For example a pH of 4 has a concentration of H+ ions of 10^-4 = 0.0001mol/dm3. To convert from a concentration of H+ ions to pH we must use the log button on your calculator. We need to do -log(concentration) so for example if the concentration is 0.00001 we need to key in to our calculator -log(0.00001) and we get the value 4, we need to write the pH to two decimal places so we write 4.00. If you have to put in the “base” for your log it is 10.

Find the pH of a strong acid is relatively simple, as they fully dissociate the concentration of H+ is straight forward.

• For monoprotic acids i.e HCL       [Acid] = [H+]

• For diprotic acids i.e H2SO4        2x[Acid] = [H+]

• For triprotic acods is H3PO4        3x[Acid] = [H+]

To calculate the pH of a solution of 0.3mol/dm3 of H2SO4, we first have to calculate the concentration of H+, as it is diprotic we must x2 so [H+] = 0.6mol/dm3. We must then convert concentration to pH so -log(0.6) = 0.22

Ionic Product of Water.

We have a very simple definition of neutral which is either the concentration of H+ ions is equal to the concentration of OH- ions or simply,

ure water is neutral because [H+]=[OH-] this is because both of these ions are produced in an equilibrium. It is an endothermic process.

In pure water the concentration of [H+] ions will vary with temperature so the pH will also vary, the concentration of H+ ions ad OH- ions will always be equal so will be neutral even if it is not pH 7. As the process of dissociation is endothermic if if the temperature is increased the concentration of H+ ions increases so the pH will decrease.

Like all equilibrium an equilibrium expression and constants can be calculated, it is given its own unit Kw which is the ionic product of water.

The concentration of water is omitted as it is such a large value changes in the concentration of H+ ions would not affect the value for Kw. As pure what is neutral and the concentration of the H+ and OH- ions are equal we can rearrange the equation in the following way.

By using the above equation and given the value of Kw at different temperatures we would be able to calculate the pH of pure water at different temperatures.

At 10C Kw= 0.3x10^-14

√Kw = √(0.3x10^-14) = 5.47x10^-8 This gives [H+]

-log(5.47x10^-8) = 7.26 -log[H+] gives pH, give to 2 decimal places

At room temperature the Kw of water is 1x10^-14, this is the most common value of Kw that you will use

-log(√1x10^-14)= 7.00

Strong bases

To calculate the pH of a strong base we need to use the ionic product of water, Kw, and the concentration of OH- ions.

Initially we must calculate the concentration of OH- ions, they are reasonabley simple to calculate.

[MOH] = [OH-]

2x[M(OH)2] = [OH-]

3x[M(OH)3] = [OH-]

Once the concentration of OH- ions has been calculated it needs to be substituted in to the ionic product of water equation along with the value of Kw depending on the temperature, most questions will use room temperature so 10^-14.

Example of Calculating pH of Strong Base

To calculate the pH of 7.4g of Ca(OH)2 in 250cm3 of water. First we need to calculate the concentration of Ca(OH)2, so we need both the volume and number of moles.

Mol.Ca(OH)2 = 7.4/74 = 0.1mol

Volume = 250cm3 = 0.25dm3

[Ca(OH)2] = 0.1/0.25 = 0.4mol/dm3

Now we have to calculate the concentration of OH- ions.

2x[Ca(OH)2] = [OH-]

[OH-] = 2×0.4 = 0.8

We now need to substitute in the values of Kw and [OH-] in to the ionic product of water.

Kw = [H+][OH-]

10^-14 = [H+] x 0.8

[H+] = (10^-14÷0.8) = 1.25x10^-14

Now we have the concentration of H+ ions we can convert to pH using -log.

-log(1.25x10^-14) = 13.90

Weak Acids

Weak acids do not completely dissociate so require an additional step in order to calculate the pH. Weak acids are in an equilibrium,

As there is equilibrium we can work out the position by using an equilibrium expression and equilibrium constant. The equilibrium constant for weak acids is known as Ka.

During dissociation each molecule of acid dissociates in to an H+ and its conjugate base, the ethanoate in the above example, in equal amounts so when dealing with just a weak acid [H+]=[Base] so the equation can be changed similar to the ionic product of water above.

As with all pH calculations we just need the concentration of H+, so we can further rearrange so H+ is the subject

Ka can be given as a value or it can be given as a pKa to convert pKa to Ka we use the same method as pH to [H+], we need to do 10 to the power of -pKa. This means by simply substituting numbers in to the above equation we can calculate the concentration of H+ and then convert to pH.

The pKa of ethanoic acid is 4.76, so its ka is 10^(-4.76) = 1.8 x10^-5, we can now use this value for Ka to calculate the concentration of H+ ions.

Working out pH of a weak acid

We will calculate the pH of 0.25mol/dm3 of ethanoic acid, pKa = 4.76

Ka = 10^(-4.76) = 1.8x10^-5

[H+]=√([CH3COOH]xKa)

[H+]=√(0.25x1.8x10^-5) = 0.002

-log(0.002) = 2.67

Buffer Solutions

A buffer solution is a solution which resists a change in pH when acid or base is added. It is produced with a weak acid and its salt or a weak base and its salt. For example it could be made with ethanoic acid and sodium ethanoate.

To calculate the H+, and then the pH, for a buffer solution the Ka, concentration of acid and concentration of salt is needed, as when we add the salt we are change the concentration of the conjugate base and can no longer treat it the same as concentration of H+. The below Ka expression is rearranged to make H+ the subject

Calculating pH of a buffer solution

To calculate the pH of buffer solutions we need to calculate the concentration of each and add them to the Ka expression. Imagine we had the following solutions and mixed them together, we will try to calculate the pH.

