Chapter 16 Review
Acid Base and Solubility Equilibria

  1. Neutralization Reactions
    1. Strong acid strong base:  reaction goes to products all the way since reactants completely ionized 
      1. Molecular rxn:  HI(aq) + KOH(aq)  g H2O(l) + KI(aq)
      2. Ionic rxn:  H+(aq) +   I-(aq) + K+(aq) + OH-(aq)  g H2O(l) + K+(aq) + I-(aq)
      3. Net ionic rxn:  H+(aq) + OH-(aq)  g H2O(l) 
      4. When the rxn is neutralized (all reactants turned into products) the pH = 7.0 because the KI salt is neutral
      5. Review section 4.5, example 4.6, problems 3.20-22
    2. Weak acid strong base:  reaction goes to products all the way since one reactant completely ionized
      1. Molecular rxn:  HF(aq) + KOH(aq)  gH2O(l) + KF(aq)
      2. Ionic rxn:  HF(aq) + K+(aq) + OH-(aq)  g H2O(l) + K+(aq) + F-(aq)
      3. Net ionic rxn:  HF(aq) + OH-(aq)  g H2O(l) + F-(aq)
      4. When the rxn is neutralized (all reactants turned into products) the pH >7.0 because the KF salt is basic
    3. Strong acid weak base:  reaction goes to products all the way since one reactant completely ionized
      1. Molecular rxn: HBr(aq) + NH3(aq)  g NH4Br(aq)
      2. Ionic rxn: H+(aq) + Br -(aq) + NH3(aq)  g NH4+(aq) +  Br -(aq)
      3. Net ionic rxn:H+(aq) + NH3(aq)  g NH4+(aq) 
      4. When the rxn is neutralized (all reactants turned into products) the pH <7.0 because the NH4Br salt is acidic
    4. Weak acid weak base:   reaction does NOT go to products all the way since no reactant is completely ionized
      1. Molecular rxn: HF(aq) + NH3(aq)  g NH4F(aq)
      2. Ionic rxn: HF(aq) + NH3(aq)  g NH4+(aq) +  F -(aq)
      3. Net ionic rxn:  none as nothing cancels out - no spectator ions
      4. When the rxn is neutralized (all reactants turned into products) the pH is around 7.0 because the NH4F salt is acidic and basic
    5. Try example 16.1 and problem 16.1
  2. Common Ion Effect, (section 16.2)
    1. Consider a solution that begins with hydrofluoric acid AND sodium fluoride in water.  Always consider one way reactions first:  NaF (s) g Na+ (aq) + F- (aq).  As this reaction is one way (due to all sodium compounds being completely soluble) the [F-] final = [NaF] initial.  So we have fluoride ions and sodium ions in water.  Now consider the other reaction:  HF D H+ (aq) + F- (aq).  This reaction sets up equilibrium.  So you complete an ICE table.  The initial [H+] is zero but the initial [F-] is what you had from NaF dissociation.  So we are starting with fluoride ions initially.
    2. Look at this from a Le Chatelier's perspective.  Since we are starting with fluoride ions already in solution, we don't need to make as many to reach equilibrium.  So the reaction will proceed to the right LESS than it would without the fluoride ions present initially.  Thus LESS HF dissociates, the common ion suppresses the weak acid dissociation. The solution is LESS acidic with the NaF present.
    3. When we have a weak acid (HF) and the conjugate base (F-) in solution we can use the Henderson Hasselbach equation (section 16.4) also called the buffer equation:  pH = pKa + log [c. base / weak acid]  This can only be used if we have a conjugate pair initially present!!!
    4. Example:  Calculate the pH of a solution containing 0.15M acetic acid and 0.13M potassium acetate. Given Ka for acetic acid is 1.8 x 10-5.  There are two methods for solving this problem:  the traditional ICE table method or the Henderson Hasselbach equation.  Let's compare both.
      1. ICE table:  First write the reactions considering the one way first.  
        1. All potassium compounds are soluble so KCH3COO  g K+ (aq) + CH3COO-(aq).  This reaction is one way.  All the potassium acetate dissolves and dissociates into ions.  So we end up with 0.13M acetate ions CH3COO-.  We also have 0.13M potassium ions but we don't care since these ions are neutral and don't contribute to pH.  
