AP Chemistry Chapter 4 1
Properties of Aqueous Solution Solutions Definition: Any substance (solid, liquid or gas) EVENLY distributed throughout another substance. Solutions have 2 parts: 1) Solvent the substance doing the dissolving 2) Solute the substance being dissolved Question: What does DISSOLVE mean? Answer: A process by which a solvent separates the components of a solute 2
*NOT A CHEMICAL REACTION* The process of dissolving is called Solvation. An important kind of solution is called an aqueous solution. Aqueous Solution A solution where H 2 O is the solvent. 3
Electrolytes and Nonelectrolytes Imagine that you are in a pool and it starts to rain and thunder. Why are you told? "Get out of the pool!" Lightening Electricity passes Electricity passes hits the through the through water water you Why??? The pool water conducts electricity!! 4
Compound that conducts an electric current in aqueous solution or the liquid state are ELECTROLYTES. All IONIC compounds are ELECTROLYTES. BaSO 4 (ionic) does not form an aqueous solution, but it is an electrolyte as molten BaSO 4. Compounds that do not conduct on electric current in either aqueous solution or in the molten state are NON-ELECTROLYTES. 5
Types of Electrolytes 1) Weak Electrolytes - only a Fraction of the solute that dissolves exist as ions. 2) Strong Electrolytes - almost all of the solute that dissolves exists as ions. YouTube Video: https://www.youtube.com/watch?v=1xwn ovm6jls 6
Precipitation Reactions Precipitate = an insoluble solid that is formed by a reaction in solution. Question: What will happen if Pb(NO 3 ) 2 (aq) is combined with KI(aq)? Answer: A double replacement reaction occurs. (aka: exchange or metathesis reaction) Pb(NO 3 ) 2 (aq) + KI(aq) PbI 2 + KNO 3 (Not balanced!) 7
Oppositely charged ions attract strongly and form a solid precipitate. Within the Pb(NO 3 ) 2 (aq), exists Pb 2+ (aq) & NO 3 1- (aq) Within the KI(aq), exists K 1+ (aq) & I 1- (aq) 8
Solubility Rules 1. Salts containing Group I elements are soluble (Li +, Na +, K +, Cs +,Rb + ). Salts containing the ammonium ion (NH 4 + ) are also soluble. 2. Salts containing nitrate ion (NO 3 - ) are generally soluble. 3. Salts containing Cl -, Br -, I - are generally soluble. Important exceptions to this rule are halide salts of Ag +, Pb 2+, and (Hg 2 ) 2+. 4. Most silver salts are insoluble. 5. Most sulfate salts are soluble. Important exceptions to this rule include BaSO 4, PbSO 4, Ag 2 SO 4 and SrSO 4. 6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al 3+ are insoluble. 7. Most sulfides of transition metals are highly insoluble. Thus, CdS, FeS, ZnS, Ag 2 S are all insoluble. 8. Carbonates are frequently insoluble. Group II carbonates (Ca, Sr, and Ba) are insoluble. 9. Chromates are frequently insoluble. 10. Phosphates are frequently insoluble. 11. Fluorides are frequently insoluble. 12. Acetates are soluble. Exception: Al(C 2 H 3 O 2 ) 3 is slightly soluble. 9
What is the solid precipitate that has formed? PbI 2, KNO 3 OR Both? To predict, we must use a Solubility Table. What does the Solubility Table predict? NO 3 1- with any positive ion will be SOLUBLE (aqueous solution) Pb 2+ and I 1- WILL NOT be SOLUBLE (Forms a SOLID!) SO, PbI 2 is the SOLID! Ionic Equations Let s write the previous chemical reaction as it truly exist. Solutions are made of ions. 10
The reactants and products would exist as: 2K 1+ (aq) + 2I 1- (aq) PbI 2 (s) + + Pb 2+ (aq) + 2NO 1-3 (aq) 2K 1+ (aq) + 2NO 1-3 (aq) This is an Ionic Equation. Net Ionic Equation Question: What ions exist on both sides as ions? Answer: K 1+ (aq) and NO 3 1- (aq) They are SPECTATOR IONS. ("they just watch") If we drop out the spectator ions we are left with: 11
This, 2K 1+ (aq) + 2I 1- (aq) PbI 2 (s) + + Pb 2+ (aq) + 2NO 1-3 (aq) 2K 1+ (aq) + 2NO 1-3 (aq) Becomes this, Pb 2+ (aq) + 2I 1- (aq) PbI 2 (s) This is a Net Ionic Equation. Let s practice: 1) BaCl 2 (aq)+na 2 CO 3 (aq) 2) Pb(NO 3 ) 2 (aq)+(nh 4 ) 2 SO 4 (aq) 3) BaS(aq)+FeSO 4 (aq) 12
Acids, Bases and Acid/Base Reactions Arrhenius Definitions: Acid - species that produces H 1+ ion in H 2 O solution Base - species that produces OH 1- ion in H 2 O solution Acids Two types exist The difference % of dissociation Strong Acids - dissociates completely to make H + ions and anions. Example: HCl(aq) H 1+ (aq) + Cl 1- (aq) 13
