A level buffers an application including acids and bases

A level buffers an application including acids and bases, preparation, working and pH calculation

1. Acids and Bases

2. Brønsted-Lowry Acids and Bases The Brønsted -Lowry Theory discusses acid-base reactions in terms of proton-transfer . Note that as a H atom only has one proton and one electron, a proton is exactly the same as a H + ion. Brønsted-Lowry Acid – proton DONOR – they release hydrogen ions (H + ), when in water hydroxonium ions are formed H 3 O + . HA ( aq ) + H 2 O  H 3 O + ( aq ) + A + ( aq ) Brønsted-Lowry Base – proton ACCEPTOR – when in solution they take hydrogen ions (H + ) from water molecules. B ( aq ) + H 2 O  BH + ( aq ) + OH - ( aq )

3. Conjugate Acid – Base Pairs The Brønsted -Lowry Theory identifies acids and bases as conjugate pairs , which transform into each other with the loss or gain of a proton . It views all acid-base reactions as equilibria. e.g. HA H + + A - The general form of an acid can be written HA. When it donates a proton it becomes the conjugate base A - In the reverse reaction, when Base A - accepts a proton, it becomes it’s conjugate acid HA

4. Conjugate Acid – Base Pairs NH 3( aq ) + H 2 O (l) NH 4 + ( aq ) + OH - ( aq ) HCO 3 - ( aq ) + S 2- ( aq ) HS - ( aq ) + CO 3 2- ( aq ) Base Acid Base Acid Acid Base Acid Base conjugate acid-base pairs

5. Strong Acids and Bases Strong acids dissociate/ionises almost completely in water – most H + will be released e.g. HCl  H + + Cl - Other strong acids include HNO 3 (nitric acid), H 2 SO 4 ( sulphuric acid) Strong bases ionises almost completely in water e.g. NaOH  Na + + OH -

6. Weak Acids and Bases Most other acids and bases are weak. They are not completely ionized and exist in an equilibrium reaction with the hydronium ion and the conjugate base. Weak acids only dissociate slightly in water – only a small amount of H + will be released. The equilibrium lies to the left CH 3 COOH CH 3 COO - + H + Weak bases only slightly dissociate in water The equilibrium lies to the left NH 3 + H 2 O NH 4 + + OH - Only about 2% of ethanoic acid is dissociated.

7. Acid – base equilibria An acid cannot lose a proton, there needs to be something to accept it – base. Protons are transferred between acids (HA) and bases (B) HA ( aq ) + B ( aq ) BH + ( aq ) + A - ( aq ) If more acid or base is added the equilibrium will shift to counter act the change. If HA or B is added the equilibrium will shift to the right If BH+ or A- is added the equilibrium will shift to the left

8. Dissociation of Water Water can also dissociate slightly The equilibrium below exists in water – lies well over to the left H 2 O + H 2 O H 3 O + + OH - or H 2 O H + + OH - An equilibrium constant expression can be written for this dissociation: The concentration of ions is extremely small, so the concentration of H 2 O remains essentially constant. This gives: constant

9. K w The constants are combined to give the ionic product of water – Kw The value of K w = 1.0 x 10 -14 mol 2 dm -6 at 298 K (25 °C). Like any equilibrium constant, K w varies with temperature. In pure water [H + ] = [OH - ] Therefore in pure water – K w = [H + ] 2 constant

10. Determination of pH The pH of a solution is defined as the negative logarithm of the molar hydrogen-ion concentration. pH = -log 10 [H + ] pH scale goes from (very acidic) to 14 (very basic). pH 7 is neutral. pH can be calculated if the hydrogen ion concentration is known. e.g. The hydrogen ion concentration of a solution is 0.005 mol dm-3. What is the pH of the solution. pH = -log 10 [H + ] = -log 10 (0.005) = 2.3 [H + ] – hydrogen ion concentration (moldm -3 )

11. Calculating [H + ] Hydrogen ion concentration can also be calculated from pH. The inverse of the pH equation can be used [H + ] = 10 -pH e.g. A solution of hydrochloric acid has a pH of 2.0. What is the hydrogen ion concentration? [H + ] = 10 -pH = 10 -2.0 = 0.01 mol dm -3

