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Acid-Base Equilibria and Solubility Products

Acid-Base Equilibria and Solubility Products. Buffer Solutions. A buffer is a solution that contains both A weak acid or a weak base A salt containing its conjugate base or acid

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Acid-Base Equilibria and Solubility Products

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  1. Acid-Base Equilibria and Solubility Products

  2. Buffer Solutions • A buffer is a solution that contains both • A weak acid or a weak base • A salt containing its conjugate base or acid • A buffer solution has the ability to maintain almost constant pH, despite the addition of small amounts of acid or base • In a living organism, buffers are critical • ex.; blood pH remains more or less constant due to a buffer consisting of the H2CO3/HCO3- conjugate acid-base pair

  3. Buffer Solutions • In a buffer solution, the acid and base must not neutralize each other • therefore we use a conjugate acid-base pair • a transfer of a proton between the two produces no net change • Imagine a buffer solution produced by adding CH3COOH and CH3COONa to pure water • In pure water, hydrolysis occurs to a very small extent, i.e., the reactants mainly remain intact • In a buffer solution, these reactions become even less important because of le Chatelier’s Principle: • The CH3COO-(aq) suppresses the hydrolysis of CH3COOH(aq) • The CH3COOH(aq) suppresses the hydrolysis of CH3COO-(aq)

  4. Buffer Solutions • A buffer solution like our CH3COO-/CH3COOH pair is capable of keeping the pH more or less constant as, upon addition of H+(aq) or OH-(aq),the following reactions are produced • i.e., the CH3COO- neutralizes the H+(aq) or the CH3COOH neutralizes the OH-(aq), and the pH of the buffer solution does not change appreciably when adding acids or bases • The buffer capacity is the ability of the buffer solution to neutralize both acids and bases

  5. Buffer Solutions • Example: Which of the following solutions are buffer systems? (a) KF/HF, (b) KCl/HCl, (c) Na2CO3/NaHCO3 • Solution: • (a) HF is a weak acid, and F- is its conjugate base, therefore this is a buffer solution. • (b) HCl is a strong acid, therefore its conjugate base, Cl-, can not neutralize an acid. This is not a buffer system. • (c) CO32- is a weak base, and HCO3- is its conjugate acid, therefore this is a buffer system.

  6. Buffer Solutions • Example: Calculate the pH of the following buffer system: NH3 (0.30M) / NH4Cl (0.36 M). What happens to the pH if 20.0 mL of 0.050 M NaOH is poured into 80.0 ml of the buffer solution? • Solution:

  7. Buffer Solutions • Solution: If we poured 20.0 mL of 0.050 M NaOH in 80.0 ml of the buffer solution: • We add (0.020 L)(0.050 mol/L) = 0.0010 mol of OH- • We have (0.080 L)(0.30 mol/L) = 0.0240 mol of NH3 • We have (0.080 L)(0.36 mol/L) = 0.0288 mol of NH4+ • The OH- will consume NH4+ and produce NH3

  8. Buffer Solutions • N.B. If in the previous example we added the same quantity of OH- to 80.0 mL pure water: • [OH-] = (0.0010 mol)/(0.100 L) = 0.01 M • If [OH-] = 0.01 M, [H+] = 1.0 x 10-12 M, therefore the pH = 12.00 • In pure water, it goes from pH = 7.00 to pH = 12.00 • In the buffer solution, it goes from pH = 9.17 to pH = 9.21 • A buffer solution is very effective at maintaining a constant pH

  9. The Henderson-Hasselbach Equation • For a weak acid, HA, • Defining the pKa as -log Ka of a weak acid, we obtain the Henderson-Hasselbach equation

  10. The Henderson-Hasselbach Equation • N.B. A buffer solution is especially effective when [HA]  [A-] or when • Thus a buffer system is most effective when pH  pKa

  11. Calculate the pH of a 1.00 L buffer solution which is 0.87 M CH3COOH and 0.47 M NaCH3COO. 0.10 mol of HCl are added to this solution. Calculate the new pH. The ionization constant for CH3COOH is 1.8 x 10-5.

  12. Calculate the pH of a 1.00 L solution of 0.537 M CH3COOH. 1.00 L of 0.197 M NaOH is added to the solution. Calculate the new pH. The ionization constant for CH3COOH is 1.8 x 10-5.

