Acids and Bases 2 PPT PDF

MED-102 General Chemistry Acids and Bases 2  LOBs covered Calculate the equilibrium concentrations of all species in weak acid dissociation Calculate the pH of a weak acid solution Calculate the Ka of a weak acid Calculate the equilibrium concentrations of all species in...

MED-102 General Chemistry Acids and Bases 2  LOBs covered Calculate the equilibrium concentrations of all species in weak acid dissociation Calculate the pH of a weak acid solution Calculate the Ka of a weak acid Calculate the equilibrium concentrations of all species in weak base dissociation Calculate the pH of a weak base solution Calculate the pH of a buffer solution Calculate the new pH of a buffer solution on addition of extra acid or base Perform calculations using the Henderson-Hasselbach equation  Equilibria in Solutions of Weak Acids Consider a weak acid in water HA(aq) + H2O(l) H3O+(aq) + A-(aq) Acid-dissociation constant, Ka [H3O+ ][A − ] Ka = [HA] Weak Acid-Dissociation Constants  Weak Acid-Dissociation Constants – Revision Slide As the value of Ka increases, the strength of the acid increases as well. In this Table, the weak acid having the largest Ka value is hydrofluoric acid, Ka = 6.8  10-4. The weak acid that has the smallest Ka value is phenol, Ka = 1.3  10-10. This is a very weak acid.  Weak Acids – Equilibrium Concentrations We have 0.10 M solution of HCN (Ka = 4.9×10-10) Find equilibrium concentrations and pH Set up conventional equilibrium problem HCN(aq) + H2O(l) H3O+(aq) + CN-(aq) t=0 0.10 M 0 0 t= 0.10 – x x x  Weak Acids – Equilibrium Concentrations Set up Ka expression + − −10 [H3O ][CN ] ( x)( x) Ka = 4.9 10 = = [HCN] 0.10 − x Make approximation: [H3O+ ][CN− ] ( x)( x) x 2 Ka = 4.9 10−10 = =  [HCN] 0.10 − x 0.10 x = 7.0×10-6  pH Calculation Calculate pH + −6 pH = − log[H3O ]total = − log(7.0 10 ) = 5.15 WATCH: https://www.youtube.com/watch?v=_gl3uwA5BgU  Determining Ka values We can easily determine the acid-dissociation constant Ka by making a pH measurement at equilibrium Let us assume that we have a weak acid HA, and that we know the initial concentration of the acid and the solution pH e.g. HF HF(aq) + H2O(l) H3O+(aq) + F-(aq) t = 0 0.10 M 0 0 t =  0.10 – x M x x Equilibrium pH = 2.08 means that [H3O+] = x = 10-pH = 10-2.08 = 0.00832 M Ka = x2 / (0.10 – x) = (0.00832)2 / (0.10 – 0.00832) = 7.6 × 10-4 (6.8 × 10-4)  Equilibria in Solutions of Weak Bases Consider a weak base in water B(aq) + H2O(l) BH+(aq) + OH-(aq) Base-dissociation constant, Kb Weak Base-Dissociation Constants  Weak Base-Dissociation Constants – Revision Slide As the value of Kb increases, the strength of the base also increases. In the Table above, Ethylamine has the largest Kb constant, Kb = 5.6  10-4. This is a relatively stronger weak base. Urea, with Kb = 1.5  10-14, is the weakest of all the weak bases shown above.  Weak-Base Equilibrium 0.15 M NH3 NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) t=0 0.15 M 0 0 t= 0.15 – x x x Kb = [NH4+][OH-]/[NH3] = 1.8 × 10-5 = x2/(0.15 – x)  x2/0.15 x = [OH-] = 1.6 × 10-3 M and since [H3O+][OH-] = 1.0 × 10-14 [H3O+] = 6.25 × 10-12 and pH = 11.20  Relationship Between Ka and Kb Ka × Kb = Kw = 1.0 × 10-14 The Ka of a weak acid multiplied by the Kb of its conjugate base gives Kw. This allows us to calculate the Kb of the conjugate base if we know the Ka of the weak acid. Acid-Base Properties of Salts Strong acid + strong base = neutral salt HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Strong acid + weak base = acidic salt HCl(aq) + NH3(aq) → NH4Cl(aq) Weak acid + strong base = basic salt HF(aq) + NaOH(aq) → NaF(aq) + H2O(l) 5-Minute Break  Common Ion Effect Assume we have a solution of acetic acid, a weak acid that dissociates only partially CH3CO2 H(aq) + H2O(l ) CH3COO− (aq) + H3O+ (aq) What will happen if we add sodium acetate to this solution? (Think equilibrium theory) What happens to the pH? What happens to the degree of dissociation?  