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Acid- base Theory. Applied to aqueous solutions only. Acids : Substances that produce hydrogen ions, H + (aq) when dissolved in water. . 1. Arrhenius (1880s) : -. E.g. HCl. Bases : Substances that produce hydroxide ions, OH (aq) when dissolved in water. . E.g. NaOH.
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Acid-baseTheory Applied to aqueous solutions only Acids : Substances that produce hydrogen ions, H+(aq) when dissolved in water. 1. Arrhenius (1880s) : - E.g. HCl Bases : Substances that produce hydroxide ions, OH(aq) when dissolved in water. E.g. NaOH
2. Brnsted / Lowry (1923) : - Applied to aqueous and non-aqueous solutions Acids : Proton donors E.g. Molecule – HCl, H2O, CH3COOH Cation – H3O+, NH4+ Anion – HSO4-
Bases: Proton acceptors e.g. Molecule - H2O, NH3 Anion - Cl, OH, CH3COO, SO42 Each contains at least one lone pair to form a dative bond with a proton (H+)
3. Lewis (1930s) : - More widely applied to systems with and without solvents Acids : electron pair acceptors, e.g. H+, BF3, AlCl3, BeCl2 (electron-deficient species)
H+(aq) + OH(aq) BF3 + NH3 Lewis acids Lewis bases Bases: electron pair donors e.g. NH3, OH species containing lone pair(s)
HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) Loss of H+ A conjugate acid-base pair is a pair of species that can be inter-converted by transfer of proton. acid Conjugate base
Gain of H+ HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) A conjugate acid-base pair is a pair of species that can be inter-converted by transfer of proton. Conjugate acid base
Gain of H+ HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) A conjugate acid-base pair is a pair of species that can be inter-converted by transfer of proton. base Conjugate acid
Loss of H+ HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) A conjugate acid-base pair is a pair of species that can be inter-converted by transfer of proton. Conjugate base Acid An acid-base equilibrium system consists of TWO inter-convertible conjugate acid-base pairs.
Acid Base Conjugate base Conjugate acid HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) H2O(l) + NH3(aq) OH(aq) + NH4+(aq) A species can behave as an acid or a base depending on the situation.
H2SO4(aq) NH4+(aq) H2O(l) HSO4(aq) SO42(aq) NH2(aq) NH3(aq) OH(aq) O2(aq) 1. Give the conjugate acid and conjugate base of each of the following species
HCl(aq) + H2O(l)Cl(aq) +H3O+(aq) H2O(l) + NH3(aq)OH(aq) + NH4+(aq) CH3COOH(aq) + H2O(l)CH3COO(aq) + H3O+(aq) H2O(l) + H2O(l)OH(aq) + H3O+(aq) HSO4(aq) + H2O(l)SO42(aq) + H3O+(aq) Concept of Conjugate Acid-base Pairs Acid Base Conjugate base Conjugate acid
stronger + stronger weaker + weaker acid base conjugate conjugate base acid Relative Strengths of Conjugate Acid-base Pairs Stronger acid/base produces weaker conjugate base/acid more completely. The forward reaction is more complete. The equilibrium position lies to the right.
weaker + weaker stronger + stronger acid base conjugate conjugate base acid Relative Strengths of Conjugate Acid-base Pairs Weaker acid/base produces stronger conjugate base/acid less completely. The forward reaction is less complete. The equilibrium position lies to the left.
Relative strength Relative strength Acid Conjugate base
HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) stronger acid stronger base weaker base weaker acid For a strong acid The equilibrium position lies far to the right. Strong acids at the top left of Table 1 almost ionize completely to give H3O+(aq) ,
HCl(aq) + H2O(l) Cl(aq) + H3O+(aq) stronger acid stronger base weaker base weaker acid For a strong acid ∴ the strongest acid present in aqueous solutions is hydronium ions, H3O+(aq)
CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+(aq) weaker acid weaker base stronger base stronger acid For a weak acid The equilibrium position lies far to the left.
CH3(aq) + H2O(l) CH4(aq) + OH-(aq) Stronger base stronger acid weaker acid weaker base For a strong base The equilibrium position lies far to the right. Strong bases at the bottom right of Table 1 react almost completely with water to give OH(aq),
CH3(aq) + H2O(l) CH4(aq) + H3O+(aq) weaker acid weaker base stronger base stronger acid For a strong base ∴ the strongest base present in aqueous solutions is hydroxide ions, OH(aq).
HSO4(aq) + H2O(l) H2SO4(aq) + OH(aq) weaker base weaker acid stronger acid stronger base For a weak base The equilibrium position lies far to the left.