Some common strong acids are: HI, HCl, HBr, HNO 3, HClO 3, HClO 4, H 2 SO 4 All other acids are considered to be weak. Weak acids - partial dissociation to H 1+ ions in H 2 O The general formula for the dissociation of weak acid is: HA (aq) H 1+ (aq) + A 1- (aq) HA, H 1+ and A 1- are all present in solution. 14
Bases Two types exist (strong and weak) Strong Bases - dissociate completely to make OH 1- (aq) ion and cations Some common strong bases are: 1. Group 1 metal hydroxides: LiOH, NaOH, KOH, RbOH, CsOH. 2. Heavy Group 2 metal hydroxides: Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 Example, NaOH(s) Na 1+ (aq) + OH 1- (aq) Weak Bases - Produce OH 1- differently! Does not go to completion. 15
Base + H 2 O Positive ion + OH 1- Example: NH 3 (aq) + H 2 O(l) NH 4 1+ (aq) + OH 1- (aq) Common class of weak bases called AMINES: contain the "N" atom Example: CH 3 NH 2 (aq) + H 2 O(l) CH 3 NH 3 1+ (aq) + OH 1- (aq) 16
ACID/BASE REACTIONS The outcome of acid/base reactions depends on the nature of the acid/base. 1) Strong acid / Strong base Example: NaOH (aq) & HCl(aq) NaOH(aq) Na 1+ (aq) + OH 1- (aq) HCl(aq) H 1+ (aq) + Cl 1- (aq) The ionic equation would be: Na 1+ (aq) + OH 1- (aq) Na 1+ (aq) + Cl 1- (aq) + + H 1+ (aq) + Cl 1- (aq) H 2 O(l) The net ionic equation would be: H 1+ (aq) + OH 1- (aq) H 2 O(l) Neutralization Reaction 17
2) Weak acid / Strong Base Two-step process: [acid dissociates] #1: HB(aq) H 1+ (aq) +B 1- (aq) [OH 1- neutralizes H 1+ ] #2: H 1+ (aq) + OH 1- (aq) H 2 O(l) HB(aq) + OH 1- (aq) B 1- (aq) + H 2 O(l) 3) Strong acid / Weak base Two-step process: #1: [dissolving of base] NH 3 (aq) +H 2 O(l) NH 4 1+ (aq) + OH 1- (aq) #2: [OH 1- neutralizes H 1+ ] H 1+ (aq) + OH 1- (aq) H 2 O(l) NH 3 (aq) + H + (aq) NH 4 + (aq) 18
Let s try one: HNO 2 + KOH Net ionic eqn? 19
Reacting Species Strong acid/ H 1+ /OH 1- Strong base Net Ionic Eqn. H 1+ (aq) + OH 1- (aq) H 2 O Weak acid/ HA/OH 1- A 1- (aq)+h 2 O HA(aq) + OH 1- (aq) Strong base Strong acid/ H + / B B(aq) + H 1+ (aq) HB 1+ (aq) Weak base Neutralization Reactions with Gas Formation H 2 S(g) Formation Acid + Sulfide salts 2HCl(aq) + K 2 S(aq) H 2 S(g) + 2KCl(aq) CO 2 (g) Formation Acid + carbonate Acid + hydrogen carbonate HCl(aq) + NaHCO 3 (aq) H 2 CO 3 (aq) + 2NaCl(aq) Then, H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) 20
OXIDATION REDUCTION REACTIONS Common - involves an exchange of electrons between 2 species Also known as REDOX reactions Many chemical reactions are actually redox reactions One species loses e- (oxidized) One species gains e- (reduced) The species that loses electrons The species that gains electrons OXIDIZED REDUCED 21
Example: Redox Rxn. Zinc(s) + HCl (aq)? YouTube Video: https://www.youtube.com/watch?v=whqu dwvoe9m The net ionic eqn: Reduced Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Oxidized Zn(s) Zn 2+ (aq) + 2e - (oxidation half-rxn) 2H 1+ (aq) + 2e - H 2 (g) (reduction half-rxn) Note: 1) No net in the # of e - 2) Oxidation and reduction must occur together. 22
Oxidation Number Charge of the element Rules to determine oxidation #: 1) The ox. # of an element in an elementary state = 0 Example: Na 0, O 2 0 2) Ox. # of an element in a monotomic ion = charge of that ion 3) Certain compounds always have the same ox. # (ex. Groups 1&2) * Many vary, some notable exceptions are: Oxygen: usually O 2-, but it can be O 1- as in H 2 O 2. 23
Hydrogen: Usually H 1+, but it can be H 1- as in NaH. 4) The sum of the ox. # in a neutral compound = 0 * In a polyatomic ion, the ox. # is equal to the charge of the ion. Example: What is the ox. # of "S" in K 2 SO 3? K +1 0=2((+1) + x + 3(-2) 0-2 0=+2 + x + -6 0= x - 4 +4 = x 24
Single Replacement Reactions AKA : Single Displacement Reactions This type of reaction occurs between a metal and either an acid or a metal salt. The general form of the reaction is: A + BC B + AC How to determine if the reaction will occur (or not) is by using the Activity Series. The Activity Series list metals in order of decreasing ease of oxidation. 25
