1.  a pH value tells us how much H + is in a solution. It can be defined by: pH = - log[H + ] (log is to base 10). (Also note, square brackets are used to denote concentration values); For example: pH = -log[10 -2 ] = 2  The pH scale gives a measure of how acidic or alkaline a solution is. It normally runs from 0 (most acidic) to 14 (most alkaline). A neutral substance is pH 7 at 25°C. Indicators  An indicator changes colour depending on the properties of the substance it is added to. Three indicators are commonly used to show whether a solution is acidic or alkaline: 1) Litmus; 2) Phenolphthalein; 3) universal indicator;  Litmus and phenolphthalein are single indicators (they only contain one colour-changing substance) whereas universal indicator is a mixed indicator (it contains several different colour-changing substances).

2.  Litmus paper can be red or blue.  The table shows its colours in acidic, neutral and alkaline solutions.

3.  Phenolphthalein changes colour sharply at about pH 8. For most purposes, this means that it is pink in alkaline solutions and colourless in acidic solutions.  Universal indicator shows a range of colours depending on the pH of the solution (see below).

5.  The pH curve below shows what happens to the pH when a strong acid (such as hydrochloric acid) is added to 25 cm 3 of a strong alkali (such as sodium hydroxide).  The acid and the alkali started off at the same concentration. Note that the pH falls: slowly at first as acid is added to the alkali rapidly at the end-point (the point where the alkali is completely neutralised) slowly again once excess acid is being added In this example, 25 cm 3 of acid was needed to neutralise the alkali.

6.  The pH curve below shows what happens to the pH when a strong alkali is added to 25 cm 3 of a strong acid.  As before, they both started off at the same concentration. Note that the pH rises: slowly at first as alkali is added to the acid rapidly at the end-point (the point where the acid is completely neutralised) slowly again once excess alkali is being added

7.  The concentration of an acid or alkali can be calculated by carrying out an experiment called a titration. The apparatus needed includes a: 1.pipette to accurately measure a certain volume of acid or alkali 2.pipette filler to use the pipette safely 3.conical flask to contain the liquid from the pipette 4.burette to add small, measured volumes of one reactant to the other reactant in the conical flask

8. 1. Use the pipette and pipette filler to add 25 cm 3 of alkali to a clean conical flask. 2. Add a few drops of indicator and put the conical flask on a white tile (so you can see the colour of the indicator more easily). 3. Fill the burette with acid and note the starting volume. 4. Slowly add the acid from the burette to the alkali in the conical flask, swirling to mix. 5. Stop adding the acid when the end-point is reached (the appropriate colour change in the indicator happens). 6. Note the final volume reading. 7. Repeat steps 1 to 5 until you get consistent readings

9.  The difference between the reading at the start and the final reading gives the volume of acid (or alkali) added.  This volume is called the titre.  For example, if the reading at the start is 1.0 cm 3 and the final reading is 26.5 cm 3, then the titre is 25.5 cm 3 (26.5 – 1.0).  Note that the titre will depend upon the volume of liquid in the conical flask, and the concentrations of the acid and alkali used.  It is important to repeat the titration several times to check that your titre value is consistent so that your calculations are reliable.  A single indicator like litmus or phenolphthalein gives a sharp end-point where the colour changes suddenly.

10.  You should be able to use titration results to calculate the concentration of an acid or alkali.  If several runs have been carried out, any irregular titres should be ignored before calculating the mean titre.  Example: 27.5 cm 3 of 0.2 mol/dm 3 hydrochloric acid is needed to titrate 25.0 cm 3 of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?  You can check your answer using this quick method. N 1 *V 1 =N 2 *V 2 ; N 1 - unknown concentration; N 2 - known concentration; V 1 - volume of unknown; V 2 - volume of known; unknown concentration = 0.2 × 27.5 / 25.0 = 0.22 mol/dm 3

11.  You should be able to use titration results to calculate the concentration of an acid or alkali.  If several runs have been carried out, any irregular titres should be ignored before calculating the mean titre.  Example: 27.5 cm 3 of 0.2 mol/dm 3 hydrochloric acid is needed to titrate 25.0 cm 3 of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?  You can check your answer using this quick method. N 1 *V 1 =N 2 *V 2 ; N 1 - unknown concentration; N 2 - known concentration; V 1 - volume of unknown; V 2 - volume of known; unknown concentration = 0.2 × 27.5 / 25.0 = 0.22 mol/dm 3

12.  You should be able to use titration results to calculate the concentration of an acid or alkali.  If several runs have been carried out, any irregular titres should be ignored before calculating the mean titre.  Example: 27.5 cm 3 of 0.2 mol/dm 3 hydrochloric acid is needed to titrate 25.0 cm 3 of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?  You can check your answer using this quick method. N 1 *V 1 =N 2 *V 2 ; N 1 - unknown concentration; N 2 - known concentration; V 1 - volume of unknown; V 2 - volume of known; unknown concentration = 0.2 × 27.5 / 25.0 = 0.22 mol/dm 3

13.  Example 1: Given the equation NaOH(aq) + HCl(aq) ==> NaCl(aq) + H 2 O; 25.0 cm 3 of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.200 mol dm -3 (0.2M) hydrochloric acid. Using a suitable indicator it was found that 15.0 cm 3 of the acid was required to neutralise the alkali. Calculate the molarity of the sodium hydroxide and its concentration in g/dm 3. (4.80 g/dm 3 )  Calculate the molarity of the sodium hydroxide and its concentration in g/dm 3.  moles = molarity x volume (in dm 3 = cm 3 /1000)  moles HCl = 0.200 x (15.0/1000) = 0.003 mol  moles HCl = moles NaOH (1 : 1 in equation)  so there is 0.003 mol NaOH in 25.0 cm 3  scaling up to 1000 cm 3 (1 dm 3 ), there are...  0.003 x (1000/25.0) = 0.12 mol NaOH in 1 dm 3  molarity of NaOH is 0.120 mol dm -3 (or 0.12M)  since mass = moles x formula mass, and Mr(NaOH) = 23 + 16 + 1 = 40  concentration in g/dm3 = molarity x formula mass  concentration in g/dm3 is 0.12 x 40 = 4.80 g/dm3

14.  Example 2: Given the equation 2KOH(aq) + H 2 SO 4 (aq) ==> K 2 SO 4 + 2H 2 O (l). 20.0 cm 3 of a sulphuric acid solution was titrated with 0,0500 mol dm -3 potassium hydroxide. If the acid required 36.0 cm 3 of the alkali KOH for neutralisation what was the concentration of the acid? (4.41 g/dm 3 )  moles = molarity x volume (in dm 3 = cm 3 /100)  mol KOH = 0.0500 x (36.0/1000) = 0.0018 mol  mol H 2 SO 4 = mol KOH / 2 (because of 2 : 1 ratio in equation above)  mol H2SO4 = 0.0018/2 = 0.0009 (in 20.0 cm 3 )  scaling up to 1000 cm 3 of solution = 0.0009 x (1000/20.0) = 0.0450 mol  mol H 2 SO4 in 1 dm 3 = 0.0450  so molarity of H 2 SO 4 = 0.0450 mol dm -3 (0.045M)  since mass = moles x formula mass, and Mr(H 2 SO 4 ) = 2 + 32 + (4x16) = 98  concentration in g/dm 3 is 0.045 x 98 = 4.41 g/dm 3