        2. Second consider the acetic acid in water (weak acid!) CH3COOH  D H+(aq) + CH3COO-(aq).  Since this is an equilibrium reaction, set up an ICE table.  Initial acid is 0.15M, H+ is 0, acetate ion is 0.13M from the potassium acetate.  The change is - x, + x and + x.  So the equilibrium concentrations are 0.15 - x, x, and 0.13 + x.  
        3. Solve using Ka = (x)(0.13 + x) / (0.15 - x) and approximate that x is terribly small so when we add or subtract it with another number it won't matter.  Now Ka = (x)(0.13)/ (0.15) = 1.8 x 10-5.  
        4. Solve for x  and get 2.077 x 10-5.  Check the approximation:  2.077 x 10-5 / 0.15 x 100% = less than 5% so the approximation is good.
        5. Now x = the [H+] so get pH = -log x = 4.68
      2. Now try the Henderson Hasselbach equation.  You must RECOGNIZE that acetic acid and potassium acetate are a conjugate acid base pair.  The acetate ion is the conjugate base of acetic acid. Potassium is just a spectator ion. We also need pKa which is -log Ka = 4.74.  Now use the buffer equation:  pH = 4.74 + log [0.13 / 0.15] = 4.68
    5. The procedure is the same if you have a weak base and the conjugate acid - just remember the buffer equation needs pKa not pKb.  But you can find Kb given Ka since KaKb = Kw
    6. For fun what is the pH of 0.15 M acetic acid alone in water???  Well since we don't have the conjugate base present initially we can NOT use the Henderson Hasselbach equation and must solve by ICE tables.  
      1. CH3COOH  D H+(aq) + CH3COO-(aq).  Since this is an equilibrium reaction, set up an ICE table.  Initial acid is 0.15M, H+ is 0, acetate ion is 0.  The change is - x, + x and + x.  So the equilibrium concentrations are 0.15 - x, x, and  x.  
      2. Solve using Ka = (x)(x) / (0.15 - x) and approximate that x is terribly small so when we add or subtract it with another number it won't matter.  Now Ka = (x)(x)/ (0.15) = 1.8 x 10-5.  
      3. Solve for x  and get 1.643x 10-3.  Check the approximation:  1.643x 10-3 / 0.15 x 100% = less than 5% so the approximation is good.
      4. Here x = the [H+] so get pH = -log x = 2.79
      5. WOW  the pH of the acid alone is much lower than with the common ion acetate present from potassium acetate in the above example.  It's TRUE - the common ion suppresses weak acid ionization - less acid dissociates when a common ion is present initially.  With less dissociation, there is less H+ so the solution is less acidic and the pH is not as low.
    7. Try example 16.2-3 and problems 16.3-5
  3. Buffers (section 16.3)
    1. A buffer is a mixture initially containing a weak acid and its conjugate base (or a weak base and its conjugate acid).  A buffer resist a change in pH meaning it can "fight" off the addition of an external acid or base and not have the pH change much.  This is important when we need a solution to stay at the same pH such as body fluids, sensitive environments, etc.
    2. How does it work?  Well when an external base is added the buffer's weak acid reacts with it till it is gone.  When an external acid is added the buffer's weak base reacts with it till it is gone.  We need a weak acid/base pair so they SET UP equilibrium - thus the equilibrium just shifts back and forth when stressed.  Also a weak acid does not react with its own conjugate base - they set up equilibrium.  A strong acid or strong base can NOT be used as a buffer because they don't set up equilibrium!  Don't you tell me HCl or NaOH can be part of a buffer system.
    3. Examples:  Which pairs can be buffers?   HCl and HF,   HF and KF,   NH3 and NH4Cl,   H2CO3 and KHCO3,   CH3COOK and CH3COOH, HNO3 and NaNO3?  Answer:  All but HCl and HF and HNO3 and NaNO3.
    4. For a buffer system we can find the pH by using the buffer equation presented above:  pH = pKa + log [base / acid]
    5. Calculate the pH of a 0.105M KF and 0.255M HF solution.  Given Ka for HF is 7.1 x 10-4.  You must RECOGNIZE that HF is a weak acid and soluble KF contains F- ions which is the conjugate base.  Sub into the buffer equation.   pH = 3.15 + log [0.105 / 0.255] = 2.76.  Why is this pH below the pKa???  Because there is MORE acid than base in this buffer solution, so the pH will be more acidic than the pKa.  
