Doc Brown's Chemistry Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5.3 5.3 Definition of a strong acid, theory, examples and pH calculations of strong acids What is a strong acid? How to calculate the pH of a strong acid solution given its concentration. GCSE/IGCSE reversible reactions–equilibrium notes * GCSE/IGCSE notes on acids and bases Equilibria Part 5 sub–index: 5.1 Lewis and Bronsted–Lowry acid–base theories * 5.2 self–ionisation of water and pH scale * 5.3 strong acids–examples–calculations * 5.4 weak acids–examples & pH–K_{a}–pK_{a} calculations * 5.5 strong bases–examples–pH calculations * 5.6 weak bases– examples & pH–K_{b}–pK_{b} calculations * 5.7 A level notes on Acids, Bases, Salts, uses of acid–base titrations – upgrade from GCSE! Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. K_{c} and K_{p} equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces
5.3 Definition, examples and pH calculations of strong acids Note: H^{+}(aq) = aqueous hydrogen ion = aqueous proton = oxonium ion = hydroxonium ion
Appendix 1. What is the real aqueous hydrogen ion concentration in dilute sulfuric acid? e.g. take 0.500 molar H_{2}SO_{4} (aq), if fully ionised, you would expect ...
BUT, what is the reality? The 1st dissociation is complete: H_{2}SO_{4(aq)} ==> H^{+}_{(aq)} + HSO_{4}^{–}_{(aq)}
The 2nd ionisation HSO_{4}^{–}_{(aq)} H^{+}_{(aq)} + SO_{4}^{2–}_{(aq)} is not complete,
Prior to the 2nd ionisation, theoretically, the initial concentrations of hydrogensulfate ions and hydrogen ions will be equal, and both 0.500 mol dm^{–3}. On the 2nd ionisation, the hydrogensulfate ion will provide the extra hydrogen ions and the only sulfate ions, but in doing so, the hydrogensulfate ion is reduced. If we call the 'equal' extra hydrogen ion concentration and the final sulfate ion concentration x, then the total hydrogen ion concentration is (0.5 + x) and the hydrogensulfate ion concentration is reduced to (0.5 – x)
This gives the quadratic equation 0 = x^{2} + 0.512x – 0.006 On solving this using the quadratic equation formula, gives roots of –0.523 and +0.0114 Therefore x must be 0.0114 (the 'extra' H^{+}), which then gives (0.5 + 0.0114) ... a total hydrogen ion concentration [H^{+}(aq)] of 0.511 mol dm^{–3}, not 1.000 mol dm^{–3} and the real calculated pH = –log10(0.511) = 0.29, not pH 0.00 and significantly higher You may think the extra hydrogen ion concentration is very low, but this is because the HSO_{4}^{–} ion is a weak acid AND the 2nd ionisation is actually heavily suppressed by the 1st ionisation – think Le Chatelier's equilibrium principle as regards the concentration effect. To fully understand this calculation its handy to have studied the weak acid calculations page Equilibria Part 5 sub–index: 5.1 Lewis and Bronsted–Lowry acid–base theories * 5.2 self–ionisation of water and pH scale * 5.3 strong acids–examples–calculations * 5.4 weak acids–examples & pH–K_{a}–pK_{a} calculations * 5.5 strong bases–examples–pH calculations * 5.6 weak bases– examples & pH–K_{b}–pK_{b} calculations * 5.7 A level notes on Acids, Bases, Salts, uses of acid–base titrations – upgrade from GCSE! |
Enter specific words e.g. topic, formula, compound, reaction, anything! |