Acids, Bases and Buffers

 
 

Acids and bases

Acids can be defined in different ways – as proton donors or electron acceptors.

The Bronsted-Lowry definition of an acid is a proton donor, which means they release hydrogen ions, H+, when added to water. In reality, the hydrogen ion will immediately bind to water to form a hydronium ion, H3O+.

 
 

Some acid can release more than one proton:

  • Sulfuric acid, H2SO4, is a diprotic acid because it releases two hydrogen ions in solution.
  • Phosphoric acid, H3PO4, is a triprotic acid and releases three hydrogen ions when added to water.
 
 

Keep this in mind when doing acid calculations. If you have x moles of H2SO4, you’ll have 2x moles of hydrogen ions in solution. Likewise, for x moles of H3PO4, you’ll have 3x moles of hydrogen ions when it dissociates.

A Bronsted-Lowry base is a proton acceptor. When they have been added to water, they will bind to hydrogen ions and remove them from the solution.

 
 

Conjugate pairs

Conjugate pairs are two species that differ by a hydrogen ion. For example, the conjugate acid H2SO4 is linked to the conjugate base HSO4-. The conjugate acid HCl is linked to the conjugate base Cl-.

 
 

To form the conjugate base, just get rid of a proton and add a negative charge. And to figure out the conjugate acid, just add hydrogen and get rid of the negative charge.

In the case where you have two acids together in solution, the strongest acid will act as the acid whereas the weaker acid will act as the base. Let’s say we add hydrochloric acid, HCl, and ethanoic acid, CH3COOH together in a beaker. We know that HCl is the stronger acid so it will lose a proton, which will be accepted by ethanoic acid because it is acting as a base.

 
 

Reactions of acids

Acids react with metals to form a salt and hydrogen gas. The metal atoms lose electrons to form metal ions – they have been oxidised. The hydrogen ions gain electrons to form hydrogen gas – they have been reduced.

 
 

Acids react with metal carbonates to form a salt, water and carbon dioxide.

 
 

Acids react with other bases, such as metal hydroxides and metal oxides to form a salt and water.

 
 

All of these reactions are examples of neutralisation reactions.

pH

pH is a measure of hydrogen ion concentration – it tells us how acidic or alkaline a solution is. The pH scale is logarithmic, meaning that a decrease in 1 pH unit means a 10x increase in hydrogen ion concentration. The scale ranges from 0 to 14. There are acids and bases which exceed this scale, such as fluoroantimonic acid, which is 100,000 billion billion billion times more acidic than hydrochloric acid and has a pH of -31. This substance is so strong it will eat through skin, bones, and any container used to store it.

If you know the hydrogen ion concentration of a solution, you can calculate the pH using the equation:

 
 

If you have the pH and want to work out the concentration of hydrogen ions, you can use the equation:

 
 

Worked example – calculating pH from H+ concentration

A solution of hydrochloric acid has a hydrogen ion concentration of 0.05 mol dm-3. What is the pH of the solution?

  • pH = -log 0.05
  • pH = 1.3

Worked example – calculating H+ concentration from pH

A solution of ethanoic acid has a pH of 4.8. Calculate the hydrogen ion concentration of the solution.

  • [H+] = 10-4.8
  • [H+] = 1.58 x 10-5

pH of strong acids

Strong acids completely dissociate in solution, which means that every single acid molecule breaks apart into a hydrogen ion and its conjugate base.

 
 

For monobasic acids, this means that the moles of acid and hydrogen ions are equal. For example, if you have 0.6 mol of HCl, you will also have 0.6 mol of H+ since they exist in a 1:1 ratio. As moles and volume are both the same, concentration will also be equal.

Worked example – calculating pH from concentration of a strong acid

You have a 0.1 mol dm-3 solution of HCl. Calculate the pH of the solution.

  • Hydrochloric acid is a strong acid, so a 0.1 mol dm-3 solution will produce a 0.1 mol dm-3 of H+.
  • pH = -log 0.1
  • pH = 1

Ionic product of water, Kw

The ionic product of water, Kw, is just another equilibrium constant that gives a value to the position of equilibrium for the dissociation of water.

 
 

The expression for the ionic product of water is:

 
 

The units for Kw are always mol2dm-6. The Kw expression is useful because it allows us to work out the concentration of hydroxide ions in a solution, so long as we know the hydrogen ion concentration. At standard temperature (298 K or 25oC), Kw is 1 x 10-14. So if you have a hydrogen ion concentration of 0.1 mol dm-3 (or 1 x 10-1), that means the hydroxide ion concentration must be 1 x 10-13. And if you have a hydroxide ion concentration of 1 x 10-5, the hydrogen ion concentration must be 1 x 10-9.

