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JF Basic Chemistry Tutorial : Acids & Bases Shane Plunkett [email protected] Acids and Bases • Three Theories • pH and pOH • Titrations and Buffers Recommended reading • M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change 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 +H2O ↔ H+ (H3O+) + X- – A base reacts with a protonic acid to give water (and a salt) e.g. HCl + NaOH NaCl + H2O • BrØnsted-Lowry Theory – acids are proton donors; bases are proton acceptors e.g. HCN + H2O ↔ H3O+ + 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 + H2O • The strength of an acid depends on the extent to which is dissociates and is measured by its dissociation constant 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, H2SO4, HNO3 • Weak acid – partially dissociates in water • Some weak acids are…carboxylic acids such as CH3COOH, C2H5COOH • 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 O2- ions • Weak base – partially dissociates in water • Some weak bases…nitrogen-containing compounds, such as NH3 • Strengths can be determined by the acid or base dissociation constant Acids • • • • Act as proton donors Electron pair acceptors Strong acids dissociate fully in water. Weak acids partially dissociate. • Ka: acid dissociation constant HA + H2O H3O+ + AKa = [H3O+][A-] [HA] • Higher Ka values mean stronger acids Bases • • • • • Act as proton acceptors Electron pair donors Strong bases dissociate fully in water Weak bases partially dissociate Kb: base dissociation constant pH and pOH [H3O+] can vary greatly logarithmic scale used pH = -log [H3O+] pOH = -log [OH-] pH > 7 basic pH = 7 neutral pH < 7 acidic Can also express dissociation constants in terms of logs: pKa = -log Ka • the higher the Ka the lower the pKa • Similarly for bases • • • • • • • Ionic product of water 2H2O H3O+ + OHKw is the ion-product constant. Kw is the product of the molar concentrations of H3O+ and OH- ions at a particular temperature +][OH-] Kw = [H O 3 ________ [H2O] is constant [H2O]2 Kw = [H3O+][OH-] = 1.0 × 10-14 at 25°C [H3O+] = [OH-] Solution is neutral pKw = -log Kw [H3O+] > [OH-] Solution is acidic [H3O+] < [OH-] Solution is basic • Can incorporate pH and pOH Kw = [H3O+][OH-] -log Kw = -log [H3O+][OH-] pKw = pH + pOH = 14 (at 25°C) 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 + H2O ↔ H3O+ + InKIn = [H3O+][In-] [HIn] • If pH goes from low to high, the indicator goes from HIn to In- and a colour change is noted 14 pH 14 7 pH 1 14 7 pH 1 Vol of acid added to base Strong acid-base curve 7 1 Vol of acid added to base Strong acid-weak base Vol of acid added to base Strong base-weak acid Buffers • Allow pH to be maintained over small additions of acid or base A H O 3 • Made up of a weak acid and its K a conjugate base, e.g. HA HA + H2O A- + H3O+ HA H 3O K a acid conjugate base A • The equilibrium will shift to the right on addition of a small log H O logK log HA 3 a amount to base and shifts to the A left on addition of small amounts A of acid pH pKa log HA • Henderson-Hasselbalch equation allows determination of pH in buffer systems: Questions What is the pH of a 0.14M aqueous solution of HCN (pKa = 9.31)? Ka = 4.9 × 10-10 pKa = -log Ka HCN + H2O H3O+ + CNKa = [H3O+][CN-] [HCN] Initial conc HCN H3O+ CN- 0.14 0 0 x x Equil. Conc 0.14 – x Ka = [x][x] [0.14-x] = 4.9 ×10-10 x2 0.14-x . = 4.9 × 10-10 x2 = 4.9 × 10-10 (0.14 – x) 2 -10 -11 x2 = 6.86 × 10-11 – 4.9 ×10-10x x + 4.9×10 x – 6.86×10 = 0 Solve for x x = 8.25×10-6 [H3O+] = 8.25×10-6 pH = -log [H3O+] Question pH = -log (8.25×10-6) pH = 5.08 The pH of 0.1M CH3COOH is 2.87. What is the value of the acid dissociation constant, Ka? pH = -log [H3O+] [H3O+] = 1.35×10-3 = 2.87 CH3COOH + H2O H3O+ + CH3COOKa = [H3O+][CH3COO-] [CH3COOH] CH3COOH H3O+ CH3COO- Initial conc 0.1 0 0 Equil. Conc 0.1 – x x x Ka = [1.35×10-3][1.35×103] [0.1] Ka = 1.8 × 10-5 Question Estimate the pH of 10-7M HCl(aq) pH = -log [H3O+] Two contributions to H3O+ concentration: HCl and H2O (aqueous soln) pH of pure water = 7 pH = -log [H3O+] If -log [H3O+] = 7 [H3O+] = 1 × 10-7 HCl + H2O H3O+ + Cl- HCl is a strong acid. dissociates completely in water [H3O+] = 10-7 Total [H3O+] = 1.0 × 10-7 + 10-7 = 2.0 × 10-7 pH = -log (2.0 × 10-7) pH = 6.7 Question 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 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 Ka for acetic acid is 1.8 × 10-5 pH = pKa + 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. CH3COOH + OH- CH3COO- + H2O CH3COOH OH- CH3COO- Initial conc 0.3 0 0.3 Change 0.3 – 0.02 0.02 0.3 + 0.02 Equil. Conc. 0.28 0.02 0.32 pH = pKa + log [A-] / [HA] pH = - log (1.8 × 10-5) + log (0.32) / (0.28) pH = 4.74 + 0.058 pH = 4.8