    6. How will this buffer work?  Well if NaOH (strong base) were added, HF would react and neutralize the NaOH by making water and NaF (the conjugate base which is part of the buffer already so no big change in pH).  If HCl (strong acid) were added, F- would react and neutralize the HCl by making chloride ions and HF (the weak acid which is already part of the buffer so no big change in pH)
    7. The BEST buffer is when both the weak acid and base are present in roughly equal molar amounts - so that you have enough of both to keep the pH stable.  Look at what happens to the buffer equation when the concentrations are exactly equal!  If [weak acid] = [c base] then pH = pKa + log [c base] / [weak acid] simplifies to pH = pKa + log 1  and since log of one is zero we get pH = pKa.  But that is only true when the concentrations are equal.  So a buffer's pH is usually around the pKa.
    8. Calculations for when acid or base is added to a buffer:
      1. What happens if 10.0 mL of 0.50M HCl is added to a buffer system of 500.0mL of 0.10M formic acid (HCHO2) and 500.0mL of 0.20 sodium formate (NaCHO2)?  Ka for HCHO2 is 1.8 x 10-4. 
        1. First consider the one way reaction:  HCl in water will ionize 100%.  How many moles of HCl are being added?  0.0100L (0.50 mol/L) = 0.0050 moles HCl which will give 0.0050 moles H+ and Cl- which is a spectator ion
        2. Sodium formate is a salt which is soluble:  It will dissociate 100%. Na+ is also a spectator ion. How many moles of the c. base are there?  0.5000L (0.20mol/L) = 0.10 mol CHO2-(aq)
        3. How many moles of the weak acid are there?  0.5000L (0.10mol/L) = 0.050 moles HCHO2
        4. Now how will this effect the buffer?  The conjugate base (formate ion) will neutralize the added HCl.  
            H+(aq)    +    CHO2-(aq)     g     HCHO2
          initial 0.0050 mol 0.10 moles 0.050 moles
          change  - 0.0050 - 0.0050 + 0.0050
          final 0.0050 - 0.0050 = 0 moles (limiting) 0.10-0.0050=0.095 moles 0.05 + 0.0050=0.055 moles
        5. We use moles in the table since the volume is changing by the addition of HCl. 
        6. Now calculate the resulting molarities after the addition of HCl.  The volume is now 1010.0 mL.  [CHO2-] = 0.095 moles / 1.010L = 0.094M and [HCHO2] = 0.055mol / 1.010L = 0.054M
        7. Now calculate the pH of this buffer system with the new molarities.  You can do an ICE table for the acid in water reaction or Henderson Hasselbach.  pH = 3.745 + log [0.094 / 0.054] = 3.99
        8. What was the starting pH of this buffer?  pH = 3.745 + log [0.20 / 0.10] = 4.05
        9. WOW the addition of strong acid barely changed the pH - that's a buffer!
        10. Try problems 16.6 - 8
    9. How to select your buffer?  Look at a table with several weak acids and their Ka values.  Calculate pKa values.  Find the closest pKa value to your desired pH.  Consider these:  HF with Ka = 7.1 x 10-4 and NH4+ with Ka = 5.6 x 10-10.  Calculate pKa values:  HF is 3.15 and ammonium ion is 9.25.
      1. Make a solution with a pH of 9.25.  This is exactly pKa for ammonium ion.  So add equal moles to water of NH4Cl for example and NH3.
      2. Make a solution with a pH of 2.95.  HF is close but not exact.  Use the buffer equation with your desired pH.  2.95 = 3.15 + log [NaF / HF].  Solve for the ratio [NaF / HF] = 0.631.  So add molar amounts in that ratio.  ie if you add 2 moles NaF you need 3.17 moles HF because 2 / 3.17 = 0.631. This makes sense because the desired pH (2.95) is less than (more acidic) than 3.15 so you need a bit more of the acid than the conjugate base for this buffer solution.  
      3. Example 16.4-5 and problems 16.9-11
  4. Henderson Hasselbach equation - already introduced above 
  5. Titrations (section 16.5)
    1. Quantitative procedure for adding an acid to a base (or vice versa) usually for the purpose of neutralizing the solution or identifying an unknown concentration or unknown chemical.  Very useful with several applications.  Sometimes used for other types of reactions as well like redox.  Remember that neutralization is complete at the equivalence point, which is where moles H+ = moles OH-. 
