Chapter 6 Problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24 Comments on “Lab Report & Pop Rocks”

6.6 From the equations HOCl  H + + OCl - K = 3.0 x HOCl + OBr -  HOBr + OCl - K= 15 Find K for HOBr  H + + OBr — HOBr + OCl -  HOCl + OBr - Flip K’ = 1/K= 1/15 K= ? K= 3.0 x /15 K= 0.2 x 10 -8

6.9 The formation of Tetrafluorethlene from its elements is highly exothermic. 2 F 2 (g) + 2 C (s)  F 2 C=CF 2 (g) (a) if a mixture of F 2, graphite, and C 2 F 4 is at equilibrium in a closed container, will the reaction go to the right or to the left if F 2 is added? (b) Rare bacteria … eat C 2 F 4 and make teflon for cell walls. Will the reaction go to the right or to the left if these bacteria are added?

6.9 The formation of Tetrafluorethlene from its elements is highly exothermic. 2 F 2 (g) + 2 C (s)  F 2 C=CF 2 (g) (C) will the reaction go to the right or to the left if graphite is added? (d) will reaction go left or right if container is crushed to one-eighths of original volume? (e) Does “Q” get larger or smaller if vessel is Heated?

6-15. What concentration of Fe(CN) 6 4- is in equilibrium with 1.0 uM Ag + and Ag 4 Fe(CN) 6 (s). Ag 4 Fe(CN) 6  4Ag + + Fe(CN) 6 4- K sp = [Ag + ] 4 [Fe(CN) 6 4- ] 8.5 x = [1.0 x ] 4 [Fe(CN) 6 4- ] [Fe(CN) 6 4- ] = 8.5 x M = 8.5 zM

6-16. Cu 4 (OH) 6 (SO 4 )  4 Cu OH - + SO 4 2- I’d first set up an ICE table: Cu 4 (OH) 6 (SO 4 )  4 Cu OH+ SO 4 2- ISome-1.0 x M- C-x+ 4xFixed at 1.0 x M+ x ESome –x + 4x Fixed at 1.0 x M + x Ksp = [Cu 2+ ] 4 [OH - ] 6 [SO 4 2- ] = 2.3 x Ksp = [4x] 4 [1.0 x ] 6 [x] = 2.3 x X = 9.75 x M Cu 2+ = 4x = x M

Chapter 6 Chemical Equilibrium

Equilibrium Constant Solubility product (K sp ) Common Ion Effect Separation by precipitation Complex formation

Separation by Precipitation

Complete separation can mean a lot … we should define complete. Complete means that the concentration of the less soluble material has decreased to 1 X M or lower before the more soluble material begins to precipitate

Separation by Precipitation EXAMPLE: Can Fe +3 and Mg +2 be separated quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH - concentrations is permissible.

Fe 3+ Mg 2+ Fe 3+ Mg 2+ Fe 3+ Mg 2+ Add OH -

Fe 3+ Mg 2+ Fe(OH) 3 (s) What is the [OH - ] when this happens equilibrium

EXAMPLE: Separate Iron and Magnesium? K sp = [Fe +3 ][OH - ] 3 = 2 X K sp = [Mg +2 ][OH - ] 2 = 7.1 X Assume [Fe +3 ] eq = 1.0 X M when “completely” precipitated. What will be the [OH - equilibrium required to reduce the [Fe +3 ] to [Fe +3 ] = 1.0 X M ? K sp = [Fe +3 ][OH - ] 3 = 2 X

EXAMPLE: Separate Iron and Magnesium? K sp = [Fe +3 ][OH - ] 3 = 2 X (1.0 X M)*[OH - ] 3 = 2 X

Dealing with Mg 2+ Find [OH - ] to start precipitating Mg 2+ Conceptually – Will assume a minimal amount of Mg 2+ will precipitate and determine the respective concentration of OH - Evaluate Q If Q>K Q<K Q=K “Left” “Right” “Equilibrium”

Fe 3+ Mg 2+ Fe(OH) 3 (s) [OH - ] equilibrium Is this [OH - ] (that is in solution) great enough to start precipitating Mg 2+? = 1.3 x

Fe 3+ Mg 2+ Fe(OH) 3 (s) [OH - ] equilibrium Is this [OH - ] (that is in solution) great enough to start precipitating Mg 2+? = 1.3 x

EXAMPLE: Separate Iron and Magnesium? What [OH - ] is required to begin the precipitation of Mg(OH) 2 ? [Mg +2 ] = 0.10 M K sp = [OH - ] = 8.4 X M [Mg 2+ ] eq = M Really, Really close to 0.1 M (0.10 M)[OH - ] 2 = 7.1 X

EXAMPLE: Separate Iron and Magnesium? [OH - ] to ‘completely’ remove Fe 3+ = 1.3 X M [OH - ] to start removing Mg 2+ = 8.4 X M “All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!! equilibrium

