Presentation on theme: "Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay,"— Presentation transcript:
Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc. Modified by S.A. Green, 2006
Aqueous Equilibria Control of solution composition Chap 16: What is the pH after adding X to water? Chap 17: How can we control the pH of a solution? How can we determine acid/base concentration? Chap 4: Solubility of ionic compounds. Chap 17: How soluble are they? Buffers Titrations K sp =solubility product
Aqueous Equilibria The Common-Ion Effect Consider a solution of acetic acid: If acetate ion (CH 3 O 2 - ) is added to the solution, Le Châtelier says the equilibrium will shift to the left <––. Where can we get CH 3 O 2 - ions?
Aqueous Equilibria The salt completely dissociates. So, adding the salt = adding CH 3 O 2 - (acetate) ions. What happens to the pH? The Common-Ion Effect Where can we get CH 3 O 2 - ions? Add them in the form of a salt:
Aqueous Equilibria Calculate the pH of a 1.0 L solution containing 0.30 moles of acetic acid, C 2 H 3 O 2 H, and 0.30 moles of sodium acetate C 2 H 3 O 2 Na, at 25°C. K a for acetic acid at 25°C is 1.8 The Common-Ion Effect
Aqueous Equilibria Use the same method as in Chapter 16: ICE + equilibrium expression But, adjust initial concentrations to account for the salt. The Common-Ion Effect
Aqueous Equilibria Use the ICE table: [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 − ], M Initial0.300 Change–x–x+x+x+x+x Equilibrium0.30 – xx0.30+x The Common-Ion Effect
Aqueous Equilibria Suppose x is small relative to 0.30: K a = x 1.8 = x = [H + ] pH=–log( 1.8 )=4.74 Check: is approximation ok? The Common-Ion Effect
Aqueous Equilibria Note: When [HA] = [A - ] then: [H 3 O + ] = K a The Common-Ion Effect: Buffering When [H 3 O + ] is small compared to [HA] & [A - ] then [H 3 O + ] = K a times the ratio [HA]/[A - ]
Aqueous Equilibria Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.
Aqueous Equilibria Buffers If a small amount of hydroxide is added to an equimolar solution of HF and NaF, for example, the HF reacts with the OH – to make F – and water.
Aqueous Equilibria Buffers If acid is added, the F – reacts to form HF and water.
Aqueous Equilibria Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A −
Aqueous Equilibria Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] –log K a = –log [H 3 O + ] + − log pKapKa pH acid base
Aqueous Equilibria Buffer Calculations So pK a = pH –log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.
Aqueous Equilibria Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a for lactic acid is1.4
Aqueous Equilibria pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH.
Aqueous Equilibria When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.
Aqueous Equilibria Addition of Strong Acid or Base to a Buffer 1.Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2.Use the Henderson–Hasselbalch equation to determine the new pH of the solution.
Aqueous Equilibria Calculating pH Changes in Buffers A buffer is made by adding mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.
Aqueous Equilibria Calculating pH Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 – pH = pK a = –log (1.8 ) = 4.74
Aqueous Equilibria Calculating pH Changes in Buffers The mol NaOH will react with mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH - (aq) C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Before reaction0.300 mol mol After reaction0.280 mol0.320 mol≈0.000 mol
Aqueous Equilibria Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = log (0.320) (0. 200) pH = pH = 4.80
Aqueous Equilibria Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).
Aqueous Equilibria Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. Equiv point: moles acid=moles base
Aqueous Equilibria Titrations Types: 1. “strong-strong” Strong acid + strong base Strong base + strong acid 2. “weak-strong” Weak acid + strong base Weak base + strong acid Why titrate? 1.How much acid or base is present. 2.What is pK a (pK b ) of weak acid (base). (we will not do this) Usual question asked: What is the pH after X mL of Y has been added?
Aqueous Equilibria Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.
