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Chem. 31 – 6/22 Lecture. Announcements I Laboratory Stuff –Cl Lab Report – Due today –AA Lab: I have posted information about your tap water that can.

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Presentation on theme: "Chem. 31 – 6/22 Lecture. Announcements I Laboratory Stuff –Cl Lab Report – Due today –AA Lab: I have posted information about your tap water that can."— Presentation transcript:

1 Chem. 31 – 6/22 Lecture

2 Announcements I Laboratory Stuff –Cl Lab Report – Due today –AA Lab: I have posted information about your tap water that can make dilutions of it more successful –IC Lab: Supplemental Information - Posted (We are switching columns – new column is faster but poorer for F - ) –Spectroscopic Lab: Working on Supplemental Information (how to use new spectrometers

3 Announcements II Tomorrow: –Additional Problem 2.1 due –Quiz on Chapter 6 (6-2 to what is covered today) Today’s Lecture –Chapter 6 (Basic Equilibrium) Complex Ions (mostly done) Acid, Bases and Salts –Chapter 7 (Advanced Equilibrium – correcting errors in Basic Equilibrium) Introduction (normally done by demonstration – may do on Thursday

4 Some Questions 1.In the reaction: Ca 2+ + Y 4- ↔ CaY 2- (where Y 4- = EDTA), which species is the Lewis acid? 2.List two applications in which the formation of a complex ion would be useful for analytical chemists. 3.List two applications in the lab in which you used or are using complex ions. 4.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.

5 One More Question 1.Cu 2+ reacts with thiosulfate (S 2 O 3 2- ) to form a complex which is most stable when two moles of thiosulfate to one mole of Cu 2+ are present. The  2 value is found to be 2.00 x 10 6. If a solution containing both Cu 2+ and S 2 O 3 2- is prepared and found to contain 1.7 x 10 -3 M free (uncomplexed) S 2 O 3 2- at equilibrium, what is the ratio of complexed to free Cu? Assume that little CuS 2 O 3 forms.

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

7 Br ø nsted-Lowry Acids - examples HCO 2 H(aq) + H 2 O(l) ↔ HCO 2 - + H 3 O + acid base conjugate conjugate base acid CH 3 NH 2 (aq) + H 2 O(l) ↔ CH 3 NH 3 + + OH - base acid conjugate conjugate acid base H 2 SO 4 + CH 3 CO 2 H(l) ↔ HSO 4 - + CH 3 CO 2 H 2 + acid base conjugate conjugate base acid

8 Br ø nsted-Lowry Acids Note: for most acids, the reaction with water is simplified: Example: HNO 2 (nitrous acid) HNO 2 ↔ H + + NO 2 -

9 Autoprotolysis and the pH Scale Autoprotolysis refers to proton transfer in protic solvents like water: H 2 O(l) ↔ H + + OH - K = K w = [H + ][OH - ] = 1.0 x 10 -14 (T = 25°C) In pure water [H + ] = [OH - ] = K w 0.5 = 1.0 x 10 -7 M pH = -log[H + ] = 7.0 Acidic is pH 7

10 Strong Acids Strong acids completely dissociate in water ( except at very high concentrations ) –HX(aq) → H + + X - (no HX(aq) exists) K a > 1 Major strong acids: HCl, HNO 3, H 2 SO 4 Note: –For H 2 SO 4, 1 st dissociation is that of a strong acid, but 2 nd dissociation is that of a weak acid (K a ~ 0.01)

11 Weak Acids Partially dissociate in water Most have H that can dissociate HX(aq) ↔ H + + X - (HX(aq) exists) Example: HNO 2 ↔ H + + NO 2 - Degree of dissociation given by K a value K a = [H + ][NO 2 - ]/[HNO 2 ] Metal cations can be acids through the reaction: M n+ + H 2 O(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”)

12 Bases Strong Bases: completely dissociate to give OH - in water –Examples: KOH (s) → K + + OH - (No KOH(aq)) Ca(OH) 2 (s) → Ca 2+ + 2OH - Weak Bases: react partially in water to give OH - - NH 3 (aq) + H 2 O (l) ↔ NH 4 + + OH - - strength of weak base given by K b for above reaction

13 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

14 Ionic Compounds in Water Conjugate bases of weak acids are basic. NO 2 - + H 2 O(l) ↔ HNO 2 (aq) + OH - Conjugate bases of weaker weak acids are stronger bases. K b = K w /K a CN - is a stronger base than NO 2 - because K a (HCN) = 6.2 x 10 -10 and K a (HNO 2 ) = 7.1 x 10 -3

15 Acidity of Ionic Compounds Determine if the ionic compounds are acidic or basic in the following examples: 1.NaCl 2.NH 4 Cl 3.NaCH 3 CO 2 4.Fe(NO 3 ) 3 5.NH 4 CN

16 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 –I had planned a demonstration, which I can do on Thursday if time

17 Demonstration – Slide 1 Summary of Observation: –Two saturated solutions of MgCO 3 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. –3.5 mL of 0.002 M HCl needed for saturated MgCO 3 and 6.0 mL needed in 0.1 M NaCl Saturated MgCO 3 Saturated MgCO 3 in NaCl(aq)

18 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? MgCO 3 (s)  Mg 2+ + CO 3 2- K sp = 3.5 x 10 -8 T = 25°C K sp = 3.5 x 10 -8 = [Mg 2+ ][CO 3 2- ] since [Mg 2+ ] = [CO 3 2- ] (assuming no other reactions), [CO 3 2- ] = (3.5 x 10 -8 ) 0.5 = 1.87 x 10 -4 M n(HCl) = (2 mol HCl/mol CO 3 2- )(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 CO 3 2- dissolved than expected. Also, the NaCl increased the solubility of MgCO 3

19 Demonstration – Slide 3 What was the affect of the NaCl? –More CO 3 2- (and Mg 2+ ) was found to dissolve in the 0.10 M NaCl Why? –The Na + and Cl - ions stabilize CO 3 2- and Mg 2+ ions

20 Ionic Strength Effects Spheres Surrounding Ions Mg 2+ Low Ionic Strength CO 3 2- HO H  HO H HO H Ion – dipole interaction HO H HO H HO H Mg 2+ CO 3 2- HO H HO H HO H  High Ionic Strength Na + Stronger ion – ion interaction replaces ion - dipole Cl - HO H


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