1 / 20

Chem. 31 – 10/15 Lecture

Chem. 31 – 10/15 Lecture. Announcements. Turn in Cl lab report Website: addition of tap water information page HW2 Passed out in lab First assignment due + quiz next Wed. Today’s Lecture Complex Ions (Ch. 6) Acid Base Chemistry (Ch. 6)

lacey-conley
Download Presentation

Chem. 31 – 10/15 Lecture

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chem. 31 – 10/15 Lecture

  2. Announcements • Turn in Cl lab report • Website: addition of tap water information page • HW2 • Passed out in lab • First assignment due + quiz next Wed. • Today’s Lecture • Complex Ions (Ch. 6) • Acid Base Chemistry (Ch. 6) • Advanced Equilibrium: introduction and effects of ionic strength (Ch. 7)

  3. Some Questions • In the reaction: Ca2+ + Y4-↔ CaY2- (where Y4- = EDTA), which species is the Lewis acid? • List two applications in which the formation of a complex ion would be useful for analytical chemists. • List two applications in the lab in which you used or are using complex ions. • AgCN is a sparingly soluble salt. However, a student observed that adding a little of a NaCN solution to a saturated solution of AgCN did not result in more precipitation of solid. Addition of more NaCN solution resulted in total dissolution of the AgCN. Explain what is happening.

  4. One More Question • Cu2+ reacts with thiosulfate (S2O32-) to form a complex which is most stable when two moles of thiosulfate to one mole of Cu2+ are present. The b2 value is found to be 2.00 x 106. If a solution containing both Cu2+ and S2O32- is prepared and found to contain 1.7 x 10-3 M free (uncomplexed) S2O32- at equilibrium, what is the ratio of complexed to free Cu? Assume that little CuS2O3 forms.

  5. Acids, Bases and Salts Definitions of Acids and Bases - Lewis Acids/Bases (defined before, most general category) - Brønsted-Lowry Acids/Bases: acid = proton donor base = proton acceptor (must have free electron pair so also is a Lewis base) - definitions are relative

  6. Brønsted-Lowry Acids - examples HCO2H(aq) + H2O(l) ↔ HCO2- + H3O+ acid base conjugate conjugate base acid CH3NH2(aq) + H2O(l) ↔ CH3NH3+ + OH- base acid conjugate conjugate acid base H2SO4 + CH3CO2H(l)↔ HSO4- + CH3CO2H2+ acid base conjugate conjugate base acid

  7. Brønsted-Lowry Acids Note: for most acids, the reaction with water is simplified: Example: HNO2 (nitrous acid) HNO2↔ H+ + NO2-

  8. Autoprotolysis and the pH Scale Autoprotolysis refers to proton transfer in protic solvents like water: H2O(l) ↔ H+ + OH- K = Kw = [H+][OH-] = 1.0 x 10-14 (T = 25°C) In pure water [H+] = [OH-] = Kw0.5 = 1.0 x 10-7 M pH = -log[H+] = 7.0 Acidic is pH < 7; basic is pH > 7

  9. Strong Acids • Strong acids completely dissociate in water (except at very high concentrations) • HX(aq) → H+ + X- (no HX(aq) exists) • Ka > 1 • Major strong acids: HCl, HNO3, H2SO4 • Note: • For H2SO4, 1st dissociation is that of a strong acid, but 2nd dissociation is that of a weak acid (Ka ~ 0.01)

  10. Weak Acids • Partially dissociate in water • Most have H that can dissociate • HX(aq) ↔ H+ + X- (HX(aq) exists) • Example: HNO2 ↔ H+ + NO2- • Degree of dissociation given by Ka value • Ka = [H+][NO2-]/[HNO2] • Metal cations can be acids through the reaction: Mn+ + H2O(l) ↔ MOH(n-1)+ + H+ (although for +1 and some +2 metals the above reactions favor reactants so strongly the metals can be considered “neutral”)

  11. Bases Strong Bases: completely dissociate to give OH- in water Examples: KOH (s) → K+ + OH- (No KOH(aq)) Ca(OH)2 (s)→ Ca2+ + 2OH- Weak Bases: react partially in water to give OH- - NH3 (aq) + H2O (l) ↔ NH4 + + OH- - strength of weak base given by Kb for above reaction

  12. Ionic Compounds in Water • First step should be dissociation to respective ions: example: NaCl(s) → Na+ + Cl- • In subsequent steps, determine how anion/cation react: - anions usually only react as bases - cations may react as acids - see if ions are recognizable conjugate acids or bases - polyprotic acids are somewhat different

  13. Ionic Compounds in Water Conjugate bases of weak acids are basic. NO2- + H2O(l) ↔ HNO2 (aq) + OH- Conjugate bases of weaker weak acids are stronger bases. Kb = Kw/Ka CN- is a stronger base than NO2- because Ka(HCN) = 6.2 x 10-10 and Ka(HNO2) = 7.1 x 10-3

  14. Acidity of Ionic Compounds Determine if the ionic compounds are acidic or basic in the following examples: • NaCl • NH4Cl • NaCH3CO2 • Fe(NO3)3 • NH4CN

  15. Chapter 7“Adjustments” to Equilibrium Theory There are two areas where the general chemistry equilibrium theory can give wrong results: When the solution has high concentrations of ions When multiple, interacting equilibria occur Go to Demonstration

  16. Demonstration – Slide 1 Summary of Observation: Two saturated solutions of MgCO3 are prepared. One is prepared in water and the other is prepared in ~0.1 M NaCl. 5.0 mL of each solution was transferred (and filtered) into a beaker. mL of 0.002 M HCl needed will need to wait for demonstration but: Left = 3.5 mL right = 6.0 mL (past trial) SaturatedMgCO3 SaturatedMgCO3 in NaCl(aq)

  17. Demonstration – Slide 2 Did the moles of HCl used match expectations? and Why did the solution containing NaCl need more HCl? First Question: How many mL of HCl were expected? MgCO3(s)  Mg2+ + CO32-Ksp = 3.5 x 10-8 T = 25°C Ksp = 3.5 x 10-8 = [Mg2+][CO32-] since [Mg2+] = [CO32-] (assuming no other reactions), [CO32-] = (3.5 x 10-8)0.5 = 1.87 x 10-4 M n(HCl) = (2 mol HCl/mol CO32-)(1.87 x 10-4 mmol/mL)(5.0 mL) = 0.001875 mmol HCl Calculate V(HCl) = 0.001875 mmol HCl/[HCl] = 0.001875 mmol HCl/0.002 mmol/mL = 0.935 mL Actual V(HCl) > 1 mL Conclusions It takes more HCl than expected, so more CO32- dissolved than expected. Also, the NaCl increased the solubility of MgCO3

  18. Demonstration – Slide 3 What was the affect of the NaCl? More CO32- (and Mg2+) was found to dissolve in the 0.10 M NaCl Why? The Na+ and Cl- ions stabilize CO32- and Mg2+ ions

  19. Ionic Strength EffectsSpheres Surrounding Ions H H H H O O O O H H H H H H H H H H O O O O O O H H H H H H Low Ionic Strength High Ionic Strength Ion – dipole interaction d+ CO32- CO32- Na+ Stronger ion – ion interaction replaces ion - dipole Mg2+ Mg2+ d- Cl-

  20. Ionic Strength Definition : m = 0.5*SCiZi2 where i is an ion of charge Z and molar concentration C. Examples: 0.10 M NaCl 0.010 M MgCl2 0.010 M Ce(SO4)2

More Related