For a solution of pure water, the concentration of hydrogen ions and hydroxide ions are equal (i.e. there is a 1:1 ratio). In this case, the Kw expression can be simplified to Kw = [H+]2.

Just like other equilibrium constants, Kw only changes value depending on temperature. Changing the concentration of hydrogen ions or hydroxide ions will only cause the position of equilibrium to shift, without changing its value.

Worked example – using Kw to find the pH of a strong base

Calculate the pH of a 0.05 mol dm-3 solution of potassium hydroxide, KOH, at 298 K.

  • At 298 K, Kw = 1 x 10-14
  • Kw = [H+] [OH-]
  • 1 x 10-14 = [H+] x 0.05
  • [H+] = 2 x 10-13
  • pH = - log 2 x 10-13
  • pH = 12.7

The acid dissociation constant, Ka

Unlike strong acids, weak acids only partially dissociate into hydrogen ions.

 
 

Imagine a beaker filled with ethanoic acid. Some of those acid molecules will be completely intact and exist as CH3COOH. Others will have releases their hydrogen ions to form the conjugate base CH3COO-.

Ka is (yet another) equilibrium constant that gives a value to the position of equilibrium between the intact acid and its dissociated ions. The larger the value of Ka, the more the position of equilibrium lies to the right and the more it is releasing hydrogen ions.

The expression for the acid dissociation constant, Ka, is:

 
 

Since there is a 1:1 ratio between hydrogen ions and the conjugate base, A-, this expression can be simplified as:

 
 

Remember that this equation is only used for weak acids. For strong acids, remember that [HA] = [H+].

Worked example – using Ka to find the pH of a weak acid

The Ka of ethanoic acid at 298 K is 1.8 x 10-5 mol dm-3. Calculate the pH of a 0.05 mol dm-3 ethanoic acid solution at this temperature.

  • Ka = [H+]2
  • 1.8 x 10-5 = [H+]2 / 0.05
  • [H+]2 = 9 x 10-7
  • [H+] = 9.5 x 10-4
  • pH = - log [H+]
  • pH = 3

The dissociation constant, pKa

pKa is just the negative log of Ka, in the same way that pH is the negative log of [H+].

To calculate pKa from the Ka, or vice versa, you need to use the following equations:

 
ka.jpg
 

Worked example – calculating pKa from Ka

The Ka of carbonic acid at 298 K is 4.4 x 10-7 mol dm-3. What is its pKa?

  • pKa = - log Ka
  • pKa = - log 4.4 x 10-7
  • pKa = 6.36

Worked example – using pKa to find the pH of a weak acid

Methanoic acid is a weak acid that is an ingredient in insecticides. It has a pKa of 3.75 at 298K. Find the pH of a 1.5 x 10-4 mol dm-3 solution of methanoic acid at 298 K.

This question combined elements of the last two worked examples. From the pKa, we can work out the Ka. Since we have the Ka and [HA], we can pop those values into the Ka expression for weak acids and rearrange to find [H+] and therefore pH.

  • Ka = 10-pKa
  • Ka = 1.78 x 10-4
  • Ka = [H+]2 / [HA]
  • 1.78 x 10-4 = [H+]2 / 1.5 x 10-4
  • [H+]2 = 2.67 x 10-8
  • [H+] = 1.63 x 10-4
  • pH = - log[H+]
  • pH = 3.8

Buffers

A buffer is a solution that minimises pH changes when small quantities of acid or base are added. It is made up of a weak acid and its conjugate base e.g. ethanoic acid, CH3COOH and the ethanoate ion, CH3COO-. An equilibrium is established between the two, by which the weak acid dissociates into its conjugate base and a hydrogen ion, which can react together to reform the weak acid.

 
 

If the solution becomes too acidic, the conjugate base will react with excess hydrogen ions, removing them from the solution. This shifts the position of equilibrium to the right.

 
 

If the solution becomes too alkaline, the acid will react with hydroxide ions to form water and the conjugate base. This shifts the position of equilibrium to the left.

 
 

The hydrogen ions from the weak acid-conjugate base can also react with excess hydroxide ions to form water.

 
 

Making buffer solutions

Buffer solutions can be made in two ways. The first involves mixing an excess of weak acid with a strong alkali. They will react to form the conjugate base and water.