12. Monoprotic Acids Strong acids ionise fully in solution Strong acids such as hydrochloric acid ( HCl ) and nitric acids (HNO 3 ) are monoprotic Monoprotic – each molecule of acid releases one proton when it dissociates. One mole of acid produces one mole of hydrogen ions. Therefore the hydrogen ions concentration is the same as the acid concentration. e.g. Calculating the pH of a 0.05 mol dm -3 solution of hydrochloric acid. HCl is a monoprotic acid therefore [ HCl ] = [H + ] [H + ] = 0.05 mol dm -3 , pH = -log 10 (0.05) = 1.3

13. Diprotic Acids For strong diprotic acids, each molecule of acid will release 2 protons when dissociated. Each mole of diprotic acid produces 2 moles of hydrogen ions. e.g. Calculating the pH of a 0.01 mol dm -3 solution of sulfuric acid (H 2 SO 4 ) Sulfuric acid is a strong diprotic acid therefore 2[H 2 SO 4 ] = [H + ] [H 2 SO 4 ] = 0.01 mol dm -3 [H + ] = 0.02 mol dm -3 pH = -log 10 (0.02) = 1.7

14. Calculating pH of Strong Base Strong bases fully ionise in water e.g. Sodium hydroxide ( NaOH ) NaOH  Na + + OH - For one mole of strong base one mole of OH- ions are released. The concentration of the base is the same as the concentration of OH - ions [Strong base] = [OH - ] To determine the pH the concentration of hydrogen ions is needed [H + ] This can be found by using the K w equation. K w = [H + ][OH - ] If [OH - ] and K w is known at a certain temperature, [H + ] can be calculation – pH found

15. Base pH Calculation Example Example: Determine the pH of 0.20 moldm -3 of NaOH at 298 K. Firstly remember from the earlier slide that K w = 1.0 x 10 -14 mol 2 dm -6 at 298 K. Put all known values in the K w equation, K w = [H + ][OH - ]. 1.0 x 10 -14 = [H + ] x 0.20 Rearrange the equation to find [H + ]. Calculate pH using [H + ] value. pH = -log 10 [H + ] = log 10 (5 x 10 -14 ) = 13.3  

16. K a K a is the acid dissociation constant – to find the K a 2 assumptions must be made Weak acids only dissociate slightly in aqueous solution – [H + ] is not the same as the [acid] The K a has to be used to find the pH The equation of dissociation: HA ( aq ) H + ( aq ) + A - ( aq ) Only a very small amount of HA dissociates, it is assumed that [HA ( aq ) ] equilibrium = [HA ( aq ) ] start Also assume that the acid dissociates much more than water therefore the H + in solution are from the acid – [H + ( aq ) ] = [A - ( aq ) ]  

17. Using K a to find pH Calculating the pH of a weak acid using K a K a is an equilibrium constant – applies to a particular acid at a specific temperature Example: Use the following information to calculate the hydrogen ions concentration and pH of 0.030 mol dm -3 hydrogen fluoride at 298 K. Hydrogen fluoride K a at 298 K = 6.6 x 10 -4 mol dm -3 Write down the expression for K a  

18. Using Ka to find pH Rearrange K a expression to find [H + ] Insert data given to find [H + ] = 1.98 x 10 -5 Calculate pH from [H + ] pH = -log 10 [H + ] pH = -log 10 (1.98 x 10 -5 ) = 4.70  

19. Using Ka to find the Concentration The K a can be used to calculate the concentration of a weak acid When pH/[H + ] and K a and given Example: Calculate the molar concentration of a propanoic acid solution. The pH of the solution is 3.30. Ka of propanoic acid at 298 K = 1.30 x 10 -5 mol dm -3 . Calculate [H + ] from the pH. [H + ] = 10 -pH [H + ] = 10 -3.30 = 5.01 x 10 -4 Write expression for K a and rearrange to find [CH 3 CH 2 COOH]  

20. Using Ka to find the Concentration Insert data into equation to calculate molar concentration [CH 3 CH 2 COOH] = 0.0193 mol dm -3  

21. pK a pK a is a measure of acid strength The smaller the pK a the stronger the acid The pK a can be calculate from the K a It is done in the same way as calculating the pH from [H + ] pK a = -log 10 ( K a ) K a = 10 -pKa

22. pK a Calculations Example: Propanoic acid has a K a of 1.30 x 10 -5 mol dm -3 . What is its pK a ? pK a = -log 10 ( K a ) pK a = -log 10 (1.30 x 10 -5 ) = 4.89 Example: HF has a pK a of 3.18. What is the K a of HF? K a = 10 -pKa K a = 10 -3.18 = 6.60 x 10 -4 You may be asked to calculate a concentration of pH from a pK a value. Must convert pK a to K a first

