Lect w9 152 - buffers and ksp_alg
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Lect w9 152 - buffers and ksp_alg

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  • Figure: 16-04 Title: Buffer Containing a Base Caption: A buffer can consist of a weak base (which can neutralize added acid) and its conjugate acid (which can neutralize added base).
  • Figure: 16-03a,b Title: Buffering Action Caption: (a) When an acid is added to a buffer, a stoichiometric amount of the weak base is converted to the conjugate acid. (b) When a base is added to a buffer, a stoichiometric amount of the weak acid is converted to the conjugate base.
  • Tier 1
  • Tier 2
  • Tier 2
  • Tier 2
  • Figure: 16-13-08UN Title: Precipitation Reaction Caption: A precipitate will form if the reaction quotient is larger than the equilibrium constant.

Lect w9 152 - buffers and ksp_alg Lect w9 152 - buffers and ksp_alg Presentation Transcript

  • General Chemistry II CHEM 152 Unit 2 Week 9
  • Week 9 Reading Assignment Chapter 16 – Sections 16.2 (buffers), 16.3 (buffers), 16.4 (titrations), 16.5 (K sp )
  • Equilibrium in Aqueous Solutions Acids and Bases We have explored the properties of acids and bases as we add them independently to water. But what happens when you mix them? View slide
  • Titrations You have covered the topic of mixing acids and bases for titrations in lab. We will assume that you understand this topic and cover some other topics of mixing acids and bases . View slide
    • What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)?
    • NH 3 (aq) + H 2 O  NH 4 + (aq) + OH - (aq)
    • Here we are adding an ion COMMON to the equilibrium ( COMMON ION EFFECT )
    Mixing Conjugate Pairs Let us first calculate the pH of a 0.25 M NH 3 solution: [NH 3 ] [NH 4 + ] [OH - ] initial 0.25 0 0 change -x +x +x equilib 0.25 - x x x
    • NH 3 (aq) + H 2 O  NH 4 + (aq) + OH - (aq)
    pH of Aqueous NH 3 Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives pOH = 2.67 and so pH = 14.00 - 2.67 = 11.33 for 0.25 M NH 3
    • What is the pH of a solution with 0.25 M NH 4 Cl and 0.25 M NH 3 (aq)?
    • NH 3 (aq) + H 2 O  NH 4 + (aq) + OH - (aq)
    • We expect that the pH will decline on adding NH 4 Cl. Let’s test that!
    pH of NH 3 /NH 4 + Mixture [NH 3 ] [NH 4 + ] [OH - ] initial 0.25 0.25 0 change -x +x +x equilib 0.25 - x 0.25 + x x
  • pH of NH 3 /NH 4 + Mixture [ OH - ] = x = (0.25 / 0.25) K b = 1.8 x 10 -5 M This gives pOH = 4.74 and pH = 9.26 pH drops from 11.33 to 9.26 on adding a common ion. K b  1.8 x 10 -5 = [NH 4 + ][OH - ] [NH 3 ] = x(0.25 + x) 0.25 - x
  • pH of NH 3 /NH 4 + Mixture
    • What is the pH when 1.00 mL of 1.00 M HCl is added to 1.00 L of a mixture that has [NH 3 ]= 0.25 M and [NH 4 + ]= 0.25 M
    • (pH = 9.26)
    H 3 O + + NH 3  NH 4 + + H 2 O The reaction occurs completely because K is large. H 3 O + + NH 3  NH 4 + + H 2 O Before rxn 0.001 0.25 0.25 Change -0.001 –0.001 +0.001 After rxn ~0 0.249 +0.251
  • pH of NH 3 /NH 4 + Mixture Equilibrium will be re-established: K a =5.6 x 10 -10 NH 4 + + H 2 O  H 3 O + + NH 3 Equilibrium 0.251-x x 0.249+x [H 3 O + ] = 5.64 x 10 -10 M  pH = 9.25 The pH has not changed much (pH = 9.26 to 9.25) on adding HCl to the mixture! Why? What would happen if we add NaOH? [ H 3 O  ]  [NH 4 + ] [NH 3 ] • K a  0.251 0.249 • ( 5 . 6 x 10 -10 )
  • Formation of a Buffer
    • The NH 3 /NH 4 + mixture is a typical example of a BUFFER . A buffer is a solution that resists changes in the pH when limited amounts of base or acid are added to it.
