 # Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria.

## Presentation on theme: "Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria."— Presentation transcript:

Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria

Solubility Equilibria All ionic compounds dissolve in water to some degree We can apply the concepts of equilibrium to salts that dissolve to a small extent (“insoluble”) We can use the equilibrium constant for the process to measure relative solubilities in water

Solubility Product Equilibrium constant for the dissociation of a solid salt into its aqueous ions: solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s)  nM m+ (aq) + mX n− (aq) The solubility product would be K sp = [M m+ ] n [X n− ] m

Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature (in grams or moles) The molar solubility is the number of moles of solute that will dissolve in a liter of solution. The molarity of the dissolved solute in a saturated solution.

Calculating K sp from Solubility 1. Copper (I) bromide has a measured solubility of 2.0 x 10 -4 M at 25˚C. Calculate K sp. 2. Calculate K sp for bismuth sulfide (Bi 2 S 3 ) which has a solubility of 1.0 x 10 -15

Calculating Solubility from K sp The Ksp for copper (II) hydroxide is 1.6 x 10 -19 at 25˚C. Calculate the solubility, [Cu 2+ ], [OH - ].

K sp and Relative Solubility Molar solubility is related to K sp However, you cannot always compare solubilities of compounds by comparing their K sp ’s In order to compare K sp ’s, the compounds must have the same dissociation stoichiometry

The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt For example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) addition of Cl − shifts the equilibrium to the left

1.Calculate the solubility of Ag 2 CrO 4 in pure water (Ksp = 9.0 x 10 -12 ) 2.Calculate the solubility of Ag 2 CrO 4 in a 0.1M solution of AgNO 3

The Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH (and the greater [OH - ], the lower the solubility of the ionic hydroxide OH - acts as a common ion, M(OH) n (s)  M n+ (aq) + nOH − (aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s)  2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq)  HCO 3 − (aq) + H 2 O(l)

Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur Q = K sp, the solution is saturated, no precipitation Q < K sp, the solution is unsaturated, no precipitation Q > K sp, the solution would be above saturation, the salt above saturation will precipitate Some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions

precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added

Selective Precipitation A solution containing several different cations can often be separated by addition of a solution that will form an insoluble salt with one of the ions, but not the others

If 10.0 mL of 0.50 M Pb(NO 3 ) 2 and 20.0 mL 1.0 M NaI are mixed, will a precipitate form?

Chapter 18 Electrochemistry

Redox Reaction Elements change oxidation number  e.g., single displacement, and combustion, some synthesis and decomposition Oxidation--oxidation number increases Reduction--oxidation number decreases  Both must occur in a reaction--two half reactions oxidizing agent is reactant molecule that causes oxidation  contains element reduced reducing agent is reactant molecule that causes reduction  contains the element oxidized

Rules for Assigning Oxidation States 1. Free elements have an oxidation state = 0 2. Monatomic ions have an oxidation state equal to their charge. 3. The sum of the oxidation states of all the atoms in a compound is 0. 4. The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion. 5. The oxidation number of fluorine is always -1 in compounds with other elements.

Rules for Assigning Oxidation States 6. Chlorine, bromine and iodine always have oxidation numbers of -1 except when bonded to O or F. 7. The oxidation number of oxygen is almost always -2; the oxidation number of hydrogen is almost always +1. Exceptions: --When oxygen is in the form of a peroxide (O 2 2- ), the oxidation number is -1. --When hydrogen forms a binary compound with a metal, the oxidation number is -1 and the compound is called a hydride.

Oxidation and Reduction Oxidation occurs when an atom’s oxidation state increases during a reaction Reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O -4 +1 0 +4 –2 +1 -2 oxidation reduction Reducing agentOxidizing agent

Identify the element that is oxidized and the element that is reduced in each of the following reactions. What is the oxidizing and the reducing agent in each reaction? 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O

Common Oxidizing Agents

Common Reducing Agents

Balancing Redox Reactions 1. Assign oxidation numbers --determine element oxidized and element reduced 2. Separate the reaction into oxidation and reduction half- reactions. 3. Balance half-reactions by mass a. First balance elements other than H and O b. Balance O using H 2 O c.Balance H using H + 4. Balance each half-reaction by charge by adding electrons to the reactants side of the reduction and the product side of the oxidation. 5. Multiply half-reactions by integers to make # electrons the same in both half-reactions 6.Add half-reactions and cancel the electrons to produce a balanced equation. 7.For reactions that occur in acidic solutions, skip to step 9. 8.For reactions that occur in basic solutions, add the same # of OH - as H + to both sides of the equation. 9. Check that reaction is balanced for mass and charge.

Download ppt "Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria."

Similar presentations