H Lone pair + N H Empty 1s orbital H H + H Ammonium ion N H H H Formation of ammonium ion
pH and Its Calculation Definition : pH is the negative power of the molarity of Hydronium ion
H2O(l) + H2O(l) H3O+(aq) + OH(aq) Self-ionization of Water ∵ [H2O(l)] 55.5 M It is in large excess treated as a constant Kc[H2O(l)]2 = Kw = [H3O+(aq)][OH(aq)]
ionic product of water Temperature dependent Kw = [H3O+(aq)][OH(aq)] At 298 K
In neutral solution, [H3O+(aq)] = [OH(aq)] ∴ pH = log10(1.00107) = 7.00
In acidic solution, [H3O+(aq)] > [OH(aq)] [H3O+(aq)] > 1.00107 mol dm3 pH < 7.00
In alkaline solution, [H3O+(aq)] < [OH(aq)] [H3O+(aq)] < 1.00107 mol dm3 pH > 7.00
Temp/oC 18.0 25.0 40.0 75.0 Kw / mol2 dm6 0.610 1014 1.00 1014 2.92 1014 16.9 1014 Q.2 Given the following Kw values at different temperatures,
Temp/oC 18.0 25.0 40.0 75.0 Kw / mol2 dm6 0.610 1014 1.00 1014 2.92 1014 16.9 1014 (a) An increase in T increases the value of Kw the equilibrium position shifts to the right the system absorbs heat by shifting to the right the forward reaction is endothermic
2.(b) In neutral solution, [H3O+(aq)] = [OH(aq)] pH = log10(1.71107) = 6.77
Example 1 Calculate the pH of (a) 0.200 M HCl and (b) 0.200 M KOH at 298 K Given : Kw at 298 K = 1.001014 mol2 dm6 • Assumptions : • (i) HCl ionizes completely when dissolved in water. • (ii) [H3O+(aq)] from H2O(l) is negligible. ∵ [HCl(aq)] = 0.200 M ∴ [H3O+(aq)] = 0.200 M assumption (i) pH = log10[H3O+(aq)] = log100.200 assumption (ii) = 0.699
assumption (ii) (b) Assumptions : (i) KOH dissociates completely when dissolved in water. (ii) [OH(aq)] from H2O(l) is negligible. ∵ [KOH(aq)] = 0.200 M ∴ [OH(aq)] = 0.200 M assumption (i) At 298 K pH = log10(5.001014) = 13.3
= 1.001014 mol2 dm6 Alternatively, log10[H3O+(aq)][OH(aq)] = 14.0 log10[H3O+(aq)] + (log10[OH(aq)]) = 14.0 pH + pOH = 14.0 (at 298 K) pH = 14.0 – pOH = 14.0 – (log100.200) = 14.0 – (0.699) = 13.3
Q.3 Calculate the pH values of the following solutions at 298 K. Given : Kw at 298 K = 1.001014 mol2 dm6 (a) Assumptions : (i) HCl ionizes completely when dissolved in water. (ii) [H3O+(aq)] from H2O(l) is negligible. [H3O+(aq)] = [HCl(aq)] = 2.00 M pH = log102.00 = 0.301
(b) Assumptions : (i) NaOH dissociates completely when dissolved in H2O. (ii) [OH(aq)] from H2O(l) is negligible. [OH(aq)] = [NaOH(aq)] = 2.00 M pH = log10(5.0010-15) = 14.3 pH values can be > 14 or < 0
2H2O(l) OH(aq) + H3O+(aq) (c) [H3O+(aq)] from H2O should be considered 1.0010-7 1.0010-7 1.0010-7 1.0010-7 + 1.0010-9 (1.0010-7 – x) (1.0010-7 + 1.0010-9 – x) Kw = (1.0010-7 – x)(1.0010-7 + 1.0010-9 – x)= 1.0010-14 x = 5.0010-10 M or 2.0010-7 M(rejected) [H3O+(aq)] = (1.0110-7 5.0010-10) = 1.005107 M pH = 6.997 < 7 (acidic)
At very low acid / base concentrations (<106 M), the [H3O+(aq)] / [OH(aq)] from H2O(l) cannot be ignored. At very high acid / base concentrations (>6 M), the ionization of strong acids / bases is no longer complete.
[H3O+] = 1106 M Variation of [H3O+(aq)] with [HCl(aq)]
[OH] = 1106 M Variation of [OH(aq)] with [NaOH(aq)]
Measurement of pH TWO ways : - Universal indicator / pH paper pH meter
Universal indicator / pH paper Universal indicator : - a mixture of dyeswhich shows different colours at different pH values. pH paper : - paper coated with universal indicator
Procedures of using universal indicator Apply a few drops of universal indicator to the sample. Apply a few drops of sample to a piece of pH paper. This method gives only a rough estimation of the pH of the sample.
pH meter Probe : A glass electrode coupled with a reference electrode Glass electrode : The electrode potential varies linearly with the pH of the solution in which it is immersed. At 298 K, The electrode potential is given by E1 = C – 0.0592pH, where C is a constant Reference electrode : An electrode with a fixed electrode potential, E2 E.g. Ag /AgCl electrode