AcidBase Equilibria and Solubility Products Buffer Solutions A

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 • A buffer solution has the ability to maintain almost constant p. H, despite the addition of small amounts of acid or base • In a living organism, buffers are critical • ex. ; blood p. H remains more or less constant due to a buffer consisting of the H 2 CO 3/HCO 3 - conjugate acid-base pair

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 CH 3 COOH and CH 3 COONa 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 CH 3 COO-(aq) suppresses the hydrolysis of CH 3 COOH(aq) • The CH 3 COOH(aq) suppresses the hydrolysis of CH 3 COO-(aq)

Buffer Solutions • A buffer solution like our CH 3 COO-/CH 3 COOH pair is capable of keeping the p. H more or less constant as, upon addition of H+(aq) or OH-(aq), the following reactions are produced • i. e. , the CH 3 COO- neutralizes the H+(aq) or the CH 3 COOH neutralizes the OH-(aq), and the p. H 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

Buffer Solutions • Example: Which of the following solutions are buffer systems? (a) KF/HF, (b) KCl/HCl, (c) Na 2 CO 3/Na. HCO 3 • 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) CO 32 - is a weak base, and HCO 3 - is its conjugate acid, therefore this is a buffer system.

Buffer Solutions • Example: Calculate the p. H of the following buffer system: NH 3 (0. 30 M) / NH 4 Cl (0. 36 M). What happens to the p. H if 20. 0 m. L of 0. 050 M Na. OH is poured into 80. 0 ml of the buffer solution? • Solution:

Buffer Solutions • Solution: If we poured 20. 0 m. L of 0. 050 M Na. OH 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 NH 3 • We have (0. 080 L)(0. 36 mol/L) = 0. 0288 mol of NH 4+ • The OH- will consume NH 4+ and produce NH 3

Buffer Solutions • N. B. If in the previous example we added the same quantity of OH - to 80. 0 m. L 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 p. H = 12. 00 • In pure water, it goes from p. H = 7. 00 to p. H = 12. 00 • In the buffer solution, it goes from p. H = 9. 17 to p. H = 9. 21 • A buffer solution is very effective at maintaining a constant p. H

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

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 p. H p. Ka

• Calculate the p. H of a 1. 00 L buffer solution which is 0. 87 M CH 3 COOH and 0. 47 M Na. CH 3 COO. 0. 10 mol of HCl are added to this solution. Calculate the new p. H. The ionization constant for CH 3 COOH is 1. 8 x 10 -5.

• Calculate the p. H of a 1. 00 L solution of 0. 537 M CH 3 COOH. 1. 00 L of 0. 197 M Na. OH is added to the solution. Calculate the new p. H. The ionization constant for CH 3 COOH is 1. 8 x 10 -5.

Strong Acid-Strong Base Titration • ex. ; or • The equivalence point is the point where equimolar quantities of acid and base have reacted (in this case, p. H= 7. 00)

Weak Acid-Strong Base Titration • ex. ; or • At the equivalence point, the OH- has neutralized all of the CH 3 COOH • All of the CH 3 COOH is converted to CH 3 COO • Because CH 3 COO-(aq) is a weak base, the equivalence point is at a p. H greater than 7

Strong Acid-Weak Base Titration • ex. ; or • At the equivalence point, the H+ has neutralized all of the NH 3 • All of the NH 3 is converted into NH 4+ • Because NH 4+ is a weak acid, the equivalence point is at a p. H lower than 7

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 (p. H zone where the colour chages) corresponds to the p. Ka of the indicator • For a given titration, we want to choose an indicator where the changeover zone corresponds to the p. H of the equivalence point

Solubility Equilibrium • Consider a salt that is insoluble in water, for example, Ba. SO 4(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

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

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

Molar Solubility and Solubility

Molar Solubility and Solubility • Example: The molar solubility of barium fluoride (Ba. F 2) is 7. 5 x 10 -3 M. What is the solubility product of this compound? • Solution: If the concentration of Ba. F 2 is 7. 5 x 10 -3 M and Ba. F 2 dissociates in water:

Molar Solubility and Solubility • Example: Calculate the molar solubility of lead carbonate (Pb. CO 3) using its solubility product (Ks = 3. 3 x 10 -14). • Solution: Let x be the molar solubility of Pb. CO 3, At equilibrium, the concentrations of Pb 2+(aq) and CO 32 -(aq) are each x, therefore • The molar solubility of Pb. CO 3 is therefore 1. 8 x 10 -7 M.

Molar Solubility and Solubility • Example: Calculate the solubility of silver chloride (Ag. Cl) in g/L using its solubility product (Ks = 1. 6 x 10 -10). • Solution: Let x be the molar solubility of Ag. Cl, Ag. Cl 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 Ag. Cl is (107. 9 + 35. 45) g/mol = 143. 4 g/mol. The solubility is thus

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

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 Ag. Cl, [Ag+] = [Cl-] • Suppose we have two salts present that share a common ion • For example, imagine a solution where we dissolve Ag. Cl (insoluble) and Ag. NO 3 (soluble)

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 Ag. NO 3, 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

The Common Ion Effect • Example: Calculate the solubility (in g/L) of Ag. Br: (a) in pure water and (b) in Na. Br 0. 0010 M (Ks of Ag. Br = 7. 7 x 10 -13).

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)

Equilibrium of Complex Ions • According to le Chatelier’s Principle, the formation of a complex ion like Ag(NH 3)2+ can increase the solubility of a compound such as Ag. Cl since the formation of the complex ion shifts the equilibrium Ag. Cl(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

Equilibrium of Complex Ions • We dissolve 2. 50 g of Cu. SO 4 in 900 m. L of a 0. 30 M NH 3(aq) solution. What are the concentrations of Cu 2+(aq), of Cu(NH 3)42+(aq), and of NH 3(aq) at equilibrium? The value of Kf for Cu(NH 3)42+ is 5. 0 x 1013.