The higher on the list the metal, the more easily the neutral metal will react and form an ionic compound. The lower on the list the metal, the less easily the neutral metal will react and form an ionic compound. Metal Activity Series 26
Molarity Concentration of Solutions Molarity (M) = moles of solute liters of solution Given the molarity and the volume, the moles of solute can be determined. Given the molarity and the moles of solute, the volume can be determined. If the [HCl] = 0.75M, a) Calculate the number of moles in 350. ml of this solution 27
b) What volume will hold 3.1 moles of HCl? Dilution Usually solutions are made by diluting a more concentrated solution. For example, if you needed a 0.5 M solution you could start with a 10 M solution and dilute it. How is this done? The main thing to remember when doing these calculations is that the number of moles of solute stays the same. 28
The number of moles of solute in the concentrated solution (moles conc )equals the moles of solute in the dilute solution(moles dil ). M moles conc = moles dil Remember that, mole volume moles M moles conc = moles dil volume M conc x volume conc = M dil x volume dil M conc x V conc = M dil x V dil Examples #1 What the concentration of a solution would be if it were made by diluting 50.0 ml of 0.40 M NaCl solution to 1000. ml? 29
Answer: M conc x V conc = M dil x V dil (50.0 ml)(0.40 M) = (M dil )( 1000. ml) 50.0mL 0.40M M dil 0. 020M 1000. ml Examples #2 A chemist wants to make 500. ml of 0.050 M HCl by diluting a 6.0 M HCl solution. How much of that solution should be used? Answer: M conc x V conc = M dil x V dil (6.0 M)(V conc ) = (0.050 M)(500. ml) 0.050 M 500. ml V conc 4. 2 6.0 M ml 30
Solution Stoichiometry Titrations Equivalence Point At what point are the acid and the base neutralized during a chemical reaction??? According to this acid-base reaction, HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 1 mole HCl(aq) will neutralize 1 mole NaOH(aq) 5 mole HCl(aq) will neutralize 5 mole NaOH(aq) etc The point at which just enough acid neutralizes the base is called the EQUIVALENCE POINT. The equivalence point of an acidbase reaction can be determined experimentation as well as by calculation. 31
Experimentally, the equivalence point is determined by the gradual addition (dropwise) of a base to an acid which place in a flask or beaker. When the equivalence point is reached, the amounts of both the acid and the base are measured. The unknown concentration of the base can be determined from the known concentration of acid (or vica-versa). This process is known as a TITRATION. YouTube Video: https://www.youtube.com/watch?v=g8jdc WC10vQ 32
During the neutralization reaction of HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) How can the equivalence point be determined? Acids, bases, salts and water all have no color. Therefore, the point at which: HCl disappears can not be seen. NaOH disappears can not be seen. Similar result for NaCl and H 2 O. Question : How can the equivalence point be seen? 33
Answer : An acid-base indicator is used. Changes color at a particular ph. A particular indicator is chosen based on the ph at which the equivalence point will be reached. A color will be observed. Titration Calculations Question #1: How many moles of H 2 SO 4 are required to neutralize 0.50 moles of NaOH? H 2 SO 4 (aq) +2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) Answer: 1 mole H 2 SO 4 0.50 mole NaOH = 0.25 mole H 2 SO 4 2 mole NaOH 34
Question #2: A 25 ml solution of H 2 SO 4 is neutralized by 18.0 ml of 1.0 M NaOH. What is the concentration of the H 2 SO 4 solution? H 2 SO 4 (aq) +2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) Answer: 18.0 ml = 0.0180 L 0.0180 L NaOH 1.0 mole NaOH 1 L NaOH 1 mole H 2 SO 4 2 mole NaOH = 0.0090 mole H 2 SO 4 mole 0.0090 moles Molarity = = = 0.36 M liters 0.025 L 35
Question #3: A 50.00 ml solution of KOH is titrated with a 0.8186 M HCl solution. The titration requires 27.87 ml of the HCl solution to reach the equivalence point. What is the molarity of the KOH solution? HCl(aq) +KOH(aq) KCl(aq) + H 2 O(l) Answer: 27.87 ml = 0.02787 L 0.02787 L HCl 0.8186 mole HCl 1 L HCl 1 mole KOH 1 mole HCl * * = 0.02281 mole KOH mole 0.02281moles Molarity = = = 0.4562 M liters 0.05000L 36