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Title: Lesson 7: K a /pK a and K b /pK b Learning Objectives: – Understand the concepts of K a and K b – Understand the concepts of pK a and pK b
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Main Menu Refresh Which values are correct for a 0.010 mol dm –3 solution of NaOH(aq) at 298 K? (K w = 1.0×10 –14 mol 2 dm –6 at 298 K) A. [H + ] = 1.0×10 –12 mol dm –3 and pH = 12.00 B. [OH – ] = 1.0×10 –12 mol dm –3 and pH = 12.00 C. [H + ] = 1.0×10 –12 mol dm –3 and pOH = 12.00 D. [OH – ] = 1.0×10 –12 mol dm –3 and pOH = 12.00
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Main Menu Dissociation constants express the strength of weak acids and bases Weak acids and bases do not dissociate fully. This means that we cannot deduce the concentrations of ions in their solutions from the initial concentrations, as ion concentration will depend on the extent of dissociation that has occurred. The dissociation reactions of weak acids and weak bases can be represented as equilibrium expressions, each with their own equilibrium constant. E.g. Generic weak acid HA dissociating in water: Therefore: Given that concentration of water is considered constant, we can combine with K c K a is known as the acid dissociation constant. It will have a fixed value for a particular acid at a specified temperature. K a tells us the equilibrium position direct measure of strength of the acid. E.g. Higher K a More dissociation Stronger acid
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Main Menu Base dissociation constant Dissociation/Ionisation of generic weak base B: Therefore: Given that concentration of water is considered constant, we can combine with K c K b is known as the base dissociation constant. It will have a fixed value for a particular base at a specified temperature. K b tells us the equilibrium position direct measure of strength of the base. E.g. Higher K b More dissociation Stronger base
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Solutions
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Main Menu Calculations involving K a and K b These values allow us to: Compare the strength of weak acids and bases Calculate ion concentrations at equilibrium Calculate pH and pOH These calculations are similar to the ‘ICE’ calculations you did in chapter 7: Given concentrations are the INITIAL concentrations (before dissociation) pH or pOH refers to the concentration of H + ions (or OH - ions) at EQUILIBRIUM Concentration values substituted into the expressions for K a and K b must be EQUILBIUM values If dissociation is very small (very low value for Ka or Kb), use these approximations:
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Main Menu 1 Calculation of K a and K b from pH and initial concentration
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2 Calculation of [H + ] and pH, [OH - ] and pOH from K a and K b
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Solutions
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Weak Acids: K a and pK a Weak acids dissociate to form an equilibrium HA(aq) H + (aq) + A - (aq) This has the equilibrium constant (aka acid dissociation constant), K a The values for K a are often very small (negative exponents), so we use ‘pK a ’ to make them easier to handle: pK a = -log 10 (K a ) K a = 10 -pKa
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Main Menu K a and pK a in action In order of decreasing acid strength: Smaller K a weaker acid Smaller pK a stronger acid AcidKaKa pK a Hydronium ion, H 3 O + 1.000.00 Oxalic acid, HO 2 CCO 2 H5.9x10 -2 1.23 Hydrofluoric, HF7.2x10 -4 3.14 Methanoic, CHOOH1.77x10 -4 3.75 Ethanoic, CH 3 COOH1.76x10 -5 4.75 Phenol, C 6 H 5 OH1.6x10 -10 9.80
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Main Menu Weak Bases: K b and pK b Weak bases dissociate to form an equilibrium BOH(aq) B + (aq) + OH - (aq) This has the equilibrium constant (aka base dissociation constant), K b The values for K b are often very small (negative exponents), so we use ‘pK b ’ to make them easier to handle: pK b = -log 10 (K b ) K b = 10 -pKb
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Main Menu K b and pK b in action In order of decreasing base strength: Smaller K b weaker base Smaller pK b stronger base Things about pK a and pK b to consider: They are positive with no units A change of one unit of pK a or pK b represents a 10 fold change in the value of K a or K b (scale is logarithmic base 10) pK a and pK b must be quoted at specified temperatures (K a and K b are temperature dependent) BaseKbKb pK b Diethylamine1.3x10 -3 2.89 Ethylamine5.6x10 -4 3.25 Methylamine4.4x10 -4 3.36 Ammonia1.8x10 -5 4.74 Data on pK a and pK b is found in Table 21 in the data booklet
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Main Menu Relationship between K a and K b, pK a and pK b for a conjugate pair Consider the K a and K b expressions for the conjugate acid-base pair HA and A - This relationship is true for any conjugate acid-base pair in aqueous solution. From the value of K a we can calculate K b for its conjugate base. Stronger acids have weaker conjugate bases and vice versa.
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Main Menu Inverse relationship between the strengths of acids and their conjugate bases
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Solutions
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Measuring K a /pK a and K b /pK b pK a and pK b can be determined experimentally At the point of half neutralisation: pH = pK a pOH = pK b This is a convenient artefact of the mathematics Follow the instructions herethe instructions here
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Main Menu Key Points For a weak acid: AND pK a = -log 10 (K a ) For a weak base: AND pK b = -log 10 (K b ) At the point of half-neutralisation: pK a = pH AND pK b = pOH
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