    2. When plot pH versus mL of base looks like this: 
  6. Strong Acid + Strong Base Titration (section 16.6)
    1. Let's have strong acid in the flask and strong base in the burette. What's in the burette is called the titrant.
    2. Point A is where we have only the strong acid in the flask before adding any base.  Do a normal pH calculation where [H+] = the concentration of the acid since it is a strong acid which dissociates completely:  HA  g H+ + A-
    3. Point B is the equivalence point:  it is the middle of the steep part of the curve and can be found graphically.  In this titration the pH at the equiv. pt. will be 7.0 because the product of the strong acid plus strong base is neutral ions and water.  So the pH is going to be neutral.  To find the volume of titrant needed use stoichiometry just like you did in first semester.
    4. Area C is between points A and B.  In this area not enough base has been added yet to react with all the acid.  So there is still unreacted acid in the flask, but not as much as we started with.  However much base has been added has reacted with acid.  So figure out how many moles of base have been added by multiplying base molarity times base volume which gives you moles of base.  That same number of moles of acid have been reacted.  So subtract that from the starting moles of acid to get the moles of acid left in the flask.  Divide the moles of acid remaining by the total volume at this point (volume of acid started with plus volume of base added) and that will get you the concentration of strong acid that is now left in the flask.  Then do a normal pH calculation where [H+] = the concentration of the acid remaining in the flask. 
    5. Area D is after the equivalence point B.  Here all the acid is reacted - all we have left in the flask is the excess base.  Figure out the total moles of base added.  Figure out the moles of acid started with.  Subtract to get the excess moles of base (that which was not reacted with the acid).  Divide the moles of unreacted base by the total volume.  That gets you [OH-] and you can get pOH, then get pH.
    6. Here is a great problem worked out.  Click on the mL of titrant to see the pH calculated for that volume of base.  Try several calculations. Try also problems 16.13-14
  7. Weak Acid + Strong Base (section 16.7)   
    1. Let's have weak acid in the flask and strong base in the burette. What's in the burette is called the titrant.
    2. Point A is where we have only the weak acid with no added base.  Do a normal pH calculation for a weak acid:  HA  D  H+ + A-  Set up your ICE table and use Ka to solve for [H+] then get pH.  
    3. Point B is the equivalence point:  it is the middle of the steep part of the curve and can be found graphically.  In this titration the pH at the equiv. pt. will be above 7.0 because the product of the weak acid plus strong base is a basic salt and water.  HA + NaOH  g H2O + NaA.  (NaA contains A- ions which are the conjugate base of the weak acid) This reaction is one way since we have a strong base despite the weak acid.  So the pH is going to be basic at the equiv pt.  To find the volume of base needed to reach the equiv pt use stoichiometry.  Now calculate the moles of weak acid that was in the flask - they are all gone and have been converted to the basic conjugate base ion.  So you know the moles of the basic product ion A-.  Divide by the total volume and you will have the [A-].  Now since this ion is a weak base it will react with water so you have this second reaction:  A- + H2O  D  HA + OH- which is a base reaction and Kb goes with it. So set up an ICE table, use Kb and solve for [OH-].  Get pOH and pH.  
    4. Area C is between points A and B. It is called the buffer zone because we have a mix of weak acid HA and weak conjugate base A-.  In this area not enough of the strong base has been added yet to react with all the weak acid.  So there is still unreacted weak acid in the flask, but not as much as we started with.   HA + NaOH  g H2O + NaA    However much base has been added has reacted completely with acid.  So figure out how many moles of base have been added by multiplying base molarity times base volume which gives you moles of base.  That same number of moles of acid have been reacted.  So subtract that from the starting moles of acid to get the moles of unreacted acid left in the flask. Also find the moles of A- formed since it is the conjugate base.  Divide the moles of unreacted acid remaining by the total volume at this point (volume of acid started with plus volume of base added) and that will get you the concentration of weak acid that is now left in the flask unreacted. Also find the [A-] by dividing the moles of A- by the total volume. Now we have a buffer solution so find the pH using the buffer equation.  
    5. Area D is after the equivalence point B.  Here all the weak acid is reacted - all we have left in the flask is the excess strong base and some conjugate weak base A-.  Figure out the total moles of strong base added.  Figure out the moles of acid started with.  Subtract to get the excess moles of strong base (that which was not reacted with the acid).  Divide the moles of unreacted strong base by the total volume.  That gets you [OH-] and you can get pOH, then get pH.  We ignore the weak conjugate base A- because the pH is largely defined by the strong base from the burette.  The small amount of base from the A- is negligible. So we will ignore the salt hydrolysis: (  A- + H2O  D  HA + OH-  )  because the hydroxide made in this rxn is very small compared to the hydroxide ion from the strong base.
