1. Acids, Bases and Buffers The Br Ø nsted-Lowry definitions of an acid and a base are: Acid: species that donates a proton Base: species that can accept a proton When an acid is added to water, the acid dissociates releasing H+ ions into solution. Depending on the formula and bonding, different acids can release different numbers of protons: HCl  H + + Cl - - this is monoprotic as it releases one proton H 2 SO 4  2H + + SO 4 2- - this is diprotic as it releases two protons H 3 PO 4  3H + + PO 4 3- - this is triprotic as is releases three protons Writing ionic equations 1.Write out the balanced chemical equation 2.Split up each compound into its ions – be careful with solids – they don’t dissociate, only aqueous solutions dissociate! 3.Cancel species that appear on both sides of the equation Example: Aqueous sodium carbonate and hydrochloric acid Na 2 CO 3 + 2HCl  2NaCl + CO 2 + H 2 O 2Na + + CO 3 2- + 2H + + 2Cl -  2Na + + 2Cl - + CO 2 + H 2 O CO 3 2- + 2H +  CO 2 + H 2 O Writing ionic equations 1.Write out the balanced chemical equation 2.Split up each compound into its ions – be careful with solids – they don’t dissociate, only aqueous solutions dissociate! 3.Cancel species that appear on both sides of the equation Example: Aqueous sodium carbonate and hydrochloric acid Na 2 CO 3 + 2HCl  2NaCl + CO 2 + H 2 O 2Na + + CO 3 2- + 2H + + 2Cl -  2Na + + 2Cl - + CO 2 + H 2 O CO 3 2- + 2H +  CO 2 + H 2 O

2. Conjugate acid-base pairs transform into each other by the gain or loss of a proton. Conjugate acid-base pairs are pairs of two species that transform into each other by gain or loss of a proton. The ethanoic acid loses a proton to become an ethanoate ion. The ethanoate ion can accept a proton to become ethanoic acid. The ethanoic acid is the acid, the ethanoate ion is the base. The water molecule can gain a proton to become H 3 O +. The H 3 O + loses a proton to become water. The H 3 O + is the acid, the H 2 O is the base.

3. Strong and Weak Acids [H + ] eqm = [HA] initial Calculation 1: Strong Acid A strong acid is a proton donor that completely dissociates in aqueous solution. H 2 SO 4 2H + + SO 4 2- The value of K c is extremely large, because equilibrium lies very far to the right. CH 3 COOH CH 3 COO - The value of K c is small, because equilibrium lies to the left. A weak acid is a proton donor that only partially dissociates in aqueous solution. Calculating pH: Strong acids are fully dissociated in aqueous solution, so we can say that the concentration of H+ is the same as the initial concentration of the acid itself. A sample of HCl has a concentration of 1.22x10 -3 moldm -3. What is it’s pH? HCl is a strong acid, therefore [H+] = [HCl] [H+] = 1.22x10 -3 moldm -3 pH = -log [H+] = -log (1.22x10 -3 moldm -3 ) = 0.76

4. Calculation 2: Strong base using K w K w = [H + ][OH - ] K w is known as the ionic product of water. K w has a value of 1x10 -14 at 25°. The definition of K w is represented by the following equation: A strong base fully dissociates in aqueous solution. Therefore, we can say the concentration of OH - is the same as the initial concentration of the base. We can use the K w equation in rearranged form to find the [H + ] and thereby find the pH. A solution of KOH has a concentration of 0.050moldm -3. What is its pH? = 2x10 -13 pH = -log [H+] = -log (2x10 -3 ) = 12.70 The acid dissociation constant, K a, shows the extent of the dissociation of an acid. It is given by the expression: K a : The Acid Dissociation Constant K a values can be made more manageable if expressed in logarithmic form, called pK a : Strong Acids have high K a values and low pK a values – this indicates a large extent of dissociation.

5. Calculation 3: Weak Acid using K a The concentration of a sample of nitrous acid HNO 2 is 0.055moldm-3. K a = 4.7 x10 -4. Calculate the pH. = 2.29

6. Strong Acid, Strong Base Strong Acid, Weak Base Weak Acid, Weak Base Weak Acid, Strong Base Acid Base Titration Curves

7. The equivalence point of the titration is the point at which the volume of one solution has exactly reacted with the volume of the second solution. This matches the stoichiometry of the reaction taking place. When the base is first added, there is only a very small increase in pH – the acid is in great excess. Within 1-2cm 3 of the equivalence point, the pH starts to increase more quickly – the acid is now only present in small excess. The vertical section shows the very sharp increase in pH brought about by a very small further addition of base. The equivalence point is in the middle of this vertical section. As further base is added, there is little additional change in pH, because the base is in great excess. An indicator must be chosen for a titration so that the end point is as close as possible to the pH value of the titration’s equivalence point. A suitable indicator changes colour within the vertical section of the titration curve. An indicator is a weak acid that is one colour in its acid form and a different colour in its conjugate base form. An indicator at its end point has equal amounts of acid and conjugate base present.

