1. JF Basic Chemistry Tutorial : Acids & Bases Shane Plunkett plunkes@tcd.ie Acids and Bases Three Theories pH and pOH Titrations and Buffers Recommended reading M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change

2. Acids and Bases Dealing with ions Arrhenius theory – an acid which contains hydrogen and can dissociate in water to produce positive hydrogen ions e.g. HX +H 2 O ↔ H + (H 3 O + ) + X - –A base reacts with a protonic acid to give water (and a salt) e.g. HCl + NaOH  NaCl + H 2 O BrØnsted-Lowry Theory – acids are proton donors; bases are proton acceptors e.g.HCN + H 2 O ↔ H 3 O + + CN - HCN is an acid, in that it donates a proton to water. Water is acting as a base, as it accepts that proton Lewis Theory – an acid accepts a pair of electrons; a base donates a pair of electrons e.g. HCl + NaOH  NaCl + H 2 O The strength of an acid depends on the extent to which is dissociates and is measured by its dissociation constant

3. Strong and weak acids and bases Strong acid – fully dissociates in water, i.e. almost every molecule breaks up to form H + ions Some strong acids are…HCl, H 2 SO 4, HNO 3 Weak acid – partially dissociates in water Some weak acids are…carboxylic acids such as CH 3 COOH, C 2 H 5 COOH Strong base – fully dissociates in water, i.e. almost every molecule breaks up to form OH - ions Some strong bases are….NaOH, compounds which contain OH - ions or O 2- ions Weak base – partially dissociates in water Some weak bases…nitrogen-containing compounds, such as NH 3 Strengths can be determined by the acid or base dissociation constant

4. Acids Act as proton donors Electron pair acceptors Strong acids dissociate fully in water. Weak acids partially dissociate. K a : acid dissociation constant HA + H 2 O  H 3 O + + A - K a = [H 3 O + ][A - ] [HA] Higher K a values mean stronger acids Bases Act as proton acceptors Electron pair donors Strong bases dissociate fully in water Weak bases partially dissociate K b : base dissociation constant

5. pH and pOH [H 3 O + ] can vary greatly  logarithmic scale used pH = -log [H 3 O + ] pOH = -log [OH - ] pH > 7 basic pH = 7 neutral pH < 7 acidic Can also express dissociation constants in terms of logs: pK a = -log K a  the higher the K a the lower the pK a Similarly for bases

6. Can incorporate pH and pOH Ionic product of water 2H 2 O  H 3 O + + OH - K w = [H 3 O + ][OH - ] pK w = -log K w K w = [H 3 O + ][OH - ] -log K w = -log [H 3 O + ][OH - ] pK w = pH + pOH = 14 (at 25°C) ________ [H 2 O] 2 K w = [H 3 O + ][OH - ] = 1.0 × 10 -14 at 25°C [H 2 O] is constant K w is the ion-product constant. K w is the product of the molar concentrations of H 3 O + and OH - ions at a particular temperature Solution is neutral[H 3 O + ] = [OH - ] [H 3 O + ] > [OH - ] [H 3 O + ] < [OH - ] Solution is acidic Solution is basic

7. Titrations Acid-base stoichiometric point can be determined by using an indicator Indicators are dyes whose colour depends on the pH: HIn is protonated form and In - is its conjugate base HIn + H 2 O ↔ H 3 O + + In - K In = [H 3 O + ][In - ] [HIn] If pH goes from low to high, the indicator goes from HIn to In - and a colour change is noted pH 14 7 1 Vol of acid added to base Strong acid-base curve pH 14 7 1 Vol of acid added to base Strong acid-weak base pH 14 7 1 Vol of acid added to base Strong base-weak acid

8. Buffers Allow pH to be maintained over small additions of acid or base Made up of a weak acid and its conjugate base, e.g. HA + H 2 O  A - + H 3 O + acid conjugate base The equilibrium will shift to the right on addition of a small amount to base and shifts to the left on addition of small amounts of acid Henderson-Hasselbalch equation allows determination of pH in buffer systems:

9. Questions What is the pH of a 0.14M aqueous solution of HCN (pK a = 9.31)? pK a = -log K a K a = 4.9 × 10 -10 HCN + H 2 O  H 3 O + + CN - K a = [H 3 O + ][CN - ] [HCN] HCNH3O+H3O+ CN - Initial conc0.1400 Equil. Conc0.14 – xxx K a = [x][x] [0.14-x] = 4.9 ×10 -10 x 2. 0.14-x = 4.9 × 10 -10 x 2 = 4.9 × 10 -10 (0.14 – x) x 2 = 6.86 × 10 -11 – 4.9 ×10 -10 x x 2 + 4.9×10 -10 x – 6.86×10 -11 = 0 Solve for x

10. x = 8.25×10 -6  [H 3 O + ] = 8.25×10 -6 pH = -log [H 3 O + ]pH = -log (8.25×10 -6 )pH = 5.08 The pH of 0.1M CH 3 COOH is 2.87. What is the value of the acid dissociation constant, K a ? Question pH = -log [H 3 O + ]= 2.87  [H 3 O + ] = 1.35×10 -3 CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - K a = [H 3 O+][CH 3 COO - ] [CH 3 COOH] CH 3 COOHH3O+H3O+ CH 3 COO - Initial conc0.100 Equil. Conc0.1 – xxx

11. K a = [1.35×10 -3 ][1.35×10 -3 ] [0.1] K a = 1.8 × 10 -5 Question Estimate the pH of 10 -7 M HCl (aq) Two contributions to H 3 O + concentration: HCl and H 2 O (aqueous soln) pH = -log [H 3 O + ] pH of pure water = 7 pH = -log [H 3 O + ]If -log [H 3 O + ] = 7  [H 3 O + ] = 1 × 10 -7 HCl + H 2 O  H 3 O + + Cl - HCl is a strong acid.  dissociates completely in water  [H 3 O + ] = 10 -7 Total [H 3 O + ] = 1.0 × 10 -7 + 10 -7 = 2.0 × 10 -7 pH = -log (2.0 × 10 -7 )pH = 6.7

12. Calculate the concentration of OH - for an aqueous solution with a pH of 9.60 Two contributions to OH - concentration. pH of pure water = 7.0  [OH - ] = 1.0 ×10 -7 pH of solution = 9.60 pH + pOH = 14  pOH = 14 – 9.60 = 4.4 pOH = -log [OH - ] = 4.4 [OH - ] = 3.98 × 10 -5  Total [OH - ] = 3.99 × 10 -5  4.0 × 10 -5 Question

13. 2001 Paper 1 Question 8 (b) A buffer solution is made by adding 0.3 mol of acetic acid and 0.3 mol of sodium acetate to enough water to make 1L of solution. Determine the pH of the buffer solution. The K a for acetic acid is 1.8 × 10 -5 pH = pK a + log [A - ] / [HA] A - = sodium acetate HA = acetic acid pH = -log (1.8 × 10 -5 ) + log (0.3) / (0.3) pH = 4.74 Calculate the pH of the buffer solution after 0.02 moles of sodium hydroxide is added 1 litre of buffer contains 0.3 moles of sodium acetate and 0.3 moles of acetic acid. Sodium hydroxide is a strong base. The acetic acid will react with the base added to try to maintain the pH. What the acid loses in concentration, the salt (sodium acetate) will gain.

14. CH 3 COOH + OH -  CH 3 COO - + H 2 O CH 3 COOHOH - CH 3 COO - Initial conc0.30 Change0.3 – 0.020.020.3 + 0.02 Equil. Conc.0.280.020.32 pH = pK a + log [A - ] / [HA] pH = - log (1.8 × 10 -5 ) + log (0.32) / (0.28) pH = 4.74 + 0.058 pH = 4.8