1. 100cm3 of 0.5moldm3 of CH3COOH

2. 25cm3 of 0.1moldm3 of CH3COONa

In this scenario we will have a change in volume so we need to calculate moles and then concentration of the new solution

1. mol.CH3COOH = 0.1 x 0.5 = 0.05mol

   [CH3COOH] = 0.05/0.125 = 0.4moldm3

2. mol.CH3OONa = 0.025 x 0.1 = 0.0025mol

   [CH3OONa]= 0.0025/0.125 = 0.002moldm3

   [CH3OO-] = 0.002moldm3

The Ka of ethanoic acid is 1.8x10^-5

We substitute these values in to the above rearanged equation

[H+] = (Ka x [Acid])/[Salt]

[H+] = (1.8x10^-5 x 0.4) / (0.002) = 0.0036

-log[H+] = -log(0.0036) = 2.44

Calculating mass for a buffer solution

If we are trying to create a 500cm3 buffer solution with Ethanoic acid with a final concentration of 0.5 mold/dm3 and a pH of 2.50. We need to calculate the mass of sodium ethanoate used.

First we must convert the pH to a concentration

[H+] = 10^(-2.5) = 0.00316

Now we need to rearrange the equilibrium expression so the salt is the subject. We are using the Ka from the previous example.

[Salt] = (1.8x10^-5 x 0.5) / (0.00316) = 0.028mol/dm3

The volume of the solution is 500cm3, so we need to substitute in to the equation moles = concentration x volume

mol = 0.028 x 0.5 = 0.0142

The Mr of Sodium Ethanoate is 82 Mass = Moles x Molecular Mass

0.0142 x 82 = 1.16g of Sodium Ethanoate

Acid-Base Reactions

The combinations which you can be asked about will have any of these three in excess, in terms of protons and hydroxides

• Strong Acid

• Strong Base

• Weak Acid

If strong acid is in excess the concentration of H+ is calculated by finding the amount of moles of H+ and then OH-. Then mol.H+ - mo.OH- which will give the remaining H+, this is then used to work out the concentration of H+ making sure to use the total combined volume. -log[H+] is then used to get the new pH.

If a strong base is in excess the excess OH- is calculated in a similar manner to the above but then the concentration of OH- is calculated and then the ionic product of water used to calculate the concentration of H+.

If the weak base is in excess and reacted with a strong base the concentration of H+ is calculated as above but also the concentration of base, in the new total volume, is calculated and used as the concentration of salt and these values are put in to the equilibrium expression to obtain the concentration of H+.

An example of a weak acid and strong base

0.25dm3 of 0.1mol/dm3 HCOOH is added to 0.2dm3 of 0.1mol/dm3 of NaOH, what is the finals solutions pH? The Ka of methanoic acid is 1.78x10^-4

First we need to calculate the moles of H+, OH-, and the amount in excess.

mol.acid = 0.25x0.1 = 0.025mol

mol.base = 0.2x0.1 = 0.02mol

Excess acid = mol.acid - mol.base = 0.025-0.02 = 0.005mol

Now we need to calculate the concentration of acid and put it in to the ka expression. The concentration of reacted base is the concentration of the salt.

[acid] = 0.005 / (0.25+0.2) = 0.011mol/dm3

[salt] = 0.02 / (0.25+0.2) = 0.044mol/dm3

[H+] = (Ka x [acid]) / [salt]

[H+] = 1.78x10^-4 x 0.011 / 0.044 = 0.0000445

-log[H+] = -log(0.0000445) = 4.35

pH Curves and Indicators

A pH curve can be obtained from titrating a base against an acid and using a pH probe to monitor the change of pH as the base is added. There is four shapes you will need to identify the strong acids and bases both change pH very minimally until they reach near the equivalence point and then rapidly change. Weak acids and bases have gradual change in pH towards the equivalence point and not rapid change in pH. The equivalence point is when enough acid or base has been added for neutralisation to occur.

Below are examples of the graphs you need to remember,

In the above graphs the vertical section indicates the equivalence point, in a titration with an indicator the colour change needs to happen at this equivalence point as it is a rapid change of pH and shows a clear change, when titrating with an indicator when it changes colour it is called the end-point.

Deciding what indicator is suitable to use relies on using one of the above graphs and knowing the pH of colour change. Phenolphthalein shows a colour change at pH 8.3 so to work out which of the 4 types of titration above can use phenolphthalein we add a line at pH 8.3.

Using the above graphs we can identify that phenolphthalein is only suitable for two combinations

Strong Acid and Strong Base, & Weak Acid and Strong Base can both use phenolphthalein as the indicator changes colour at the vertical line in the graph which indicates a rapid change in pH and will have an obvious colour change.

The other two graphs, Strong Acid and Weak Base, & Weak Acid and Weak Base the line at pH 8.3 cross a non-vertical section which means the pH is gradually changing at the colour change would not be obvious or not at the equivalence point.

The weak acid and weak base graph does not have a vertical section so no indicator is suitable for those titrations.

Worksheets

• coming soon!

Revision Papers

• Bronsted-Lowry Acids MS

• Strong Bases and ionic product of water MS

• Weak Acids and Buffer Solutions MS

• pH Curves and indicators MS

• MCQ MS