  13. Strong Acid-Strong Base Titration • ex.; • The equivalence point is the point where equimolar quantities of acid and base have reacted (in this case, pH= 7.00) or

  14. Weak Acid-Strong Base Titration or • ex.; • At the equivalence point, the OH- has neutralized all of the CH3COOH • All of the CH3COOH is converted to CH3COO- • Because CH3COO-(aq) is a weak base, the equivalence point is at a pH greater than 7

  15. Strong Acid-Weak Base Titration or • ex.; • At the equivalence point, the H+ has neutralized all of the NH3 • All of the NH3 is converted into NH4+ • Because NH4+ is a weak acid, the equivalence point is at a pH lower than 7

  16. Acid-base Indicators • The equivalence point of an acid-base titration is often indicated by the colour change of a coloured indicator • An indicator is usually a weak organic acid or base, where the acidic and basic forms have different colours • In an acidic medium, the solution takes the colour of the indicator’s acidic form • In a basic medium, the solution takes the colour of the indicator’s basic form • The change-over zone (pH zone where the colour chages) corresponds to the pKa of the indicator • For a given titration, we want to choose an indicator where the changeover zone corresponds to the pH of the equivalence point

  17. Solubility Equilibrium • Consider a salt that is insoluble in water, for example, BaSO4(s) • A small quantity dissolves in water, • The equilibrium constant for this reaction is • We give this equilibrium constant a special name: the solubility product, Ks

  18. The Solubility Product • The solubility product of a compound is the product of the molar concentrations of the ions which constitute it, each of these concentrations being raised to the exponent equal to its stoichiometric coefficient in the balanced equation • N.B. The smaller the value of Ks, the less soluble the compound is in water

  19. Molar Solubility and Solubility • Ks is a measure of solubility • It is often difficult to compare the solubility of two compounds using the Ks valuese as the expression for Ks is different if the stoichiometry of the dissociation is different (i.e., a different number of cations/anions are produced) • There are two other ways to express the solubility: • The molar solubility is the number of moles of solute per liter of a saturated solution • The solubility is the grams of solute per liter of a saturated solution • N.B. The solubility of a compound depends on the temperature

  20. Molar Solubility and Solubility

  21. Molar Solubility and Solubility • Example: The molar solubility of barium fluoride (BaF2) is 7.5 x 10-3 M. What is the solubility product of this compound? • Solution: If the concentration of BaF2 is 7.5 x 10-3 M and BaF2 dissociates in water:

  22. Molar Solubility and Solubility • Example: Calculate the molar solubility of lead carbonate (PbCO3) using its solubility product (Ks = 3.3 x 10-14). • Solution: Let x be the molar solubility of PbCO3, At equilibrium, the concentrations of Pb2+(aq) and CO32-(aq) are each x, therefore • The molar solubility of PbCO3 is therefore 1.8 x 10-7 M.

  23. Molar Solubility and Solubility • Example: Calculate the solubility of silver chloride (AgCl) in g/L using its solubility product (Ks = 1.6 x 10-10). • Solution: Let x be the molar solubility of AgCl, AgCl  Ag+(aq) + Cl-(aq) At equilibrium, the concentrations of Ag+(aq) and Cl-(aq) are each x, therefore • The molar solubility is therefore 1.26 x 10-5 M. The molar mass of AgCl is (107.9 + 35.45) g/mol = 143.4 g/mol. The solubility is thus

  24. Molar Solubility and Solubility • N.B. The solubility is the amount of a substance which can dissolve within a certain amount of water • It can be in g/L (solubility) • It can be in mol/L (molar solubility) • The solubility product is an equilibrium constant (and therefore unitless) • The molar solubility, the solubility, and the solubility product all relate to saturated solutions

  25. The Common Ion Effect • Up to this point, we considered that the salt in question was the only salt in the solution • Therefore, for example, in a solution of AgCl, [Ag+] = [Cl-] • Suppose we have two salts present that share a common ion • For example, imagine a solution where we dissolve AgCl (insoluble) and AgNO3 (soluble)

  26. The Common Ion Effect • In this kind of situation, the solubility product is always valid and respected, i.e., Ks = [Ag+][Cl-] • But due to AgNO3, which is very soluble, [Ag+]  [Cl-] • And, more precisely, [Ag+] > [Cl-] • The common ion effect is the shift in the equilibrium caused by the presence of a common ion

  27. The Common Ion Effect • Example: Calculate the solubility (in g/L) of AgBr: (a) in pure water and (b) in NaBr 0.0010 M (Ks of AgBr = 7.7 x 10-13).

  28. Equilibrium of Complex Ions • A complex ion is an ion containing a metallic cation bound to one or more ions or molecules • In these reactions, the metal is a Lewis acid (electron pair acceptor), the ions/molecules bound to the metal are Lewis bases (electron pair donors), and the complex ion is a Lewis salt • To find the equilibrium concentrations, we exploit the fact that the equilibrium constant is huge (and that of the reverse reaction is therefore very small)

  29. Equilibrium of Complex Ions • According to le Chatelier’s Principle, the formation of a complex ion like Ag(NH3)2+ can increase the solubility of a compound such as AgCl since the formation of the complex ion shifts the equilibrium AgCl(s) Ag+(aq) + Cl-(aq) towards the right, replenishing the free Ag+ in solution. • The formation constant, Kf, is the equilibrium constant for the formation of the complex ion • A large value of Kf implies that the complex ion is very stable

  30. Equilibrium of Complex Ions • We dissolve 2.50 g of CuSO4 in 900 mL of a 0.30 M NH3(aq) solution. What are the concentrations of Cu2+(aq), of Cu(NH3)42+(aq), and of NH3(aq) at equilibrium? The value of Kf for Cu(NH3)42+ is 5.0 x 1013.

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