23. Titration Overview Acid – base titration Acid with unknown concentration is added to conical flask Base with known concentration is added from burette Burette is used so the exact amount of base added is known Indicator added to acid to determine when it has been neutralised by base The amount of base used to neutralise the acid can be used to calculate the concentration of the acid This can also be reverse, where the base has the unknown concentration

24. Titrations To get the most accurate results possible you must: Measure the volume of the unknown as precisely as possible Repeat the titration at least 3 times – calculate the mean value, to improve reliability All results should be within 0.1 cm 3 of each other – ignore anomalous results

25. pH curves pH curve shows the change in pH as a base is titrated with an acid

26. pH Curves STRONG acid – STRONG base Graph starts at 13 - pH of strong base. As strong acid is added the pH decreases until it reaches the pH of the strong acid of around 1. WEAK acid – STRONG base Graph starts at 13 - pH of strong base. As weak acid is added the pH decreases until it reaches the pH of the weak acid of around 5. STRONG acid – WEAK base Graph starts at around 9 - pH of weak base. As strong acid is added the pH decreases until it reaches the pH of the strong acid of around 1. WEAK acid – WEAK base Graph starts at around 9 - pH of weak base. As weak acid is added the pH decreases until it reaches the pH of the weak acid of around 5.

27. pH Curves Each graph (except weak acid - weak base) has a vertical section This is when the base has been neutralised – end point If a small amount of acid is added the pH will change drastically The pH changes gradually with a weak acid – weak base titration. No sudden change in pH pH meter should be used to determine end point, as the change in colour of an indicator will be gradual

28. Indicators An indicator is used to determine the end point of a titration, due to a change in colour The change in colour needs to occur at the exact end point The chosen indicator must change colour over a narrow pH range that lies on the vertical part of the pH curve

29. Indicators Common indicators used for acid – base titrations Methyl orange Phenolphthalein Strong acid – Strong base titrations – BOTH of these indicators can be used Strong acid – Weak base – METHYL ORANGE should be used as the indicator Weak acid – Strong base – PHENOLPHTHALEIN should be used as the indicator Weak acid – Weak base – NEITHER of these indicators can be used, the change in pH is gradual so you would not be able to tell when the endpoint is. Name of indicator Colour at low pH pH of colour change Colour at high pH Methyl Orange RED 3.1 – 4.4 YELLOW Phenolphthalein COLOURLESS 8.3 – 10 PINK

30. Titration Calculations Acid/Base titration results can be used to calculate unknown concentration Example: 55 cm 3 of HCl was completely neutralised by 30 cm 3 of 0.8 mol dm -3 NaOH . Use these results to calculate the concentration of HCl . Write a balanced equation. NaOH + HCl  NaCl + H 2 O Calculate the number of NaOH moles. Number of moles of NaOH = (0.8 x 30)/1000 = 0.024 mol  

31. Titration Calculations NaOH + HCl  NaCl + H 2 O From the equation you know that 1 mol of NaOH neutralises 1 mol of HCl . Therefore 0.024 mol of NaOH neutralises 0.024 mol of HCl . Concentration of HCl can be calculated. Concentration of HCl = (0.024x1000)/55 = 0.44 mol dm -3  

32. Titration Calculations: Diprotic Acid Titration between base and diprotic acid

33. Titration Calculations: Diprotic Acid Example: 25 cm 3 of 0.35 mol dm -3 H 2 SO 4 was needed to neutralise 30 cm 3 of KOH. Use these results to calculate the concentration of KOH. Write a balanced equation 2KOH + H 2 SO 4  K 2 SO 4 + 2H 2 O Calculate how many moles of H 2 SO 4 Number of moles of H 2 SO 4 = (0.35x25)/1000 = 0.0088 mol  

34. Titration Calculations: Diprotic Acid 2KOH + H 2 SO 4  K 2 SO 4 + 2H 2 O From the equation you know 1 mole of H 2 SO 4 is needed to neutralise 2 moles of KOH. Therefore 0.0088 mol of H 2 SO 4 must neutralise (0.0088 x 2) = 0.018 mol of KOH. Calculate the concentration of KOH. Concentration of KOH = (0.018 x 1000)/30 = 0.58 mol dm -3  

