Chapter 6 Problems 6.19, 6.21, 6.24 6-29, 6-31, 6-39, 6.41 6-42, 6-48,

6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate?

6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? Ca2+ at equilibrium CaSO4  Ca2+ + SO42- Ksp= 2.4 x 10-5 = [Ca2+][SO42-] CaSO4  Ca2+ + SO42- Ksp= 2.4 x 10-5 = [0.000500][SO42-] [SO42-] = 0.048 M Sep. is NOT feasible Ag2SO4  2Ag+ + SO42- Ksp = 1.5 x 10-5 Q>K Q = [Ag+]2[SO42-] Q = [0.03]2[0.000500] Q = 4.3 x 10-5

6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? Ag2SO4  2Ag+ + SO42- Ksp = 1.5 x 10-5 K/[Ag2+]2 = [SO42-] 1.5 x 10-5/[0.0300]2 = [SO42-] 1.67 x 10-2= [SO42-] Find Ca2+ CaSO4  Ca2+ + SO42- Ksp= 2.4 x 10-5 = [Ca2+][1.67 x 10-2] [Ca2+] = 0.0014 M About 2.8 % remains in solution

6.21 If a solution containing 0.10 M Cl-, Br-, I- and Cr2O42- is treated with Ag+, in what order will the anions precipitate? AgCl  Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][Cl] AgBr  Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][Br] AgI  Ag+ + I- Ksp = 8.3 x 10-17=[Ag][I] Ag2CrO4  2Ag+ + CrO4- Ksp = 1.2 x 10-12=[Ag]2[Cl] AgCl  Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][0.1] AgBr  Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][0.1] AgI  Ag+ + I- Ksp = 8.3 x 10-17=[Ag][0.1] Ag2CrO4  2Ag+ + CrO4- Ksp = 1.2 x 10-12=[Ag]2[0.1] 1.8 x 10-9=[Ag] 5.0 x 10-12=[Ag] 8.3 x 10-16=[Ag] 3.5 x 10-6=[Ag] SOLVE for Ag+ required at equilibrium

6-24. The cumulative formation constant for SnCl2(aq) in 1.0 M NaNO3 is b2=12. Find the concentration of SnCl2 for a solution in which the concentration of Sn2+ and Cl- are both somehow fixed at 0.20 M. SnCl2(aq) Sn2+ (aq) + 2Cl- (aq)  SnCl2 (aq) b2=12 b2=12

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

Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions Pb2+ + I- PbI+ PbI+ + I- PbI2 K2 = 1.4 x 101 PbI2 + I- PbI3- K3 =5.9 PbI3+ I- PbI42- K4 = 3.6

Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions Pb2+ + I- <=> PbI+ PbI+ + I- <=> PbI2 K2 = 1.4 x 101 Pb2+ + 2I- <=> PbI2 K’ =? Overall constants are designated with b This one is b2

Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions Pb2+ + I- PbI+ PbI+ + I- PbI2 K2 = 1.4 x 101 PbI2 + I- PbI3- K3 =5.9 PbI3+ I- PbI42- K4 = 3.6

Acids and Bases & Equilibrium Section 6-7

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: NH2- (the amide ion)

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?

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: NH3

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

Equilibrium and Water Question: Calculate the Concentration of H+ and OH- in Pure water at 250C.

EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Initial liquid - Change -x +x Equilibrium Liquid-x Kw = [H+][OH-] = ? KW=(X)(X) = ?

EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Initial liquid - Change -x +x Equilibrium Liquid-x Kw = [H+][OH-] = 1.01 X 10-14 KW=(X)(X) = 1.01 X 10-14 (X) = 1.00 X 10-7

Example What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 25 oC)? “From now on, assume the temperature to be 25oC unless otherwise stated.” Kw = [H+][OH-] 1 x 10-14 = [1 x 10-3][OH-] 1 x 10-11 = [OH-]

pH ~ -3 -----> ~ +16 pH + pOH = - log Kw = pKw = 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. CO2 + H2O HCO3- + H+

Weak Acids and Bases Ka’s ARE THE SAME HA + H2O(l) H3O+ + A- HA H+ + A- Ka’s ARE THE SAME HA + H2O(l) H3O+ + A-

Weak Acids and Bases Kb B + H2O BH+ + OH-

Relation Between Ka and Kb

Relation between Ka and Kb Consider Ammonia and its conjugate acid. NH3 + H2O NH4+ + OH- Kb NH4+ + H2O NH3 + H3O+ Ka H2O + H2O OH- + H3O+

Example The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base. Kw = Ka x Kb

Example Calculate the hydroxide ion concentration in a 0.0100 M sodium hypochlorite solution. OCl- + H2O  HOCl + OH- The acid dissociation constant = 3.0 x 10-8

1st Insurance Problem Challenge on page 120

Chapter 8 Activity

Write out the equilibrium constant for the following expression Fe3+ + SCN- D Fe(SCN)2+ Q: What happens to K when we add, say KNO3 ? A: Nothing should happen based on our K, our K is independent of K+ & NO3-

K decreases when an inert salt is added!!! Why?

8-1 Effect of Ionic Strength on Solubility of Salts Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- Ksp=1.3x10-18 I C E some - - -x +x +2x some-x +x +2x A seemingly strange effect is observed when a salt such as KNO3 is added. As more KNO3 is added to the solution, more solid dissolves until [Hg22+] increases to 1.0 x 10-6 M. Why?

Increased solubility Why? Complex Ion? How else? No Hg22+ and IO3- do not form complexes with K+ or NO3-. How else?

The Explanation Consider Hg22+ and the IO3- - 2+ Electrostatic attraction - 2+

The Explanation Consider Hg22+ and the IO3- - 2+ Electrostatic attraction - 2+ Hg2(IO3)2(s) The Precipitate!!

The Explanation Consider Hg22+ and the IO3- - 2+ Add KNO3 NO3- K+ NO3- Electrostatic attraction K+ - K+ 2+ NO3- NO3- NO3- NO3- NO3- K+ K+ NO3- Add KNO3

The Explanation Consider Hg22+ and the IO3- - 2+ NO3- K+ K+ NO3- NO3- NO3- K+ - K+ 2+ NO3- NO3- NO3- NO3- K+ K+ NO3- NO3- Hg22+ and IO3- can’t get CLOSE ENOUGH to form Crystal lattice Or at least it is a lot “Harder” to form crystal lattice

The potassium hydrogen tartrate example

Alright, what do we mean by Ionic strength? Ionic strength is dependent on the number of ions in solution and their charge. Ionic strength (m) = ½ (c1z12+ c2z22 + …) Or Ionic strength (m) = ½ S cizi2

Examples (m) = ½ (c1z12+ c2z22 + …) Calculate the ionic strength of (a) 0.1 M solution of KNO3 and (b) a 0.1 M solution of Na2SO4 (c) a mixture containing 0.1 M KNO3 and 0.1 M Na2SO4. (m) = ½ (c1z12+ c2z22 + …)

Alright, that’s great but how does it affect the equilibrium constant? Activity = Ac = [C]gc AND

Relationship between activity and ionic strength Debye-Huckel Equation m = ionic strength of solution g = activity coefficient Z = Charge on the species x a = effective diameter of ion (nm) 2 comments What happens to g when m approaches zero? Most singly charged ions have an effective radius of about 0.3 nm Anyway … we generally don’t need to calculate g – can get it from a table

Activity coefficients are related to the hydrated radius of atoms in molecules

Relationship between m and g

Back to our original problem Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- Ksp=1.3x10-18 At low ionic strengths g -> 1

Back to our original problem Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- Ksp=1.3x10-18 In 0.1 M KNO3 - how much Hg22+ will be dissolved?

Back to our original problem Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- Ksp=1.3x10-18