    6. Here is a great problem worked out.  Click on the mL of titrant to see the pH calculated for that volume of base. Note they do not use the buffer equation - but do it all with ICE tables.  You can do either method. Try problems 16.15-16
  8. Strong Acid + Weak Base (section 16.8) 
    1. very similar to weak acid + strong base.  Same principles apply.  
    2. A note about acid base indicators:  Figure 16.7
      1. Indicators are used to find the equivalence point when a pH meter is not available.  The equivalence point is when the moles of H+ from the strong or weak acid = moles of OH- from the base.  The equivalence point is only at pH of 7 when it is a strong acid + strong base since the product salt is neutral.  The equivalence point is less than 7 for a strong acid + weak base because the product salt is acidic (conjugate of the weak base).  The equivalence point is more than 7 for a weak acid + strong base because the product salt is basic (conjugate of the weak acid).  Graphically the equiv pt is the middle of the steep part of a titration curve.
      2. Indicators are weak acids than have a different color from their conjugate base.  So they turn different colors during a titration.  They usually have a small range of pH where they change color.  Phenolphthalein turns color around 8.3 to 10.  It is good for titrations with a strong base so the equiv pt is at pH 7 or higher.  It is not good for weak base titrations as the equiv pt is below 7 and too far from it's color range pH.  You always pick an indicator that will change color near the equivalence point of a titration.
  9. Polyprotic Acid Titrations (section 16.9)
    1. When titrating a polyprotic acid there is an "equivalence point" for each H that comes off.  Consider the graph for a diprotic acid. 
    2. Point A is where the diprotic acid acid is in water alone at equilibrium:  H2A  +  H2O    HA-  +  H3O+
    3. Point B is where the first H is coming off with the addition of base and pKa1 where 50% is H2A and 50% is HA-
    4. Point C is the first equivalence point where we have HA-
    5. Point D is pKa2 where 50% is HA- and 50% is A2- and the second H is coming off
    6. Point E is the second equivalence point where we have A2- and all the acid is reacted
    7. Point F is where we have no acid remaining and only excess base
  10. Solubility (section 16.10)
    1. Solubility rules tell us which ionic compounds are soluble in water (dissolve completely - ionize 100% - compounds with Na+, K+, NO3-, NH4+...and others) and which are insoluble in water (do NOT dissolve completely, ionize only a small amount).  For those that are insoluble, the ions that do dissociate set up equilibria with the undissolved solid.  The reaction is always solid(s)  D  ion(aq) + ion (aq) and the equilibrium constant that goes with this type of equilibria is Ksp for solubility product. Remember solids are not in K expressions, so there is nothing in the denominator as the reactant is always a solid for Ksp problems. This makes calculations easy!!!
    2. There are 3 stages to adding a solid to solvent
      1. unsaturated - when more solid can dissolve in the solution, the [ion] is not at the max yet
      2. saturated - exactly as much solid that can dissolve has been added, no more can dissolve, the [ion] is the max
      3. past saturation - more solid than can dissolve has been added, thus some undissolved solid is sitting on the bottom of the flask in equilibrium with the ions, the [ion] is at the max
    3. Ksp expressions
      1. Write the Ksp expressions for calcium phosphate.  Answer:  you must know the formula first of all:  Ca3(PO4)2  Now write the reaction:  Ca3(PO4)2 (s)  D  3 Ca2+(aq) + 2 PO43-(aq)   And finally write Ksp = [Ca2+]3[PO43-]2
      2. Example 16.7 and problem 16.20
  11. Solubility (section 16.11) Solubility = max grams of a compound that can dissolve in 1.00 L solvent (g/L).  Molar solubility  = max moles of a compound that can dissolve in 1.00 L solvent (mol/L)  You must be able to calculate Ksp from solubility and vice versa! You should also be able to calculate Ksp from concentrations and vice versa. 
