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Practical Applications of the Knowledge of the temperature dependence of the equilibrium constant

Practical Applications of the Knowledge of the temperature dependence of the equilibrium constant. (i) M(s) + 1/2O 2 (g) → MO(s) (ii) 1/2C(s) + 1/2O 2 (g) → 1/2CO 2 (g) (iii) C(s) + 1/2O 2 (g) → CO(g) (iv) CO(g) + 1/2O 2 (g) → CO 2 (g). 9.5 The response of equilibria to pH.

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Practical Applications of the Knowledge of the temperature dependence of the equilibrium constant

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  1. Practical Applications of the Knowledge of the temperature dependence of the equilibrium constant (i) M(s) + 1/2O2(g) → MO(s) (ii) 1/2C(s) + 1/2O2(g) → 1/2CO2(g) (iii) C(s) + 1/2O2(g) → CO(g) (iv) CO(g) + 1/2O2(g) → CO2(g)

  2. 9.5 The response of equilibria to pH • Definition of pH: An indication of hydronium ion concentration, pH = - logα(H3O+), where, H3O+ is hydronium, a representation of the state of proton in aqueous solution • Acid-base equilibria in water: acidity constant Ka, which is defined as HA(aq) + H2O (l) ↔ H3O+ (aq) + A-(aq) (9.30) Quite often, we report Ka with pKa, which is defined as –log(Ka).

  3. Basicity constant Kb B(aq) + H2O (l) ↔ HB+ (aq) + OH-(aq) (9.32) One can also use pKb to represent Kb, which is pKb = -logKb • Autoprotolysis constant Kw 2H2O(l) ↔ H3O+ (aq) + OH- (aq) Kw = α(H3O+)*α(OH-) ( 9.33) pKw = -log(Kw) = -log(α(H3O+)*α(OH-)) = -logα(H3O+) - log α(OH-) similar to pH = -logα(H3O+), one may introduce pOH = - log α(OH-) then pKw = pH + pOH (9.34)

  4. The Thermodynamic View of Strong Acids and Bases

  5. The pH of weak acids • HA(aq) + H2O (l) ↔ H3O+ (aq) + A-(aq) • Because [H3O+] = [A-] • [H3O+] ≈ (Ka[HA])1/2 pH = - log([H3O+] ) = - ½ log(Ka[HA]) = -1/2 log(Ka) - ½ log([HA]) recall log(Ka) = - pKa pH = ½ pKa - ½ log([HA]) (9.36)

  6. Example: Calculate the pH of 0.100M CH3COOH (aq) • Solution: CH3COOH is a weak acid, i.e. only small portion of it will dissociate. therefore assumptions for the equation 9.36 hold pKa can be found from table 9.1 Discussion: how about a strong acid such as nitric acid?

  7. The pH of weak bases • B(aq) + H2O (l) ↔ HB+ (aq) + OH-(aq) since [HB+] = [OH-] [OH-] = (Kb[B])1/2 thus pOH = -log([OH-] ) = - ½ log(Kb) - ½ log([B]) using eqn 9.34, pKw = pH + pOH pH = pKw – pOH = pKw + ½ log(Kb) + ½ log([B]) pH = pKw - ½ pKb + ½ log([B]) (9.37)

  8. 9.5 Acid-base titrations • Case 1. Using strong base to titrate weak acid. First step: examine the source of H3O+: Source 1: HA(aq) + H2O (l) ↔ H3O+ (aq) + A-(aq) Source 2: 2H2O(l) ↔ H3O+ (aq) + OH- (aq) • Approximation 1: the amount of A- from source is very small comparing with the amount of HA (Because it is a weak acid). • Approximation 2: The amount of hydronium ion coming from source 2 is negligible, though HA is a weak acid. • Facts after adding VB amount of strong base 1. The total volume of the solution becomes V = VA + VB 2. Salt (A-) in the solution is formed due to the acid and base reaction. The contribution from source 1 is negligible. We represent it with [A-] = S. HA (aq) + OH-(aq) ↔ A-(aq) + H2O(l) 3. The concentration of HA (A’) left in the solution equals (A0VA – SV)/V

  9. Recall • So [H3O+] = Ka A’/S pH = - log([H3O+]) = - log(Ka) - log(A’/S) pH = pKa - log(A’/S) = pKa - log([Acid]/[Salt]) (9.41) • Eqn 9.41 indicates that when concentrations of the acid and salt are equal, pH = pKa. Hence the pKa of a substance can be measured in the lab. • Eqn 9.41 is also known as the Henderson-Hasselbalch equation

  10. pH at three special situations 1. At the beginning of the titration: calculated with eqn 9.36 pH = (1/2)pKa – (1/2)log[A0] 2. At the stoichiometric point: OH- ions are from the following process: A-(aq) + H2O(l) ↔ HA(aq) + OH-(aq) so: [OH-] = (Kb*S)1/2 [H3O+] = Kw /(Kb*S)1/2 Because Kb = Kw/Ka [H3O+] = Kw /(S*Kw/Ka)1/2 = (Kw*Ka)1/2/S1/2 pH = -log [H3O+] = ½ pKw + ½ pKa + ½ logS (9.44)

  11. 3. Beyond stoichiometric point: here the major source of OH- is the strong base, i.e. [OH-] = [Base] since [H3O+] = Kw/[OH-] [H3O+] = Kw/[Base] pH = -log [H3O+] = -logKw + log[Base] pH = pKw + log[Base] Example: Estimate the pH at the stoichiometric point of a titration of 25.00mL of 0.200M CH3COOH(aq) with 0.300M NaOH(aq). Solution: Step 1: write down the reaction CH3COOH(aq) + NaOH(aq) ↔ CH3COONa + H2O Step 2: how many mL of 0.300M NaOH(aq) is required to reach the stoichometric point? Step 3: calculate the concentration of salt. Step 4: plug into eqn 9.44.

  12. A summary of the pH curve during the titration of a weak acid with strong bases

  13. The “exact” form of the pH curve Following five conditions are used to derive the “exact” form of pH during a titration • Electrical neutrality: [H3O+] + [M+] = [A-] + [OH-] • Conservation of A groups: [A-] + [HA] = A0VA/(VA + VB) • Concentration of M+ group: [M+] = B0VB/ (VA + VB) • Deprotonation of HA: [HA] Ka = [H3O+] [A-] • Autoprotolysis: Kw = [H3O+] [OH-] (derivation will be discussed in class)

  14. pH buffers

  15. When the molar concentrations of salt and the weak acid are equal, the variation of pH with respect to the addition of strong acids or bases is slow. This is the basis of buffer action. It is because: • The abundant supply of salt ions can remove any H3O+ ions brought by the addition of strong acid. • The numerous weak acid molecules (HA) can supply H3O+ ions to react with any strong base added to the system. Example: calculate the pH of an aqueous buffer solution which contains 0.100mol L-1 NH3 and 0.200 mol L-1 NH4Cl. Solution: First:Identify which reagent is the weak acid and which one corresponds to the conjugate salt. NH4+(aq) + H2O  NH3(aq) + H3O+(aq) then pH = pKa - log([NH4+]/[NH3]) = 9.25 – log2 = 8.949 = 8.95

  16. How does a titration indicator work? • The indicator must have two distinct colors in the acidic and the conjugate base form. • HIn(aq) + H2O(l)  In-(aq) + H3O+(aq) Kin = • Log = pKIn - pH • In a weak acid-strong base titration, pKIn > 7 is required because the stoichiometric point lies at pH > 7 • In a strong acid-weak base titration, pKIn < 7 is required because the stoichiometric point lies at pH < 7 • End point:

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