Common Ion Effect – Revision Slide Assume we have a solution of acetic acid, a weak acid that dissociates only partially CH3CO2 H(aq) + H2O(l ) CH3COO− (aq) + H3O+ (aq) What will happen if we add sodium acetate to this solution? (Think equilibrium theory) If we add sodium acetate, we are essentially adding acetate (CH3COO-) ions. We are adding a product to the equilibrium mixture. By Le Châtelier’s Principle, the equilibrium will shift to the left, to consume part of the added acetate. This will decrease the degree of dissociation of acetic acid. It will also decrease the concentration of H3O+ ions. This will cause the pH to increase slightly.  Buffer Solutions A buffer solution contains a weak acid and its conjugate base (e.g. acetic acid – sodium acetate) A buffer solution can also be formed from a weak base and its conjugate acid (e.g. NH3 – NH4Cl) CH3CO2H(aq) and CH3COO− (aq) Buffer solutions keep the pH relatively constant on addition of extra acid or base Buffers are important in biological systems Blood contains bicarbonate and phosphate buffers that keep the pH near 7.4. This is optimum for proper operation of the biochemical reactions that take place in blood (enzyme activity)  Acetic acid – acetate buffer 0.10 M acetic acid and 0.10 M acetate CH3CO 2H( aq) + H 2O(l ) CH 3COO − ( aq) + H3O+ (aq) Eq'm 0.10 − x 0.10 + x x [H3O + ][CH3COO − ] Ka = [CH3COOH] [CH3COOH] (0.10 − x) 0.10 [H3O + ] = Ka = Ka  Ka = Ka = 1.8 10−5 M [CH3COO− ] (0.10 + x) 0.10 pH = pK a = − log(1.8 10−5 ) = 4.74  Addition of base to a buffer The added OH- ions will neutralize some acetic acid, producing more acetate: [CH3COOH] = 0.10 mol − 0.01 mol = 0.09 mol [CH3COO− ] = 0.10 mol + 0.01 mol = 0.11 mol [CH3COOH] 0.09 [H3O+ ] = Ka = Ka = 1.5 10−5 M [CH3COO− ] 0.11 pH = − log(1.5 10−5 ) = 4.82  Addition of acid to a buffer The added acid will react with acetate ions, producing more acetic acid: [CH3COOH] = 0.10 mol + 0.01 mol = 0.11 mol [CH3COO− ] = 0.10 mol − 0.01 mol = 0.09 mol + [CH3COOH] 0.11 [H3O ] = Ka = Ka = 2.2 10−5 M [CH3COO− ] 0.09 pH = − log(2.2 10−5 ) = 4.66  Buffer Capacity This is defined as the maximum amount of acid or base that can be added to a buffer without causing a large pH change The buffer capacity increases with increasing amounts of acid and conjugate base – therefore, it is higher for more concentrated buffer solutions Increasing the volume of the buffer solution also leads to increased buffer capacity  Henderson – Hasselbalch Equation Hydronium ion concentration + [Acid] [H3O ] = K a [Base] -log of both sides  Uses of the Henderson-Hasselbalch equation It can be used to calculate the pH of a buffer provided we know the acid and conjugate base concentrations It can be used to calculate the amounts of acid and conjugate base required to make a buffer of particular pH  Summary for Revision We treat weak acid dissociation and weak base dissociation as equilibrium problems. The calculations for these species are entirely analogous to those for standard equilibrium problems. We can calculate the pH of a weak acid solution if we are given its initial concentration and Ka value. We can calculate the Ka value of a weak acid if we are given its initial concentration and the pH. Similar types of calculations can be made for weak bases. In these types of calculations, since the Ka/Kb values are relatively small, we can assume that the concentration of the weak acid/weak base do not change significantly after dissociation. This simplifies the calculations, by avoiding the use of a quadratic formula. Ka of an acid multiplied by the Kb of its conjugate base equals Kw. When we have a weak acid solution at equilibrium, and we add a salt that contains its conjugate base, then the degree of dissociation of the weak acid decreases, and the pH of the solution increases. This is known as the Common Ion effect. Buffer solutions resist large changes to the pH on addition of extra acid or base. A buffer solution can be formed from a weak acid and a salt containing its conjugate base. It can also be formed from a weak base and a salt containing its conjugate acid. A buffer has a finite capacity to resist large pH changes. If this capacity is exceeded, then the buffer solution fails. Calculations on buffer solutions can be made using the Henderson-Hasselbalch equation.