    1. A saturated solution contains [Ca2+] = 3.15 x 10-3M and [SO42-] = 1.95 x 10-3M.  Calculate Ksp for calcium sulfate.  First write the Ksp reaction for calcium sulfate:  CaSO4(s)  D  Ca2+(aq)  +  SO42-(aq)   So Ksp = [Ca2+] [SO42-] = 6.14 x 10-6
    2. The solubility of calcium hydroxide is 0.233 g/L.  Calculate Ksp.  First change to molar solubility:  0.233 g / L ( 1 mol / 74.1 g) = 3.14 x 10-3 mol/L Ca(OH)2.  Now write the appropriate rxn:  Ca(OH)2(s)  D Ca2+(aq) + 2 OH-(aq).  Set up an ICE table.  Initial concentrations in order are -----,    0,    0.  Change is   - x,       + x,      + 2x.   Equilibrium concentrations are ------,      x,     2x.  The molar solubility is the amount of solid that dissolved and is the -x in the ICE table.  So x = 3.14 x 10-3M.  Solving for Ksp = [Ca2+][OH-]2 = (x)(2x)2 = 4x3 = 1.24 x 10-7.
    3. Ksp for silver bromide is 7.7 x 10-13.  Calculate solubility.  First write the rxn.  AgBr(s) D  Ag+(aq) + Br-(aq).  Fill in the ICE table with initial concentrations:  ---,      0,      0.  Change    -x,       +x,       +x.      Equilibrium concentrations    -----,     x,    x.  Ksp = x2 = 7.7 x 10-13. Solving x = 8.77 x 10-7M  which is molar solubility as x is the amount of solid AgBr that dissolved.  Change to just solubility:  8.77 x 10-7mol / L (187.8 g / mol) = 1.6 x 10-4 g/L.
    4. Examples 16.8-10 and problems 16.21-24
  12. Common Ion Effect on Solubility (section 16.12)
    1. What happens if solid Na2S is put into a saturated CdS solution?  Well before we have some Cd2+ ions and S2- ions in solution with the maximum dissolved possible (saturated).  If I add soluble Na2S, I am adding Na+ ions and S2- ions to the already saturated solution.  As I add the common ions S2- which is already at it's max solubility in the flask, some S2- will recombine with Cd2+ and precipitate out as CdS solid.  Consider the equilibrium reaction:  CdS(s) D  Cd2+(aq) + S2-(aq).  As sulfide is added, the rxn shifts to the left and solid precipitates.  Thus lowering the solubility of CdS in this case.  Adding a common ion decreases solubility.  
    2. What is the concentration of silver ion and bromide ion if 0.0244 g of AgBr and 0.0111 g of NaBr is put into 1.00 L of water? Ksp for AgBr is 7.7 x 10-13.  Always consider the one way reaction first.  NaBr dissolves all the way so we will have 0.0111 g ( mol / 102.9 g) = 1.08 x 10-4 moles / L of sodium and bromide ions.  Now consider AgBr.  0.0244g ( mol / 187.8 g) = 1.30 x 10-4 moles / L of silver and bromide ions IF it all dissolved.  That's a big if.  Could it all have dissolved???  Let's see:  AgBr(s) D  Ag+(aq) + Br-(aq). Q = [Ag+][Br-] = (1.30 x 10-4)(1.30 x 10-4 + 1.08 x 10-4) = 3.09 x 10-8 which is larger than the real Ksp so not all of that CAN actually be dissolved.  So we need to set up an ICE table and solve for the real amount AgBr that dissolved:  AgBr(s) D  Ag+(aq) + Br-(aq). Initial is ---, 0, 1.08 x 10-4  Change is -x, +x, +x.  Equilibrium concentrations is ---, x, x +1.08 x 10-4.  Ksp = (x)(x +1.08 x 10-4) = 7.7 x 10-13 = (x) (1.08 x 10-4 )  using approximation.  Solving for x = 7.13 x 10-9M.  So [Ag+] = 7.13 x 10-9M and [Br-] = 1.08 x 10-4 M.  Less dissolved than normal with the addition of the common ion.  
    3. Example 16.11 and problem 16.25
    4. pH and Solubility 
      1. Weak base metal hydroxides are insoluble, which is why they are weak bases - they hardly dissolve at all in water making very little hydroxide ions.  Compare to strong base metal hydroxides (NaOH, KOH, LiOH) which dissolve completely making 100% hydroxide ions.  Insoluble Basic salts also set up equilibrium with their basic ion. 