8. Buffer Solutions A buffer solution is a system that minimises the pH change on addition of small amounts of acid or base. A buffer system is made from a weak acid and a salt of the weak acid – for example, methanoic acid and sodium methanoate. Alternatively, the weak acid could be partially neutralised by a an aqueous alkali such as NaOH, to give a solution containing a mixture of the salt and an excess of the weak acid. How does a buffer system work? In the ethanoic acid/sodium ethanoate buffer system: The weak acid CH 3 COOH dissociates partially: CH 3 COOH H + + CH 3 COO - The salt (conjugate base), CH3COO - Na + dissociates completely: CH3COO - Na +  CH3COO - + Na + The equilibrium mixture formed contains a high concentration of the weak acid and its conjugate base. The resulting buffer solution contains large reservoirs of the weak acid and its conjugate base, both of which control pH. The overall principle of the buffer solution is that the weak acid removes added alkali and the conjugate base removes added acid. On addition of acid:  [H + ] increased  Conjugate base CH 3 COO - reacts with H +  Equilibrium shifts left removing most of added H + ions. On addition of alkali:  [OH - ] increased  Added OH- reacts with H + to form H 2 O  CH 3 COOH dissociates to form CH 3 COO - + H +. Equilibrium shifts right restoring most of H + ions that have reacted.

9. Human Blood: Carbonic Acid-Hydrogencarbonate buffer system Human blood plasma needs to have a pH of between 7.35 and 7.45. This is controlled by the Carbonic acid-Hydrogencarbonate buffer. The weak acid H 2 CO 3 partially dissociates. Hydrogencarbonate, HCO 3 - acts a conjugate base. On addition of acid:  [H + ] increased  Conjugate base HCO 3 - reacts with H + forming H 2 CO 3  Equilibrium shifts left removing most of added H + ions. On addition of alkali:  [OH - ] increased  Added OH- reacts with H+ to form H2O  H 2 CO 3 dissociates to form HCO 3 - + H +. Equilibrium shifts right restoring most of H + ions that have reacted. Plenty of H 2 CO 3 to make more H + if [OH - ] increases Plenty of HCO 3 - to combine with H + if [H + ] increases H 2 CO3 HCO 3 - + H +

10. Calculation 4: pH of a buffer solution Buffer Solutions Calculate the pH of the following buffer solution: 50 cm 3 of 1.0 moldm -3 methanoic acid, K a = 1.78 x 10 -4 moldm -3 ; 20 cm 3 of 1.0 mol dm -3 sodium methanoate. Number moles acid = 1 x 50x10 -3 = 0.05 Number moles sodium methanoate = 1 x 20x10 -3 = 0.02 The pH of a buffer depends on the amount of dissociation, the temperature and the concentration ratio of the weak acid and its conjugate base. The expression used for pH of a buffer solution is:

11. Calculation 5: pH change of a buffer solution How would we calculate the pH change when 10cm 3 of 0.1 moldm -3 NaOH is added to the buffer solution in the previous example? What we already know about the buffer solution: Initial moles of acid = 0.05 moles Initial moles of NaCOOH = 0.02 moles Initial pH of buffer = 3.35 K a = 1.78 x 10 -4 moldm -3 When NaOH is added, a reaction occurs between the acid and the NaOH base: HCOOH + OH -  HCOO - + H 2 O So, for every one mole of NaOH added, we get one mole more of the conjugate base, NaCOOH and one mole less of acid. Initial MolesChangeFinal Moles HCOOH0.05-0.0010.049 NaCOOH0.02+0.0010.021 Moles NaOH added: 0.1 x 10x10-3 = 0.001 Therefore, the amount of acid present will go down by this amount and the amount of conjugate base will go up by this amount. We can now use a K a calculation with the new amounts present in the buffer solution to calculate the pH: = 0.03 pH= -log(4.15x10 -4 ) = 3.38 therefore pH change = 3.38 – 3.35 = 0.03

12. Calculation 6: Neutralisation 1.Calculate moles of H + before reaction. 2.Calculate moles of OH - before reaction 3.Calculate excess amount of H+ or OH- remaining after the reaction 4.Calculate excess concentration of OH - or H + 5.Find pH suing appropriate calculation. Calculate the pH when 62cm 3 of 1.05moldm -3 NaOH is added to 23cm 3 of 0.91moldm -3 HCl. Calculate the pH when 35cm 3 of 1.25moldm -3 NaOH is added to 73cm 3 of 0.92moldm -3 HCl.