Acids and bases

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Cards (90)

  • Bronsted-Lowry definition of an acid a proton donor
  • Bronsted-Lowry definition of a base a proton acceptor
  • Strong acid completely dissociates in aqueous solutionHX ---> H+ + X-
  • Weak acid only partially dissociates in aqueous solutionvery few acid molecules actually dissociateHX ⇌ H+ + X-
  • Strong base completely dissociates into ions in aqueous solutionNaOH ---> Na+ + OH-
  • Weak base only partially dissociates into ions in aqueous solutionvery few base molecules actually dissociate
  • Why is a logarithmic scale used for the pH scale? the concentration of hydrogen ions in aqueous solution cover a very wide range
  • Equation for pH pH = - log[H+(aq)]
  • Equation for hydrogen ion concentration using pH [H+(aq)] = 10-pH
  • How does an acid form a conjugate base? loses a protonbecause product is able to accept a proton to reform acidAcid ⇌ Proton + Conjugate Base
  • How does a base form a conjugate acid? gains a protonbecause product is able to donate/lose a proton to reform baseBase + Proton ⇌ Conjugate Acid
  • Write an equation to show how water stabilises the ions for this dissociation. HA(aq) ⇌ A-(aq) + H+(aq) HA(aq) + H2O(aq) ⇌ A-(aq) + H3O+(aq)
  • Equation for dissociation of water equilibrium. H2O(l) ⇌ H+(aq) + OH-(aq)
  • Derive the expression for Kw using Kc of water equilibrium.
  • Kw expression Kw = [H+(aq)][OH-(aq)]
  • What factor influences the value of Kw? temperature
  • What does a neutral pH mean in terms of ion concentrations? [H+(aq)] = [OH-(aq)]
  • Given that at 25oC, Kw = 1 x 10-14, show that the pH of pure water at 25oC is 7.
  • Calculate the pH of water at 50oC given that Kw = 5.476 x 10-14 mol2dm-6 at 50oC
  • Calculate the pH of the strong base 0.1 mol dm-3 NaOH. Assume complete dissociation. Kw = 1 x 10-14 moldm-3
  • Given [OH-(aq)], how can pH of a strong alkali be calculated? Rearrange the Kw expression to find [H+(aq)]Sub in values for Kw and [OH-(aq)]Use pH = -log [H+(aq)]
  • Weak acid dissociation expression for HA (aq) ⇌ H+(aq) + A-(aq)
  • Write an equation for the dissociation of ethanoic acid and its Ka expression. CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
  • What happens to the value of Ka as the strength of the acid increases? larger Ka
  • What happens to the value of pKa as the strength of the acid increases? smaller pKa
  • Equation for pKa pKa = -log Ka
  • Equation for Ka from pKa Ka = 10-pKa
  • What 2 assumptions are made to allow for this simplification? [H+(aq)]eqm = [A-(aq)]eqm because they have dissociated according to a 1:1 ratio initial concentration of the undissociated acid has remained constant (as amount of dissociation is small) So [HA(aq)]eqm = [HA(aq)]initial
  • Calculate the pH of a solution of 0.01 moldm-3 ethanoic acid. Ka = 1.7 x 10-5 moldm-3 CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)
  • Calculate the concentration of propanoic acid with a pH of 3.52 Ka = 1.35 x 10-5 moldm-3 CH3CH2COOH(aq) ⇌ H+(aq) + CH3CH2COO-(aq)
  • Given Ka and [HA(aq)], how can you work out the pH of the weak acid?
  • Given Ka and the pH of a weak acid, how can you calculate the concentration of the acid?
  • How to work out the pH of a solution with strong base and strong acid (neutralisation) Work out moles of acid added and hence moles of H+.Work out moles of base added and hence moles of OH-.Work out which one is in excessIf the acid is in excess use \ce{[H^+]=\frac{excess H^+ moles}{total volume (dm^3)}}If alkali is in excess use \ce{[OH^-]=\frac{excess OH^- moles}{total volume (dm^3)}} and [H+]=[H^+]=Kw[OH−]\frac{K_w}{[OH^-]}.pH=pH=−log[H+]-log[H^+]
  • If the acid is in excess in a strong acid and strong base neutralisation, how is the pH worked out? \ce{[H^+]=\frac{excess H^+ moles}{total volume (dm^3)}} pH=pH=−log[H+]-log[H^+]
  • If the base is in excess in a strong acid and strong alkali neutralisation, how is the pH worked out? \ce{[OH^-]=\frac{excess OH^- moles}{total volume (dm^3)}}[H+]=[H^+]=Kw[OH−]\frac{K_w}{[OH^-]}pH=pH=−log[H+]-log[H^+]
  • 15cm3 of 0.5 moldm-3 HCl is reacted with 35cm3 of 0.55 moldm-3 NaOH. Calculate the pH of the resulting mixture.
  • 35cm3 of 0.5 moldm-3 H2SO4 is reacted with 30cm3 of 0.55 moldm-3 NaOH. Calculate the pH of the resulting mixture.
  • 15cm3 of 0.5moldm-3 HCl is reacted with 35cm3 of 0.45 moldm-3 Ba(OH)2 . Calculate the pH of the resulting mixture.
  • How to work out the pH of a solution with strong base and weak acid (neutralisation) when acid is in excess Work out new concentration of excess HA:\ce{[HA]=\frac{initial moles HA - moles OH-}{total volume (dm^3)}} Work out the concentration of the salt formed:\ce{[A-]=\frac{moles OH- added}{total volume (dm^3)}} Use \ce{[H+]=\frac{K_a×[HA]}{[A-]}}pH=pH=−log[H+]-log[H^+]
  • How to work out the pH of a solution with strong base and weak acid (neutralisation) when base is in excess \ce{[OH^-]=\frac{excess OH^- moles}{total volume (dm^3)}}[H+]=[H^+]=Kw[OH−]\frac{K_w}{[OH^-]}pH=pH=−log[H+]-log[H^+]