 
 

All of the alkali should be used up in the reaction. Because we started with an excess of weak acid, we are left with a mixture of weak acid and its conjugate base.

You might be expected to do a calculation for the preparation of a buffer solution. For these, remember that the moles of strong alkali (e.g. NaOH), will be equal to the moles of conjugate base (e.g. CH3COO-), since they react in a 1:1 ratio. The moles of the weak acid at equilibrium will be the moles that you started with minus the moles that reacted with NaOH to form the conjugate base. So if you started with 0.8 mol of CH3COOH and 0.2 mol of NaOH was added, you’ll have 0.6 mol of CH3COOH (and 0.2 mol of CH3COO-) remaining at equilibrium.

The other method involves mixing a weak acid with the salt of its conjugate base. In solution, the salt will dissociate into its ions.

 
 

Blood as a buffer solution

Blood contains a natural buffer to keep pH within a narrow range (around 7.35 – 7.45). The buffer consists of carbonic acid and its conjugate base, hydrogen carbonate. The reactions that occur are:

 
 

Buffer calculations

We can rearrange the Ka expression to calculate the pH of a buffer solution, providing we know the Ka and concentrations of the weak acid and its conjugate base. Here’s how the Ka expression can be rearranged to find hydrogen ion concentration:

 
 

Worked example – calculating pH of a buffer solution

A buffer solution contains 0.1 mol dm-3 of ethanoic acid, CH3COOH, and 0.2 mol dm-3 of sodium ethanoate, CH3COO-Na+. At 298 K, the Ka of ethanoic acid is 1.74 x 10-5. Calculate the pH of the buffer solution.

  • Rearrange the Ka expression to give: [H+] = Ka x [HA] / [A-]
  • Concentration of sodium ethanoate = concentration of ethanoate ion (conjugate base)
  • [H+] = 1.74 x 10-5 x 0.1 / 0.2
  • [H+] = 8.7 x 10-6 mol dm-3
  • pH = - log 8.7 x 10-6
  • pH = 5.06

Measuring pH

pH can be measured using a pH meter – an electronic device which measures the concentration of hydrogen ions in a solution. It is a more accurate way of measuring pH than universal indicator.

To use the pH meter, you first need to calibrate the device. You do this by placing in distilled water and adjust the reading to pH 7. You repeat these in standard solutions of pH 4 (adjusting the reading to 4) and another standard solution of pH 10 (adjusting the reading to 10). Rinse the probe with distilled water after each calibration. You can then place the probe into the solution that you want to measure. Wait until the reading stops changing and then record the result. Before you measure a different solution, you’ll need to rinse the probe with distilled water.

pH curves

A pH curve shows how the pH of a solution changes during a titration. Data can be collected for the pH curve by measuring the pH (using a pH meter) of the titration mixture at fixed intervals as more and more acid/base is added. The shape of the pH curve depends on the strength of the acid and/or base in the titration.

pH curves have a vertical section in the middle which represents the point of neutralisation. At this point, the concentration of hydrogen ions is equal to the concentration of hydroxide ions. It’s called the end point (aka the equivalence point). When small amounts of acid/base are added to a titration mixture that has reached the end point, large changes in pH can occur.

  • Strong acid-strong base titrations will have a curve starting at around pH 1 and ending at around pH 13-14.

  • Strong acid-weak base titrations will have a curve starting at around pH 1 and ending at around pH 9. The end point will lie further over to the right, since a larger amount of weak base will be needed to neutralise a strong acid.

  • Weak acid-strong base titrations will have a curve starting at around pH 5 and ending at around pH 13-14. The equivalence point will lie further over to the left, since a smaller amount of strong base will be needed to neutralise a weak acid.

  • Weak acid-weak base titrations will have a curve starting at around pH 5 and ending at around pH 9. The end-point is very small here and less vertical compared to the other pH curves.

Indicators

Titrations require indicators so that we know when neutralisation has occurred. It is important that we select an indicator which changes colour over a narrow range which corresponds to the vertical part of the pH curve. Methyl orange changes colour at a more acidic pH so it is suitable for titrations involving strong acids. It changes colour from red to yellow at a pH of around 3-4. Phenolphthalein changes colour at a more alkaline pH so it is suitable for titrations involving strong bases. It changes from colourless to pink at a pH of around 8-10.

  • Strong acid-strong base titrations – methyl orange or phenolphthalein

  • Strong acid-weak base titrations – methyl orange

  • Weak acid-strong base titrations – phenolphthalein

  • Weak acid-weak base titrations – no suitable indicators so you need to use a pH meter