KSEAB II PUC Chemistry study materials
KSEAB II PUC Chemistry study resources

KSEAB II PUC Chemistry: Equilibrium and Redox Reactions

KSEAB II PUC Chemistry · Work through your exam, one topic at a time.

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Equilibrium and Redox Reactions

Domain 5: Equilibrium and Redox Reactions

This domain tests dynamic equilibrium, acid–base calculations, solubility, and the electron-balance method for redox equations.

Core rules

  • At equilibrium, forward and reverse rates are equal and concentrations are constant, not equal. Kc uses equilibrium concentrations; Kp uses partial pressures; Kp=Kc(RT)^Δn_g for gases.
  • Le Chatelier: changing concentration, pressure, or temperature shifts the position; only temperature changes K. A catalyst changes neither position nor K.
  • Weak acid: Ka=[H+][A−]/[HA]; for a weak monoprotic acid, [H+]≈√(KaC) when dissociation is small; pH=−log[H+]. Buffers use Henderson–Hasselbalch: pH=pKa+log([A−]/[HA]).
  • Ksp is an equilibrium constant; common-ion addition lowers solubility. Precipitation occurs when the ion product exceeds Ksp.
  • Redox: oxidation number tracks electron ownership; balance atoms, then oxygen with H2O, hydrogen with H+, and charge with electrons. In basic medium, add OH− to neutralise H+.

Worked example

For 0.100 M acetic acid, Ka=1.8 x 10^−5. Let x=[H+]. x^2/(0.100−x)=1.8 x 10^−5; x≈√(1.8 x 10^−6)=1.34 x 10^−3 M. pH=−log(1.34 x 10^−3)=2.87. The approximation is valid because x is about 1.3% of C.

Traps

  • Treating strong and weak acids identically; HCl is fully ionised, acetic acid is not.
  • Forgetting that Kp is unitless in the thermodynamic convention but has units in some expressions.
  • Thinking pressure changes shift equilibria with equal gas moles.
  • Mis-assigning oxidation numbers in peroxides (O −1) and hydrides (H −1).
  • Balancing atoms but not charge in redox.