Chapter 2: Acids and Bases- Acid-Base Rxn: proton is transferred from an acid to a base- Bronsted-Lowry Acid: proton donor (HB)- Bronsted-Lowry Base: proton acceptor (A-)- Conjugate Base: species formed when a proton is removed from an acid (B-)- Conjugate Acid: species formed when a proton is added to a base (HA)HB (aq) + A- (aq) B- (aq) + HA (aq)- Arrhenius Acid: material that donates a proton (H+)- Arrhenius Base: material that donates a hydroxide ion (OH-)o Base= alkali- Lewis Acid: accepts an electron pair from a LB- Lewis Base: donates an electron pair to a LA- Acidic soln pH < 7- Basic soln pH > 7- Strength of an acid or base depends on the extent of ionization in HOHo Strong Acids: HCl, HBr, H2SO4, HNO3, HI, HClO4o Strong Bases: NaOH, KOH, LiOH, CsOH, Mg(OH)2, Ca(OH)2, Ba(OH)2o Weak acids and bases ionize to a lesser extent in HOH  Weak Acids: solutes that react reversibly w/ HOH to form H3O+ ions Weak Bases: solutes that react w/ HOH to acquire a H+ ion and leave OH- behindo Many org acids and bases are insoluble in HOH, so a solvent other than HOH is required to put them into soln. Solvents are typically another org compound- Ex: diethyl ether, hexane, dichloromethane  A very weak acid can react w/ a particularly strong base, and a very weak base can react w/ a particularly strong acid.- Mineral Acid: any inorganic acid (MA) o Many of the strong MA’s are insoluble in org. solvents So mixes of HOH and org solvent are used- The strength of an acid depends on the bases w/ which it reactso Critical for est-ing whether the equilibrium lies to the L or the R- An acid reacts w/ a base, regardless of solvent, but the extent of ionization changes w/ the solvent used and w/ the base- Ka= [Conjugate Acid] [Conjugate Base] [Acid] [Base]o In gen chem, the HOH (base) is removed from the equation to simplify equations, however, when HOH is removed from the K expression, this also removes the base from the acid-base equilibrium. Therefore, if there’s a solvent other than HOH used, then the base must be put back into the expression.- Acids react w/ bases in order to ionize- What happens when HOH accepts a H+ to form H3O+?o A new sigma covalent bond is formed b/w the H atom and the O atomso One of the unshared electron pairs on O reacts w/ H+ to form the new bond to the protono Blue curved arrow indicates O donates 2 electrons to H+ to form new O—H bond The curved arrow formalism is used in chem rxns to indicate transfer electron density in order to form a new chem bond.o A proton (H+) has no electron density associated w/it, so it accepts the electron density from the electron-rich O atom. O = base H+ = acido Resulting H3O+ has 3 bonds to O, making it electron deficient, so it takes on a (+) charge.o Further in this rxn, one electron pair on O is donated to the electron deficient H+ to form a new bond. O accepts H+ by donating 2 electrons to form a new bond.Chapter 2: Acids and Bases- Bases may be defined as 2 electron donors to an electron deficient center.o If the base donates electrons to a H+, it’s a Bronsted-Lowry baseo If the base donates electrons to an atom other than H, it’s formally a Lewis baseo To donate electrons, a base must have an excess of electron density. Molecules containing O or N react as bases b/c they have excess electron density. The more easily an atom is able to donate electrons, the more basic that atom should be.o If all bases donate 2 electrons, the electron flow is from the base to the acid, rather than from the acid to the base. An acid doesn’t donate the proton, but rather the H+ is attacked by the base to form a new bond to the proton Electron flow is always from a source of high electron density to a point of low electron densityo A base is electron rich and will donate 2 electrons to an electron deficient atom, suchas H+, to form the CA. There are no free protons they come from an acid (such as HCl or HCOOH)o A base doesn’t react w/ a free proton, but rather w/ an electron deficient H atom that’s attached to another atom. When the base donates 2 electrons to the H atom, it literally pulls the base away. The bond b/w atoms will break as the 2 electrons from the base are used to form a new bond, such as O—H and the 2 electrons in the former bond will migrate toward the other atom (like Cl- from HCl) to form the CB.- Sometimes the acid-base rxn doesn’t work very well b/c the Ka favors the acid and base (left) rather than the CA and CB (right)o If the Ka for a given A-B rxn is unfav., changing the base may make Ka more fav. Changing the base to one that’s stronger for a given acid should shift Ka toward the CA and CB.- One base may be stronger than another, and one must be able to predict relative basicity.- If one acid has a pKa = 1 and the other pKa = 8, the first is the stronger acid while the second in the stronger base.o B/c pKa corresponds to pH- Why is HI more acidic than HF?o HI is a stronger acid indicates Ka is larger If Ka is larger, than pKa is smallero The bond strength of HI is smaller. The covalent radius of I is larger than F - So the bond distance b/w H and I will be longer than the bond distance b/w H and F. A longer bond is a weaker bond.- If the H—I bond is weaker, then it should break more easily when attacked by the baseo This pushes the rxn toward the right to favor the CA and CB Means than Ka is larger for the rxn w/ HI- A weaker bond is consistent w/ a more acidic compoundo The more stable ion will be less reactive In the equilibrium rxn, the CA and CB will react to give the acid and the base Reactivity refers to the reverse rxn of the A-B equilibrium- I- is more less reactive b/c it’s larger and less able to donate electrons o I- is weaker base I- is more stable, and therefore less reactive- B/c it’s less reactive, there’s a higher concentration of I- relative to F-, which means that the equilibrium in the HI rxn is pushed more to the right, and the equilibrium for HF is pushed further to the left.Chapter 2: Acids and Baseso I- is the CB, so a high concentration means a larger Ka, so HI will be a stronger acid. Why does a more stable CB make Ka larger?- The charge in I- is dispersed over a much larger area b/c I- is larger than F-o Makes it more difficult to donate those electrons to the acido Larger size of the ion and dispersal …