    • Buffer Composition
    • Weak Acid + Conjugate Base
    • CH 3 COOH + CH 3 COO 
    • H 2 PO 4  + HPO 4 2 
    • Weak Base + Conjugate Acid
    • NH 3 + NH 4 +
    Buffer Solutions
  • H + ? OH - ? Action of a Buffer
    • [CH 3 COOH] + H 2 O  [CH 3 COO - ]+ [H 3 O + ]
    • initial
    • change
    • Equilib
    Buffer Solutions What is the pH of a buffer that has [CH 3 COOH] = 0.700 M and [CH 3 COO - ] = 0.600 M? CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + K a = 1.8 x 10 -5 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x [H 3 O + ] = 2.1 x 10 -5 pH = 4.68 Assuming that x << 0.700 and 0.600 , we have
    • Notice that the expression for calculating the H 3 O + concentration of the buffer is
    Buffer Solutions Take the negative log of both sides of this equation Henderson-Hasselbalch Equation
  • Preparing a Buffer
    • Imagine that you want to prepare a buffer solution with a pH = 4.30.
    • It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ≈ pK a ).
    • Then you get the exact pH by adjusting the ratio of acid to conjugate base.
    • You want to buffer a solution at pH = 4.30 and have access to the following acids:
    • POSSIBLE ACIDS K a
    • HSO 4 - / SO 4 2- 1.2 x 10 -2
    • CH 3 COOH/CH 3 COO - 1.8 x 10 -5
    • HCN / CN - 4.0 x 10 -10
    Preparing a Buffer Which one would you choose? If you have a 0.100 M concentration of the base, what concentration of the acid would you need?
  • If you have a 0.100 M concentration of the base, what concentration of the acid would you need?
  • Your Turn The blood of mammals is an aqueous solution that maintains a constant pH. The normal pH of human blood is 7.40 . Carbon dioxide provides the most important blood buffer. In solution, CO 2 , reacts with water to form H 2 CO 3 , which ionizes to produce H 3 O + and HCO 3 - ions: CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) K a = 4.2 x 10 -7 Use this information to determine the concentration ratio [HCO 3 - ]/[H 2 CO 3 ] in blood.
  • Determine the concentration ratio [HCO 3 - ]/[H 2 CO 3 ] in blood
  • If the pH of human blood decreases below 7.35 , a condition known as acidosis occurs; increasing the pH above 7.45 causes alkalosis . Both of these conditions can be life-threatening. CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) K a = 4.2 x 10 -7 What is an easy way to treat alkalosis? ( Apply Le Chatelier’s Principle ) Your Turn
  • More with Henderson-Hasselbalch pK a (carb. acid) = 2.0 pK a (prot. amine) = 10.5
    • Draw the structure of the amino acid at
    • pH = 1.0, b) pH = 7.0, and c) pH = 12.0
    • Start by using the Henderson-Hasselbalch equation!
  • Functional group structures Acidic forms Basic forms Which version of each functional group should exist at a particular pH?
  • @pH = 1.0 The acid form (-COOH) is more prevalent. For the carboxylic acid group: The acid form (-NH 3 + ) is more prevalent. For the amine group:
  • @pH = 1.0 Now – in your groups – Draw the prevalent structures of the amino acid at pH=7 and at pH = 12.
  • What is happening here? solution solution Solubility Equilibria
  • Solubility of Salts Not all salts are completely soluble in water. Many of them only dissolve to a small extent. When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED . When an insoluble salt is placed in water, chemical equilibrium can be established: AgCl(s)  Ag + (aq) + Cl - (aq)
  • Solubility of Salts Consider: AgCl(s)  Ag + (aq) + Cl - (aq) Write the expression of the equilibrium constant K sp ( solubility product ) associated with this process. K sp = [Ag + ][Cl - ]/ [AgCl(s)] K sp = [Ag + ][Cl - ]
  • Solubility of Salts Consider: AgCl(s)  Ag + (aq) + Cl - (aq) The concentration of [Ag + ] in a saturated solution of AgCl is 1.34x10 -5 M. This number represents the SOLUBILITY of the salt. What is the concentration of Cl ¯ ions in the system? What is the value of K sp for this salt? K sp =[Ag + ][ Cl - ] = (1.34 x10 -5 ) 2 = 1.80 x 10 -10
  • Estimate the solubility of the following salts: PbSO 4 Your Turn K sp = 1.8 x10 -8 K sp = (x)(x) x = 1.3 x 10 -4 M
  • Determine the molar solubility of PbCl 2 ? K sp = 1.7 x 10 -5
  • Estimate the solubility of the following salt: Your Turn PbCl 2 K sp = 1.7 x 10 -5 K sp = (x)(2x) 2 x =1.6 x 10 -2 M
  • PbI 2 (s)  Pb 2+ (aq) + 2I - (aq) Calculate K sp for PbI 2 if [Pb 2+ ] = 0.00130 M in a saturated solution.
