CM4106 Chemical Equilibria & ThermodynamicsLesson 2Acid-Base EquilibriaA Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/
Fundamentals:1. Identify acids/ bases Acid Base 1. proton H+ donor 1. proton H+ acceptor 2. electron acceptor 2. electron donor (vacant orbital) (lone pair)2. Identify conjugate acids/ bases – H+ Acid Conjugate Base + H+ + H+ Base Conjugate Acid – H+
Fundamentals:3. Identify stronger acids/ bases (A) In any acid-base reaction, the equilibrium will favor the reaction where the stronger acid reacts with the stronger base. i.e. (In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.) (B) The stronger an acid, the weaker its conjugate base. The stronger a base, the weaker its conjugate acid. HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq) CH3CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + CH3CO2- (aq)
Fundamentals:Important relations p = -log10 At 25ºC: pH + pOH = pKw = 14 pKa + pKb = 14 [H+][OH–] = 10-14 Ka x Kb = Kw = 10-14
(I) CalculationspH, pKa, [H+], Ka pOH, pKb, [OH–], Kb Step 1: Determine what is present in the solution. (A) Acid : Strong Acid vs Weak Acid Monoprotic Acid / Diprotic Acid / Triprotic Acid Concentration of Acid (B) Base: Strong Base vs Weak Base Monoprotic base / Diprotic base / Triprotic base Concentration of Bas Take note of concentration of acid / bases (For dilute solutions, we need to take into consideration of [H+] / [OH-] from auto-ionization from water)Step 2: Use the appropriate equations for the respective species.
pH of Acid/ Base Ka = x2 Strong Acid Weak Acid ([HA] – x)Strong acids dissociate HA(aq) ⇌ H+(aq) + A-(aq)completely into ions inaqueous solution. I [HA] 0 0 C -x +x +x[H+] = [HA] E [HA] - x +x +x Kb = x2 Strong Base Weak Base ([B] – x)Strong bases dissociate B + H2O ⇌ BH+ + OH-completely into ions in I [B] - 0 0aqueous solution. C -x - +x +x[OH-] = [B] E [B] - x - x x
pH of Acid/ Base1. Determine if acid/ base is strong or weak (more common)2. For weak acids, if asked to determine Ka, pH or [H+], you can save time by using the formula: Assumption: x is negligible [H+] = Ka × c [OH–] = Kb × cExample:Calculate the pH of 0.50M HF solution at 25ºC where the Ka = 7.1 x 10-4 [H+][F-] x2 Ka = = = 7.1 x 10-4 [HF] (0.50 – x) Assumption: For weak acids, x must be very small 0.50 – x ≈ 0.5 x2 = 7.1 x 10-4 x = [H+] = 0.0188 M Assumption is valid; 0.50 x < 5% of [HF]initial pH = 1.73 (to 2 d.p.)
Calculate the pH of the following solutions at 298K0.10 mol dm−3 CH3COOH (pKa = 4.75) Weak acid solution (monoprotic acid) CH3COOH(aq) + H2O (l) ⇌ CH3COO(aq) + H3O+(aq) Initial (M) 0.10 - 0.00 0.00 Change (M) -x - +x +x Eqm (M) 0.10 - x - +x +x [CH3COO][H3O+] x2 Ka = = = 10 4.75 = 1.79 x 10-5 [CH3COOH] (0.10 – x) Assumption: For weak acids, x must be very small 0.10 – x ≈ 0.10 x2 = 1.79 x 10-5 Assumption is justified, 0.10 x < 5% of [CH3COOH]initial x = [H3O+] = 1.338 x 10-3 M Concentration: 2 s.f. pH = 2.87 ( 2 d.p.) pH: 2 d.p.
Calculate the pH of the following solutions at 298K0.30 mol dm3 ethylamine, CH3CH2NH2 (pKb= 3.27) Weak base solution (monoprotic base) CH3CH2NH2 (aq) + H2O (l) ⇌ CH3CH2NH3+(aq) + OH-(aq) Initial (M) 0.30 - 0.00 0.00 Change (M) -x - +x +x Eqm (M) 0.30 - x - +x +x [CH3CH2NH3+][OH-] x2 Kb = = = 10 3.27 = 5.37 x 104 [CH3CH2NH2] (0.30 – x) Assumption: For weak bases, x must be very small 0.30 – x ≈ 0.30 x2 = 5.37 x 10-4 Assumption is justified, 0.30 x < 5% of [CH3CH2NH2]initial x = [OH] = 0.01269 M pH = 14.00 – 1.896 pOH = - log10 [OH] pH = 12.10 (2 d.p.) = - log10[0.01269 ] Concentration: 2 s.f. = 1.896 pH: 2 d.p.
pH of salt solutionsSalt solutions can be (i) neutral (ii) weak acids or (iii) weak bases Basic SaltCH3COONa (aq) → CH3COO- (aq) + Na+ (aq)CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)conjugate base pH > 7 Acidic Salt NH4Cl (aq) → NH4+ (aq) + Cl- (aq) NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)conjugate acid pH < 7
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