Acids-and-Bases, Buffer and pH calculation

This lecture is about acid bases and buffer, properties and pH calculation,

1. Acids and Bases A-level Chemistry (0971)

2. What is an acid and a Base

3. Bronsted-Lowry Acids and Bases The Bronsted-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 the same as a H + ion. Bronsted-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 ) Bronsted-Lowry Base is proton ACCEPTOR – when in solution they take hydrogen ions (H + ) from water molecules. B ( aq ) + H 2 O  BH + ( aq ) + OH - ( aq )

4. Conjugate Acid – Base Pairs The Bronsted-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

5. 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 Conjugate Acid Conjugate Base Acid Acid Conjugate Base Conjugate Acid Base conjugate acid-base pairs

6. Identify the acid, base, conjugate acid and conjugate base for each of the following. HClO 4( aq ) + H 2 O (l) ⇌ H 3 O + ( aq ) + ClO 4 – ( aq ) H 2 SO 3( aq ) + H 2 O (l) ⇌ H 3 O + ( aq ) + HSO 3 – ( aq ) CH 3 COOH ( aq ) + H 2 O(l) ⇌ H 3 O + ( aq ) + CHCOO = ( aq ) H 2 S (g) + H 2 O (l) ⇌ H 3 O + ( aq ) + HS – ( aq ) HSO 3 – ( aq ) + H 2 O (l) ⇌ H 3 O + ( aq ) + SO 3 2– ( aq ) 02:58

7. 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 - We can simplify above equilibrium as H 2 O (l) H + ( aq ) + OH - ( aq ) We can write equilibrium expression for this dissociation:  

8. Continued... The constants are combined to give the ionic product of water – Kw so K w is called the ionic product of water , t he 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

9. Therefore, in pure water K w = [H + ] 2 Rearranging this equation the [H + ] in pure water The range of possible hydrogen ion concentrations in different solutions is very large ( 10 −15 M to 10 M). To overcome the problem of dealing with a wide range of numbers, the Danish chemist Soren Sorensen introduced the pH scale. pH = -log 10 [H + ] We can use this equation to convert [H+] to pH or pH to [H+] The negative sign is introduced to make the pH values positive in most cases

10. 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 -

11. Calculating pH of a strong Acid 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 )

12. Monoprotic Acids Strong acids ionise fully in solution 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 + ] = 2 x 0.01 mol dm -3 pH = -log 10 (0.02) = 1.7

14. Try yourself.. 04:58

15. Calculating [H + ] from pH Hydrogen ion concentration can also be calculated from pH. The inverse of the pH equation can be used [H + ] = antilog(-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

16. Try yourself Calculate the concentration of hydrogen ions in solutions having the following pH values: pH 2.90 pH 3.70 pH 11.2 pH 5.40 pH 12.9 02:59

17. 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. 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 - ] [H + ] = K w /[OH - ]

18. 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 ; Calculate pH using [H + ] value. pH = -log 10 [H + ] = log 10 (5 x 10 -14 ) = 13.3  

19. Try yourself… Calculate ph of a solution contain 0.001 00 mol dm −3 KOH ( K w = 1.00 × 10 −14 mol 2 dm −6 ) Calculate pH of an aqueous solution containing 0.200 g of NaOH per dm 3 ( K w = 1.00 × 10−14 mol 2 dm −6 ). 02:58

20. 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.

21. Dissociation constant of a weak acid A week acid is an acid which dissociates partially in its solution. Consider the dissociation of a weak acid HA + H 2 O H 3 O + + A - We can simplify it as HA ( aq ) H + ( aq ) + A - ( aq ) Applying equilibrium law  

22. Significance of K a K a is called dissociation constant of an acid. It is temperature dependent. It increase with the increase of temperature. Its units depends on the acid, for a monobasic acid its mol/dm 3 . A high value for K a (for example, 40 mol dm −3 ) indicates that the position of equilibrium lies to the right. The acid is almost completely ionized (A strong acid). A low value for K a (for example, 1.0 × 10 −4 mol dm −3 ) indicates that the position of equilibrium lies to the left. The acid is only slightly ionised and exists mainly as HA molecules and comparatively few H + (a weak acid) and A − ions.

