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Arrhenius theory

Svante August Arrhenius 1859 – 1927 Swedish. Arrhenius theory. HCl → H + ( aq .) + Cl – ( aq .). Acids generate H + in aqueous solution. Bases generate OH – in aqueous solution. NaOH → Na + ( aq .) + OH – ( aq .). NH 3 + H 2 O → NH 4 + ( aq .) + OH – ( aq .).

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Arrhenius theory

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  1. Svante August Arrhenius 1859 – 1927 Swedish Arrhenius theory HCl → H+(aq.) + Cl–(aq.) Acids generate H+ in aqueous solution Bases generate OH– in aqueous solution NaOH → Na+(aq.) + OH–(aq.) NH3 + H2O → NH4+(aq.) + OH–(aq.) Neutralization is the reaction of H+ with OH– H+(aq.) + OH–(aq.) → H2O Advantage: explains behavior in aqueous solution. — all strong acids and all strong bases have similar sets of reactions — all strong acid and strong base solutions conduct electricity Disadvantage: does not explain behavior of bases in non-aqueous solutions CHEM 114 — e.g. NH3 reacts with acids even if there is no water present.

  2. HCl → H+(aq.) + Cl–(aq.) Acids donate H+ in solution (aqueous or non-aqueous) Bases accept H+ in solution (aqueous or non-aqueous) NaOH + H+(aq.) → Na+(aq.) + H2O Brønsted-Lowry theory NH3 + H+(aq.) → NH4+(aq.) Acid-base equilibria: acids Example: acetic acid Ka = 1.8 × 10–5 AH is the acid; A– is the conjugate base CHEM 114

  3. H+ does not exist free in solution. In fact, it is bound to (usually more than one) waters. We sometimes simplify a complicated situation by writing it as a hydronium ion, H3O+ The hydronium ion hydronium H+(aq.) + H2O → H3O+(aq.) The full reaction can be written as CH3COOH + H2O → CH3COO– + H3O+ The water is acting as a proton acceptor — a base. H3O+ is a conjugate acid. CHEM 114

  4. B + H2O → BH+(aq.) + OH–(aq.) Brønsted-Lowry theory: bases B is the base; BH+ is the conjugate acid Kb = 1.8 × 10–5 Example: ammonia Strong acids are mostly dissociated, so Kais large Strong bases are mostly dissociated, so Kbis large A strong acid has a weak conjugate base A strong base has a weak conjugate acid CHEM 114

  5. Brønsted-Lowry theory: acids CHEM 114

  6. Water can undergo an acid-base reaction with itself 2 H2O → H3O+(aq.) + OH–(aq.) At 25°C, Kw = 1.0 × 10–14 Let’s do an ICE table The dissociation of water Kw = [H3O+] [OH–] = x2= 1.0 × 10–14 ⇒x= [H3O+] = [OH–] = 1.0 × 10–7 This constraint is always present in aqueous solution; [H3O+] and [OH–] are inextricably linked and their product must be 1.0 × 10–14 For example; if [H3O+] = 1 mM = 1.0 × 10–3 M, then [OH–] = 1.0 × 10–11 M We can combine Kw with the Kb of a base to get Ka of the conjugate acid. Reaction 1: NH3 + H2O → NH4+(aq.) + OH–(aq.) Kb = 1.8 × 10–5 Reaction 2: H3O+(aq.) + OH–(aq.) → 2 H2O K = 1/Kw = 1.0 × 1014 CHEM 114 NH3 + H3O+(aq.) → NH4+(aq.) + H2O K′ = KKb = 1/Ka KaKb = Kw Ka = 1/ K′ = 1/(KKb) = 5.6 × 10–10

  7. Scientific notation can be cumbersome, so we create a log scale for hydrogen ion concentration pH = –log10(aH+) ~ –log10[H+] pH example: if [H+] = 10–3 M, then = log10(10–3) = –3, pH = 3 pOH = –log10(aOH–) ~ –log10[OH–] Since Kw = [H3O+] × [OH–] = 1.0 × 10–14 ⇒pOH + pH = 14 CHEM 114

  8. pKa = –log10(Ka) We obtain pK values just like pH values pKb = –log10(Kb) Since KaKb = Kw, pKa + pKb = pKw ~ 14 @ 25°C pKa and pKb pKb = –log10(1.8 × 10–5) = 4.74 Example: ammonia: Kb = 1.8 × 10–5 pKa for ammonium ion = 14 – pKb = 9.26 CHEM 114

  9. Strong acids (Ka > 1): HCl, HBr, HI, H2SO4, HNO3, HClO4 Brønsted-Lowry theory: acids CHEM 114

  10. Question: what is the pH of a 0.05 M solution of acetic acid in water? Ka = 1.8 × 10–5 CH3COOH ⇌ CH3COO– + H+ pH calculations Note that we neglect the H+ ions in pure water (10–7 M), because they are insignificant x2 = 1.8 × 10–5 × (0.05 – x) K = x2/(0.05 – x) = 1.8 × 10–5 x2 + (1.8 × 10–5)x – (9.0 ×10–7) = 0 x = 0.00211 or –0.00213 Negative concentrations are impossible so x = 0.00211 [H+] = x = 0.00211, so pH = –log10(0.00211) = 2.68 CHEM 114

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