{aq}\right)+ {\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+ {\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)$

Overall, the pH of the solution remains roughly the same. The buffer solution has resisted change in pH.

An extremely important buffer system is the one maintained in your bloodstream. Carbonic acid, H2CO3, and hydrogen carbonate, HCO3-, work together to keep your blood at a suitable pH of around 7.4.

Other buffer solutions are found in shampoos to avoid causing rashes and skin irritation, and in the brewing industry, where brewers want to carefully control the conditions to optimise the fermentation of sugars into alcohol.

## Buffer solution calculations

Now that we understand how buffer solutions work, we can have a go at calculating their pH.

A solution contains 0.5 mol dm-3 CH3CH2COOH and 1.0 mol dm-3 CH3CH2COO-Na+. The Ka for CH3CH2COOH equals 1.35 x 10-5. Calculate the pH of the solution.

To find the pH, we need to know [H+], the concentration of hydrogen ions in solution. First of all, let’s write out the equation linking Ka and [H+]:

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

We can rearrange to find [H+]:

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

Here, CH3CH2COOH is our weak acid, HA. Only a tiny proportion of it dissociates into ions in solution, and so [HA] roughly equals the concentration given in the question: 0.5 mol dm-3. In contrast, all of the CH3CH2COO-Na+ molecules ionise into CH3CH2COO- and Na+ ions. The CH3CH2COO- ion is our A- and has a concentration of 1.0 mol dm-3, as given in the question. Substituting these values into the equation gives us the following:

$\left[{\mathrm{H}}^{+}\right]=\frac{0.5\mathrm{x}\left(1.35\mathrm{x}{10}^{-5}\right)}{1.0}=6.75\mathrm{x}{10}^{-6}\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-\mathrm{log}\left(6.75\mathrm{x}{10}^{-6}\right)=5.17$

Not too tricky, right? Now let’s take another example, this time calculating the pH of a buffer solution formed in the reaction between a weak acid and a strong base.

250 cm3 of 0.1 mol dm-3 ethanoic acid is mixed with 50 cm3 of 0.2 mol dm-3 sodium hydroxide. Calculate the pH of the buffer solution formed. The Ka for ethanoic acid = 1.76 x 10-5.

You’ll notice that this is a neutralisation reaction. The acid (ethanoic acid, CH3COOH) will react with the base (sodium hydroxide, NaOH) to form a salt (sodium ethanoate, CH3COO-Na+) and water. Before we work out the pH of the solution, we first need to find out how many moles we have remaining of acid and salt, as we need these values for our calculation.

At the start of our reaction, we have the following:

Moles of ethanoic acid = 0.250 x 0.1 = 0.025

Moles of sodium hydroxide = 0.050 x 0.2 = 0.010

Remember to change your volume into dm3.

Sodium hydroxide is the limiting reagent. There is less of it, and so it will be used up first. The acid and base will react together until all of the sodium hydroxide is used up, as shown in the table below. We can find the concentration of the important species at the end of the reaction by dividing their number of moles by the total volume:

Calculating the pH of buffer solutions. Anna Brewer, StudySmarter Original

Look at the equation for Ka:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{H}}^{+}\right]=\frac{\left[\mathrm{HA}\right]\mathrm{x}{\mathrm{K}}_{\mathrm{a}}}{\left[{\mathrm{A}}^{-}\right]}$

Ethanoic acid is our weak acid, HA. The salt product, sodium ethanoate, dissociates into our negative ion, A-. We’ve worked out the concentrations of these in the table above. We also know the Ka value for ethanoic acid given in the question. We can therefore fill in these values in our equation:

$\left[{\mathrm{H}}^{+}\right]=\frac{0.05\mathrm{x}\left(1.76\mathrm{x}{10}^{-5}\right)}{0.033}=2.667\mathrm{x}{10}^{-5}\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-\mathrm{log}\left(2.667\mathrm{x}{10}^{-5}\right)=4.57$

Well done! You made it through buffer solutions. The following flow charts show how you calculate the pH of an acidic buffer solution to help summarise your knowledge:

A flow chart showing you how to calculate the pH of buffer solutions. Anna Brewer, StudySmarter Original

Most exam boards don’t need you to know how to calculate the pH of a basic buffer solution, but if yours does, take a look at the ‘Deep Dive’ box down below to find a handy equation.