23. pKa of an Acid As K a values for many acids are very low, we can use p K a values to compare their strengths. p K a = –log 10 K a pK a is inversely relates to K a , so, for a stronger acid large K a will have a small pK a value. for example . sulfuric(IV) acid H 2 SO 3 ⇌ H + + HSO 3 − K a = 1.5 × 10 −2 ; pK a = 1.82 benzoic acid C 6 H 5 COOH ⇌ H + + C 6 H 5 COO − K a = 6.3 × 10 −5 ; pK a = 4.20 ethanoic acid CH 3 COOH ⇌ H + + CH 3 COO − K a = 1.7 × 10 −5 ; pK a = 4.77

24. A week acid is an acid which dissociates partially in its solution. Consider the dissociation of a weak acid HA + H 2 O H 3 O + + A - We can simplify it as HA ( aq ) H + ( aq ) + A - ( aq ) Applying equilibrium law Using this equation, we assume that the ionisation of the weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as that of the original acid   Calculating K a of a weak Acid

25. Try yourself… Calculate the value of K a and pKa for the following acids: 0.0200 mol dm −3 , 2-aminobenzoic acid, which has a pH of 4.30. 0.0500 mol dm −3 , propanoic acid, which has a pH of 3.10 0.100 mol dm −3 , 2-nitrophenol, which has a pH of 4.10 02:58

26. 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

27. K a and pH of weak Acids K a is the acid dissociation constant The K a can be used to find the pH. The equation of dissociation: HA ( aq ) H + ( aq ) + A - ( aq ) Only a very small amount of HA dissociates so 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 ) ]  

28. Using K a to find pH Let’s 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 As HF ( aq ) H + ( aq ) + F - ( aq ) 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 (1.98 x 10 -5 ) = 4.70  

29. Using Ka to find the Concentration 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] so = = 0.0193 mol dm -3  

30. 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 Ka = 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, you must convert pK a to K a first.

31. Buffers A buffer solution is a solution in which the pH does not change significantly when small amounts of acids or alkalis are added , or when it’s diluted. A buffer solution is used to keep pH (almost) constant. 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. A buffer solution is either a weak acid and its conjugate base or a weak base and its conjugate acid , to minimizes any change in pH when an acid or alkali is added.

32. Acidic Buffers Acidic buffer solutions contains a weak acid (CH 3 COOH) and the salt of a weak acid (CH 3 COONa). CH 3 COOH ( aq ) H + ( aq ) + CH 3 COO - ( aq ) (dissociates slightly) CH 3 COONa (s) + water  CH 3 COO - ( aq ) + Na + ( aq ) ( Salt dissociates completely) 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.

33. Acidic Buffer working 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 - )

34. Basic Buffers Basic buffer solutions contain a weak base and the salt of that weak base. For example: Ammonia (NH 3 ) and Ammonium chloride (NH 4 Cl). NH 3( aq ) + H 2 O (l) NH 4 + ( aq ) + OH - ( aq ) Weak base will hydrolyse partially. NH 4 Cl ( aq )  NH 4 + ( aq ) + Cl - ( aq ) Salt dissociates completely 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

35. 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 [OH - ] ions therefore the equilibrium shifts to the right , to form more OH - ions and to maintain its original pH. If a small amount of base is added , the concentration of OH - increases . This reacts with NH 4 + 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 - )

36. 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 ) ;  

37. Calculating the pH of a Buffer Rearrange the expression to make [H + ] the subject taking -log on both sides. -log pH = pKa – log ( ) = pKa + log ( )   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

38. Buffer Applications Biological buffers – blood needs to have a pH in the range of 7.35 to 7.45 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.

