Refresh For each of the following equilibria, write the expression for the equilibrium constant Kc and state its units: 2NO2(g) == N2O4(g) CH3CH2CO2H(l) + CH3CH2OH(l) == CH3CH2CO2CH2CH3(l) + H2O(l) H2(g) + I2(g) == 2HI(g) 2SO2(g) + O2(g) == 2SO3(g) N2(g) + 3H2(g) == 2NH3(g)

Answers

Calculate the equilibrium constant for the following Answers 28.65 48.33 1.1*109 3.5

More extreme In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction. N2(g) + 3H2(g) == 2NH3(g)   Hydrogen chloride can be oxidised to chlorine by the Deacon process: 4HCl(g) + O2(g) == 2Cl2(g) + 2H2O(g) 0.800 mol of hydrogen chloride was mixed with 0.200 mol of oxygen in a vessel of volume 10 dm3. At equilibrium it was found that the mixture contained 0.200 mol of hydrogen chloride. Calculate Kc for the reaction. A 0.04 sample of SO3 is introduced into a 3.04 litre vessel and allowed to reach equilibrium. The amount of SO3 present at equilibrium is found to be 0.0284 mole. Calculate the value of Kc for the reaction 2SO3(g) == 2SO2(g) + O2(g). Answers: 1-> 6.6mol-2dm6 2 -> 1013mol-1dm3 3 -> 3.2*10-4 mol dm-3

Acids and Bases (HL) - Lesson 7 Ka/pKa and Kb/pKb

Lesson 6: Ka/pKa and Kb/pKb Objectives: Understand the concepts of Ka and Kb Understand the concepts of pKa and pKb

Weak Acids: Ka and pKa pKa = -log10(Ka) Weak acids dissociate to form an equilibrium HA(aq)  H+(aq) + A-(aq) This has the equilibrium constant (aka acid dissociation constant), Ka The values for Ka are often very small, so we use ‘pKa’ to make them easier to handle: pKa = -log10(Ka)

Ka and pKa in action In order of decreasing acid strength: Smaller Ka  weaker acid Smaller pKa  stronger acid Acid Ka pKa Hydronium ion, H3O+ 1.00 0.00 Oxalic acid, HO2CCO2H 5.9x10-2 1.23 Hydrofluoric, HF 7.2x10-4 3.14 Methanoic, CHOOH 1.77x10-4 3.75 Ethanoic, CH3COOH 1.76x10-5 4.75 Phenol, C6H5OH 1.6x10-10 9.80

Weak Bases: Kb and pKb pKb = -log10(Kb) Weak bases dissociate to form an equilibrium BOH(aq)  B+(aq) + OH-(aq) This has the equilibrium constant (aka base dissociation constant), Kb The values for Kb are often very small, so we use ‘pKb’ to make them easier to handle: pKb = -log10(Kb)

Kb and pKb in action In order of decreasing base strength: Smaller Kb  weaker base Smaller pKb  stronger base Base Kb pKb Diethylamine 1.3x10-3 2.89 Ethylamine 5.6x10-4 3.25 Methylamine 4.4x10-4 3.36 Ammonia 1.8x10-5 4.74

Measuring Ka/pKa and Kb/pKb pKa and pKb can be determined experimentally At the point of half neutralisation: pH = pKa pOH = pKb This is a convenient artefact of the mathematics Follow on the instruction sheet

Key Points At the point of half-neutralisation: For a weak acid: AND pKa = -log10(Ka) For a weak base: AND pKb = -log10(Kb) At the point of half-neutralisation: pKa = pH AND pKb = pOH