35. Buffers A buffer is a solution that resists changes in pH when small amounts of acid or base are added, or when it’s diluted. Adding a very small amount of dilute acid to water, changes its pH drastically Sudden changes in pH in the laboratory can cause problems Using a buffer prevents this issue The buffer will not stop the pH changing completely, but reduces the changes of pH Both acidic and basic buffers can be used

36. Acidic Buffers Acidic buffer solutions contains a weak acid and the salt of a weak acid Have a pH <7 For example: Ethanoic acid (CH 3 COOH) and Sodium ethanoate (CH 3 COO - Na + ) Solution contains a large amount of undissociated ethanoic acid and lots of ethanoate ions from the dissociated salt When acid (H + ) or base (OH - ) is added to the solution the equilibrium shifts to counteract the change Weak acid – dissociates slightly CH 3 COOH ( aq ) H + ( aq ) + CH 3 COO - ( aq ) Salt – fully dissociates CH 3 COONa (s) + water  CH 3 COO - ( aq ) + Na + ( aq )

37. Acidic Buffers CH 3 COOH ( aq ) H + ( aq ) + CH 3 COO - ( aq ) If a small amount of acid is added , the concentration of H + increases . The equilibrium shifts to the left , so the excess H + react with CH 3 COO - to form CH 3 COOH, in order to remove H+ from the solution, and return the solution to its original pH. If a small amount of base is added , the concentration of OH - increases . This reacts with H+ ions to form water, removing H+ from the soliton, therefore the equilibrium is shift to the right , in order to form more H+ ions. H+ ions are formed until the solution is close to its original pH. Addition of acid (H + ) Addition of base (OH - )

38. Basic Buffers Basic buffer solutions contain a weak base and the salt of that weak base. Have a pH >7 For example: Ammonia (NH 3 ) and Ammonium chloride (NH 4 Cl) Solution contains a large amount of undissociated ammonia and lots of ammonium ions from the dissociated salt When acid (H + ) or base (OH - ) is added to the solution the equilibrium shifts to counteract the change Weak acid – some will react with water NH 3( aq ) + H 2 O (l) NH 4 + ( aq ) + OH - ( aq ) Salt – fully dissociates NH 4 Cl ( aq )  NH 4 + ( aq ) + Cl - ( aq )

39. Basic Buffers NH 3( aq ) + H 2 O (l) N 4 H + ( aq ) + OH - ( aq ) If a small amount of acid is added , the concentration of H + increases . The H+ ions react with OH - to form water, this reduces the concentration of OH - ions therefore the equilibrium shifts to the right , in order to form more OH - ions and return the solution to its original pH. If a small amount of base is added , the concentration of OH - increases . This reacts with NH4 ions forming NH 3 and H 2 O. The equilibrium is shift to the left , to remove OH - ions from the solution, until the solution is close to its original pH. Addition of acid (H + ) Addition of base (OH - )

40. Calculating the pH of a Buffer To calculate the pH you need to know the K a of the weak acid and the weak acid and salt concentrations e.g. A buffer solution contains 0.30 mol dm -3 ethanoic acid (CH 3 COOH) and 0.50 mol dm -3 sodium ethanoate (CH 3 COO - Na + ). K a = 1.7 x 10-5 mol dm -3 for ethanoic acid. What is the pH of this buffer? Write the K a expression for the weak acid: CH 3 COOH( aq ) H + ( aq ) + CH 3 COO - ( aq )  

41. Calculating the pH of a Buffer Rearrange the expression to make [H + ] the subject Add data to equation to calculate [H + ( aq ) ] Use [H + ( aq ) ] to calculate pH. pH = -log 10 [H + ( aq ) ] = -log 10 (1.02x10 -5 ) = 4.92   A few assumptions have to be made: CH 3 COO - Na + has fully dissociated – assume that the equilibrium concentration of CH 3 COO - is equal to the initial concentration of CH 3 COO - Na + CH 3 COOH has only slightly dissociated therefore the equilibrium concentration is the same as the initial concentration

42. Buffer Applications Biological buffers – blood needs to have a pH of close to 7.4 therefore it contains a buffer system, to maintain this pH, to prevent damage to cells and organs. Hair care – Alkaline shampoos can damage hair, therefore a buffer is required to prevent this – most shampoos contain a pH 5.5 buffer. Biological washing powder – enzymes in the washing powder need to be in solution which is at the optimal pH in order for them to work most effectively. Therefore buffers are added to washing powder to get the best results.