  • Consider PbI 2 dissolving in water PbI 2 (s)  Pb 2+ (aq) + 2I - (aq) Calculate K sp for PbI 2 if [Pb 2+ ] = 0.00130 M PbI 2 (s)  Pb 2+ (aq) + 2I - (aq) K sp =[0.00130][0.00260] 2 = 8.79 x10 -9 0.00130 0.00260 Your Turn
  • Solubility of Salts
    • Consider the following insoluble salt:
    • CaCO 3 (s)
    • Will the solubility of this salt increase, decrease, or remain the same after adding:
    • Sulfuric acid ( H 2 SO 4 )
    • Hydrochloric acid ( HCl )
    • Calcium Chloride ( CaCl 2 )
  • Solubility of Salts
    • Consider the following insoluble salt:
    • AgCl(s)
    • Will the solubility of these salts increase, decrease, or remain the same after adding:
    • Sulfuric acid ( H 2 SO 4 )
    • Hydrochloric acid ( HCl )
    • Calcium Chloride ( CaCl 2 )
  • The Common Ion Effect
    • Adding an ion “common” to an equilibrium causes the equilibrium to shift back to reactant.
    AgCH 3 CO 2 (s)  Ag + (aq) + CH 3 CO 2 - (aq) AgNO 3 (aq)  Ag+(aq) + NO 3 - (aq)
  • Kidney Stones Kidney stones are normally composed of: calcium oxalate ( CaC 2 O 4 ), calcium phosphate ( Ca 3 (PO 4 ) 2 ), magnesium ammonium phosphate ( MgNH 4 PO 4 ). Precipitate formed from calcium ions from food rich in calcium, dairy products, and oxalate ions from chocolate, spinach, celery, black tea: Ca +2 + C 2 O 4 2-  CaC 2 O 4 How would you try to dissolve the CaC 2 O 4 stones?
  • Precipitating an Insoluble Salt
    • Reactant Quotient, Q=[Ag + ] o [Cl - ] o
    • Q < K sp - no precipitate forms
      • unsaturated solution
      • Q > K sp - precipitate may form supersaturated solution
      • Q = K sp - no precipitate forms saturated solution
    Consider the following problem: Imagine you mix 500.0 mL of a solution of AgNO 3 1.5 x 10 -5 M with 500.0 mL of a solution of NaCl 1.5 x 10 -5 M. Will AgCl precipitate? AgCl(s)  Ag + (aq) + Cl - (aq) K sp = 1.8 x 10 -10 DID YOU CONSIDER DILUTION?
  • Precipitating an Insoluble Salt Consider the case: Hg 2 Cl 2 (s)  Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ]=0.010 M, what [Cl - ] is required to just begin the precipitation of Hg 2 Cl 2 ? That is, what is the maximum [Cl - ] that can be in solution with 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 ? K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 = [0.010][Cl - ] 2 [Cl - ] = 1.0 x 10 -8 M
  • Separating Salts by Differences in K sp
    • If two or more substances can precipitate in a solution, the substance whose K sp is first exceeded precipitates first.
    Consider a solution containing 0.020 M Cl - and 0.010 M CrO 4 2- ions in which Ag+ ions are added slowly. Which precipitates first, AgCl or Ag 2 CrO 4 ? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for AgCl = 1.8 x 10 -10
  • Separating Salts by Differences in K sp Consider a solution containing 0.020 M Cl - and 0.010 M CrO 4 2- ions in which Ag+ ions are added slowly. K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for AgCl = 1.8 x 10 -10 What is [Ag + ] when the maximum AgCl is precipitated, but none of the Ag 2 CrO 4 has precipitated? What percent of Cl - remains in solution?
  • Separating Salts by Differences in K sp If PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) K sp =1.7 x 10 -5 PbCrO 4 (s)  Pb 2+ (aq) + CrO 4 2- (aq) K sp =1.8 x 10 -4 What is the equilibrium constant for the reaction: PbCl 2 + CrO 4 2-  PbCrO 4 + 2Cl - ? What does this tell you?
  • 1. What is the pH of a solution of 0.150 M HIO 3 (K a = 0.17) and 0.350 M NaIO 3 ?
  • 2. What is the maximum mass of calcium fluoride that will dissolve in 500. mL of aqueous solution? K sp = 1.46 x 10 -10 M = 78.08 g/mol
  • 3. What is the molar solubility of CaF 2 in a solution of 0.10 M NaF? K sp = 1.46 x 10 -10