Aqueous Equilibria Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly. --> transition from acidic to basic state
Aqueous Equilibria Titration of a Strong Acid with a Strong Base At the equivalence point: moles acid = moles base. pH=7 The solution contains only water and the salt from the cation of the base and the anion of the acid.
Aqueous Equilibria Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off. Movie….
Aqueous Equilibria Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH. The pH at the equivalence point is >7. Phenolphthalein is commonly used as an indicator in these titrations.
Aqueous Equilibria Titration of a Weak Acid with a Strong Base Before the equivalence point: the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present.
Aqueous Equilibria Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
Aqueous Equilibria Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.
Aqueous Equilibria SAMPLE EXERCISE 17.7 Calculating pH for a Weak Acid–Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M HC 2 H 3 O 2 (K a = 1.8 10 –5 ). Solution Write reaction Ask: Is this strong-strong or weak-strong titration? Are we before, at, or after equivalence point? Need to find: # moles of weak acid present initially # moles of base added Concentrations of acid and conjugate base
Aqueous Equilibria SAMPLE EXERCISE 17.7 Calculating pH for a Weak Acid–Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M HC 2 H 3 O 2 (K a = 1.8 10 –5 ). Need to find: # moles of weak acid present initially # moles of base added Concentrations of acid and conjugate base The 4.50 10 –3 mol of NaOH consumes 4.50 10 –3 mol of HC 2 H 3 O 2 :
Aqueous Equilibria SAMPLE EXERCISE 17.7 continued The total volume of the solution is The resulting molarities of HC 2 H 3 O 2 and C 2 H 3 O 2 – after the reaction are therefore Equilibrium Calculation: The equilibrium between HC 2 H 3 O 2 and C 2 H 3 O 2 – must obey the equilibrium-constant expression for HC 2 H 3 O 2 : Solving for [H + ] gives Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M HC 2 H 3 O 2 (K a = 1.8 10 –5 ).
Aqueous Equilibria Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.
Aqueous Equilibria Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq)
Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2- ] where the equilibrium constant, K sp, is called the solubility product.
Aqueous Equilibria Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
Aqueous Equilibria Factors Affecting Solubility The Common-Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium shifts to the left and the solubility of the salt decreases. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq)
Aqueous Equilibria Factors Affecting Solubility pH If a substance has a basic anion, it is more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions.
Aqueous Equilibria Factors Affecting Solubility Complex Ions Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.
Aqueous Equilibria Factors Affecting Solubility Complex Ions The formation of these complex ions increases the solubility of these salts.
Aqueous Equilibria Factors Affecting Solubility Amphoterism Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Examples of such cations are Al 3+, Zn 2+, and Sn 2+.
Aqueous Equilibria Will a Precipitate Form? In a solution, If Q = K sp, the system is at equilibrium and the solution is saturated. If Q < K sp, more solid will dissolve until Q = K sp. If Q > K sp, the salt will precipitate until Q = K sp.
Aqueous Equilibria Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.
Aqueous Equilibria Chapter 17 problems A buffer is prepared by adding 5.0 g of NH 3 and 20.0 g NH 4 + to enough water to make 2.50 L of solution. (a)pH=? (b)(b) Write complete ionic equation for the reaction that occurs when a small amount of concentrated nitric acid is added.
Aqueous Equilibria Chapter 17 problems How would you make about 1 L of a buffer with pH=3.50? The following 0.10 M solutions are available: HCO 2 HC 2 H 3 O 2 H 3 PO 4 NaCHO 2 NaC 2 H 3 O 2 NaH 2 PO 4
Aqueous Equilibria Chapter 17 problems Consider the titration of 35.0 mL of 0.15 M acetic acid with 0.15 M NaOH. What is the pH after the following amounts of titrant have been added? 34.5 mL 35.0 mL 35.5 mL
Aqueous Equilibria Chapter 17 problems The molar solubility of PbBr 2 is 1.0 x mol/L. Cacluate K sp.