39. Past Paper Question The numerical value of the Ka of HBrO is 2.00 × 10 –9 . X is a solution of HBrO which contains 4.00 × 10 –3 mol of HBrO in 100cm 3 of solution. In this solution the following equilibrium is established in which there are two conjugate acid-base pairs. HBrO + H2 O BrO – + H3 O+ ( i ) Define conjugate acid-base pair. (ii) Identify the two-conjugate acid-base pairs shown in the equation above (iii) Calculate the pH of solution X. Show all your working. (iv) A solution containing 2.00 × 10 –3 mol of NaOH is added to solution X. A buffer solution is formed. Calculate the pH of this buffer solution.

40. Any Question

41. Solubility Product A-Level Chemistry (0971)

42. Recall

43. Equilibrium and Solubility Some ionic compounds are partially soluble in water, such compounds form an equilibrium when dissolved in water. Example AgCl. As shown above.

44. Solubility Product When solid silver chloride dissolves, it is in contact with saturated silver chloride solution and the following equilibrium is set up: AgCl (s) ⇌ Ag + (aq) + Cl − (aq) the equlibirum expression For AgCl(s) the concentration at equilibrium is unchanged. So we can write this equilibrium expression as: K sp is called the solubility product . Values are quoted at 298 K. K sp = [C y + ( aq ) ] a [A x − ( aq ) ] b  

45. Try Yourself Write down the dissociation of following sprangly soluble salt, also write (a). Ksp exression (b). Units of Ksp Mg(OH) 2 BaSO 4 Fe 2 S 3 PbCl 2 Al(OH) 3 02:57

46. Solubility product calculations Calculate the solubility product of a saturated solution of magnesium fluoride, MgF 2 , has a solubility of 1.22 × 10 −3 mol dm −3 . The dissociation of MgF 2 is MgF 2(s) ⇌ Mg 2+ ( aq ) + 2F − ( aq ) When 1.22 × 10−3 mol dissolves to form 1 dm3 of solution the concentration of each ion is: [Mg 2+ ] = 1.22 × 10 −3 mol dm −3 ; [F − ] = 2 × 1.22 × 10 −3 mol dm −3 = 2.44 × 10 −3 M K sp = [Mg 2+ ] [F − ] 2 ; K sp = (1.22 × 10 −3 ) × (2.44 × 10 −3 ) 2 = 7.26 × 10 −9 K sp = 7.26 × 10 −9 mol 3 dm −9

47. Solubility product calculations Calculate the solubility of copper(II) sulfide in mol dm −3 . ( Ksp for CuS = 6.3 × 10 −36 mol 2 dm −6 ) . Solution: The equation: CuS (s) ⇌ Cu 2+ (aq) + S 2− (aq) so K sp = [ Cu 2+ (aq) ][ S 2− (aq) ] From the equilibrium equation [ Cu 2+ (aq) ] = [ S 2− (aq) ] So K sp = [ Cu 2+ (aq) ] 2 Substitute the value of K sp . (6.3 × 10 −36 ) = [Cu 2+ ] 2 [Cu 2+ ] = (6.3 × 10 −36 ) = 2.5 × 10 −18 mol dm −3  

48. Try yourself… Calculate the solubility in mol dm −3 of zinc sulfide, ZnS. ( K sp =1.6 × 10 −23 mol 2 dm −6 ) Calculate the solubility of silver carbonate, Ag 2 CO 3 . ( K sp = 6.3 × 10 −12 mol 3 dm −9 ) Calculate the solubility product of the following solutions: a saturated aqueous solution of cadmium sulfide, CdS (solubility = 1.46 × 10 −11 mol dm −3 ) a saturated aqueous solution of calcium fluoride, CaF 2 , containing 0.0168 g dm −3 CaF 2 04:58

49. Partition Coefficient