EXAMPLE: Separate Iron and Magnesium? Q vs. K K sp = [Mg 2+ ][OH - ] 2 = 7.1 X Q = [0.10 M ][ 1.3 x ] 2 = 1.69 x Q<K Reaction will proceed to “Right” Mg(OH) 2 (s)  Mg OH -

Dealing with Mg 2+ Find [OH - ] to start precipitating Mg 2+ Conceptually – Will assume a minimal amount of Mg 2+ will precipitate and determine the respective concentration of OH - Evaluate Q If Q>K Q<K Q=K “Left” “Right” “Equilibrium” NO PPT

“Real Example” Consider a 1 liter solution that contains 0.3 M Ca 2+ and 0.5 M Ba 2+. Can you separate the ions by adding Sodium Carbonate? Sodium Chromate ? Sodium Fluoride? Sodium Hydroxide? Sodium Iodate? Sodium Oxylate?

An example Consider Lead Iodide PbI 2 (s) Pb I - K sp = 7.9 x What should happen if I - is added to a solution? Should the solubility go up or down?

Complex Ion Formation

Complex Formation complex ions (also called coordination ions) Lewis Acids and Bases acid => electron pair acceptor (metal) base => electron pair donor (ligand)

Pb 2+ I-I- I-I-

I-I- I-I-

I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I-

I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I-

I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I-

I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I- I-I-

Effects of Complex Ion Formation on Solubility Consider the addition of I - to a solution of Pb +2 ions Pb 2+ + I - PbI + PbI + + I - PbI 2 K 2 = 1.4 x 10 1 PbI 2 + I - PbI 3 - K 3 =5.9 PbI 3 + I - PbI 4 2- K 4 = 3.6

Effects of Complex Ion Formation on Solubility Consider the addition of I - to a solution of Pb +2 ions Pb 2+ + I - PbI + PbI + + I - PbI 2 K 2 = 1.4 x 10 1 Pb I - PbI 2 K’ =? Overall constants are designated with  This one is  

Protic Acids and Bases Section 6-7

Question Can you think of a salt that when dissolved in water is not an acid nor a base? Can you think of a salt that when dissolved in water IS an acid or base?

Protic Acids and Bases - Salts Consider Ammonium chloride Can ‘generally be thought of as the product of an acid-base reaction. NH 4 + Cl - (s) NH Cl - From general chemistry – single positive and single negative charges are STRONG ELECTROLYTES – they dissolve completely into ions in dilute aqueous solution

Protic Acids and Bases Conjugate Acids and Bases in the B-L concept CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + acid + base conjugate base + conjugate acid conjugate base => what remains after a B-L acid donates its proton conjugate acid => what is formed when a B-L base accepts a proton

Question: Question: Calculate the Concentration of H + and OH - in Pure water at 25 0 C.

EXAMPLE: Calculate the Concentration of H + and OH - in Pure water at 25 0 C. H 2 O H + + OH - Initialliquid-- Change-x+x Equilibrium Liquid-x +x K W = (X)(X) = 1.01 X K w = [H + ][OH - ] = 1.01 X (X) = 1.00 X 10 -7

Example Concentration of OH - if [H + ] is 1.0 x o C? K w = [H + ][OH - ] 1 x = [1 x ][OH - ] 1 x = [OH - ] “From now on, assume the temperature to be 25 o C unless otherwise stated.”

pH ~ > ~ +16 pH + pOH = - log K w = pK w = 14.00

Is there such a thing as Pure Water? In most labs the answer is NO Why? A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value. CO 2 + H 2 O HCO H +

6-9 Strengths of Acids and Bases

Strong Bronsted-Lowry Acid A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). HCl as example

Strong Bronsted-Lowry Base Accepts protons from water molecules to form an amount of hydroxide ion, OH -, equivalent to the amount of base added. Example: NH 2 - (the amide ion)

Weak Bronsted-Lowry acid One that DOES not donate all of its acidic protons to water molecules in aqueous solution. Example? Use of double arrows! Said to reach equilibrium.

Weak Bronsted-Lowry base Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added. example: NH 3

Common Classes of Weak Acids and Bases Weak Acids carboxylic acids ammonium ions Weak Bases amines carboxylate anion

Weak Acids and Bases HA H + + A - HA + H 2 O (l) H 3 O + + A - KaKa K a ’ s ARE THE SAME

Weak Acids and Bases B + H 2 O BH + + OH - KbKb

Relation Between K a and K b

Relation between Ka and Kb Consider Ammonia and its conjugate base. NH 3 + H 2 O NH OH - KaKa NH H 2 O NH 3 + H 3 O + KbKb H 2 O + H 2 O OH - + H 3 O +

Example The K a for acetic acid is 1.75 x Find K b for its conjugate base. K w = K a x K b

1 st Insurance Problem Challenge on page 120