There is a much simpler formula you can use to find the pH of an acidic buffer solution. It takes a bit of deriving to form it from the equation for Ka, including using the laws of logarithms:

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

Rearrange to find [H+] then take logs of both sides:

$\phantom{\rule{0ex}{0ex}}\left[{\mathrm{H}}^{+}\right]=\frac{\left[\mathrm{HA}\right]\mathrm{x}{\mathrm{K}}_{\mathrm{a}}}{\left[{\mathrm{A}}^{-}\right]}\phantom{\rule{0ex}{0ex}}\mathrm{log}\left(\left[{\mathrm{H}}^{+}\right]\right)=\mathrm{log}\left(\left[\frac{\mathrm{HA}\right]\mathrm{x}{\mathrm{K}}_{\mathrm{a}}}{\left[{\mathrm{A}}^{-}\right]}\right)$

Expand using log laws:

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

Multiply both sides by -1. The left-hand side now looks like pH, and the right-hand side features pKa:

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

In other words:

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}-\mathrm{log}\left(\frac{\left[\mathrm{HA}}{\left[{\mathrm{A}}^{-}\right]}\right]\right)$

Try it using the values from the first example above:

$\mathrm{pH}=-\mathrm{log}\left(1.35\mathrm{x}{10}^{-5}\right)-\mathrm{log}\left(\frac{0.5}{1.0}\right)=5.17$

Likewise, we can derive a similar equation for the pH of a basic buffer solution. It ends up looking like this:

$\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}-\mathrm{log}\left(\frac{\left[\mathrm{base}\right]}{\left[\mathrm{salt}\right]}\right)$

Remember that $\mathrm{pOH}+\mathrm{pH}={\mathrm{pK}}_{\mathrm{w}}$. At room temperature, ${\mathrm{pK}}_{\mathrm{w}}=14$. Therefore:

$\mathrm{pH}=14-\mathrm{pOH}\phantom{\rule{0ex}{0ex}}\mathrm{pH}=14-{\mathrm{pK}}_{\mathrm{b}}+\mathrm{log}\left(\frac{\left[\mathrm{base}\right]}{\left[\mathrm{salt}\right]}\right)$

## Buffer Solutions - Key takeaways

• A buffer solution is a solution that maintains a constant pH when small amounts of acid or alkali are added to it.
• Acidic buffer solutions are formed from a weak acid and one of its salts.
• Basic buffers solutions are formed from a weak base and one of its salts.
• Buffer solutions resist change in pH by maintaining constant hydrogen and hydroxide ion concentrations.
• At the half-neutralisation point of a reaction between a weak acid and a strong base, [HA] = [A-] and pKa = pH. The solution can be used as a buffer.
• We can calculate the pH of buffer solutions using Ka, [A-] and [HA].

## Buffer Solutions

A buffer solution is a solution that maintains a constant pH when small amounts of acid or alkali are added to it.

An example of a buffer solution is a mixture of ethanoic acid and sodium ethanoate. This is an acidic buffer solution.

Buffer solutions can be acidic or basic, depending on their components. Acidic buffer solutions are made by mixing a weak acid and one of its salts. A basic buffer solution is made in a similar way – by mixing a weak base and one of its salts.

A buffer solution is used in an EDTA titration in order to maintain a constant pH. This is necessary because the reactions between EDTA and metal ions depend on the pH.

You form an acidic buffer solution by mixing a weak acid and one of its salts. A basic buffer solution can be made in a similar way, by mixing a weak base and one of its salts.

## Final Buffer Solutions Quiz

Question

What is a buffer solution?

A solution that maintains a constant pH when small amounts of acid or alkali are added to it.

Show question

Question

How do you form an acidic buffer solution?

Mix a weak acid with one of its salts.

Show question

Question

How do you form a basic buffer solution?

Mix a weak base with one of its salts.

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Question

Which of the following can be mixed with ethanoic acid to form an acidic buffer solution?

CH3COO-Na+

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Question

Which of the following can be mixed with ammonia to form a basic buffer solution?

CH3COO-Na+

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Question

What is the half-neutralisation point of a reaction?

The point where half of the acid in solution has been neutralised by a base.

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Question

What is the relationship between half-neutralisation point, pKa and pH?

At half-neutralisation, pKa = pH. This is because [A-] = [HA].

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Question

Calculate the pH of a buffer solution containing 0.100 mol dm-3 ethanoic acid and 0.100 mol dm-3 sodium ethanoate. pKa = 4.75.

4.75

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Question

How do buffer solutions resist change in pH?

They react with added H+ or OH- ions, keeping their overall concentrations the same.

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