      2. Consider this insoluble weak base Mg(OH)2 has Ksp = 1.2 x 10-11
      3. consider the rxn that goes with Ksp:  Mg(OH)2 (s)   D Mg2+(aq) + 2 OH-(aq).  As pH increases [OH-] increases so the rxn shifts left towards the solid - less soluble.  As pH decreases [OH-] decreases so the rxn shifts right and more dissolves - more soluble.  
      4. You should be able to explain why these weak bases are more soluble in acidic solution and less soluble in basic solution. Problem 16.26
    5. Complex Ions 
      1. A complex ion is usually a transition metal with anions or molecules bonded around it and it has a charge.  
      2. Examples:  CoCl42-,  Cu(NH3)42+,   ReH72-,   Cu(H2O)62+
      3. Complex ions are in equilibrium with their metal ion and the species bonded to them:  The generic rxn looks like this:  M2+ (aq)  +  6L (aq)  D  ML62+ (aq).  Of course the charge on the metal and number of "L" varies, but you get the idea.  One specific example:  Co3+(aq) + 6 NH3 (aq)  D Co(NH3)63+ (aq).  
      4. Kf goes with this type of reaction for complex ions.  Kf is the formation of complex ions. Kf values are typically very large (10+30 and such)  Huge numbers meaning the equilibrium lies far, far on the product side favoring the complex ion. 
      5. The larger the Kf, the more stable the complex ion.  
    6. Skip the amphoterism paragraphs
  13. Precipitation - section 16.13
    1. A precipitate (solid) will form once the ions reach their maximim concentration (once Ksp is passed) Review precipitations reactions. 
    2. Note the text uses "IP" for "Q" but we will just call is Q like the chapters before
    3. Use Q to see if a solution is unsaturated, saturated, or past saturation thus precipitate is present
      1. Q is calculated the exact same way as Ksp its just we are not sure if we really are at equilibrium.  If Q < Ksp the solution is unsaturated, if Q = Ksp the solution is saturated, if Q > Ksp the solution is past saturated.
      2. 2.45 mg of magnesium carbonate is put in 1.00 L of water.  Did it all dissolve?  Or can more dissolve?  Ksp for MgCO3 is 4.0 x 10-5.  Write the rxn:  MgCO3(s)  D  Mg+2(aq) + CO32-(aq).  Figure out the [ion] if it all dissolved.  2.45 mg ( 1 g / 1000mg)(1 mol / 84.3 g) = 2.91 x 10-5 mol MgCO3.  Divide by liters (1.00) = 2.91 x 10-5M MgCO3.  If it all dissolved the [Mg2+] = [CO32-] = 2.91 x 10-5M.  So now Q = [Mg2+] [CO32-] = (2.91 x 10-5M)2 = 8.47 x 10-10.  Well this is smaller than the real Ksp so yes, it all dissolved and even more can dissolve.  It can dissolve until Ksp is reached, then no more will dissolve and solid will form.
    4. Example 16.14 and problems 16.29-30
  14. Skip section 16.14
  15. Qualitative Analysis - Read for lab. CHM 154LL students definately need to read this section for their 154LL experiments. 

 

Final Remarks!  Important! 

  1. You must be able to recognize if a chemical is a strong or weak acid, strong or weak base, or acidic, basic or neutral salt.  You must be able to write the reactions of them in water and know if the reaction is one way or sets up equilibrium. 
  2. A strong acid is not identified by pH or concentration.  A strong acid is one of the six acids that dissociate completely.  Just because the pH is 1.5 does not mean it is a strong acid - it could be a very concentrated weak acid!  Just because a pH is 5 does not mean a weak acid - it could be a very dilute strong acid.  Strong means ionize 100%.  All pH tells you is the concentration of H+ but you don't know where that H+ came from unless you are told the acid's name.
  3. Recognizing neutral, basic and acidic ions. 
    1. Neutral ions come from strong acids and bases:  Na+, K+, Li+, Cl-, Br-, I-, NO3-, ClO4-
    2. Basic ions come from weak acids:  CH3COO-, F-, HCO3-, CO32-,
    3. Acidic ions come from weak bases:  NH4+
  4. In water would these make neutral, basic or acidic solutions?
    1. KF - basic
    2. NaNO3 - neutral
    3. KCl - neutral
    4. NH4Cl - acidic
    5. Li2CO3 - basic
    6. NH4F - both so figure they cancel out and are roughly neutral