19.1 Acids & Bases 1. Compare and contrast the properties of acids & bases. 2. Describe the self-ionization of water & the concept of K w. 3. Differentiate between the Arhennius & Bronsted-Lowry models of acids & bases. 4. Identify conjugate acid-base pairs. 5. Describe monoprotic, diprotic & polyprotic acids Properties of Acids & Bases Acids Bases Taste Sour (lemon) Bitter (soap) Feel N/A Slimy Rxn w. Metal Produces H 2 Gas N/A Conductivity Conducts electricity Conducts electricity Effect on indicators Color changes Color changes Litmus Red Blue Phenolphthalein No change Pink/Purple Cabbage Juice Red Yellow/Green Neutralizes Bases Acids Parent compound Molecular Molecular or Ionic Forms by Ionization Ionization or Dissociation Neutralization When acids and bases are mixed, they neutralize to form a salt. Hydrogen & Hydroxide Ions Aqueous Solutions contain hydrogen ions (H + ) & hydroxide ph = 0 7 14 ions (OH - ) Acid contains more H + than [H + ] OH - Base contains more OH - than Acid Base H + [OH - ] Neutral solutions contain equal amounts of OH - and H + Self-ionization of Water [H + ] = 1.0 x 10 0 1.0 x 10-7 1.0 x 10-14 Some water molecules break in pure water to create H + and OH - ions o H + and OH - will occur in equal amounts in pure water H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) o [H + ] = [OH - ] = 1.0 x 10-7 mol/liter o Pure water is neutral because [H + ] = [OH - ] Arrhenius Model of Acids & Bases Svante Arrhenius Swedish Chemist (1859-1927) Proposed that:
o Acids donate H + ions in solution HCl(g) H + (aq) + Cl - (aq) Formulas of Arrhenius acids begin with an H. o Bases donate OH - ions in solution NaOH(s) Na + (aq) + OH - (aq) Formulas of Arrhenius bases end with an OH. Allows for neutralization through the creation of water: H + + OH - = H 2 O o Arrhenius acids and bases will neutralize to form a salt and water. o Explains liberation of H 2 gas when acids react with metals. Mg(s) + 2HCl(aq) MgCl 2 (aq) + H 2 (g) Bronsted-Lowry Acids Johannes Bronsted Danish Chemist (1879-1947) Thomas Lowry English Chemist (1874-1936) Proposed that: o Acids donate H + ions in solution H + is donated from a polar bond o Bases accept H + ions in solution HX(aq) + H 2 O(l) H 3 O + (aq) + X - (aq) Conjugate Acid-Base Pairs o When an acid donates a H +, it leaves an ion that can accept a H +. o That product is a base since it can accept H +. o The acid and its leftover anion is called an acid-base conjugate. o HA + B BH + + A - H + donor H + acceptor H + donor H + acceptor Acid Base Acid Base H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 - (aq) Acid Base Acid Base Bronsted-Lowry Conjugate Base Pairs NH 3 (aq) + H 2 O(l) NH + 4 (aq) + OH - (aq) H + acceptor H + donor H + donor H + acceptor Base Acid Acid Base Conjugate acid-base pairs Conjugate acid-base pairs Conjugate acid-base pairs Strong acids have weak conjugate bases. Weak acids have strong conjugate bases. Strong bases have weak conjugate acids. Weak bases have strong conjugate acids.
Amphoteric substances = substances that may behave as either an acid or a base (water, for example) Monoprotic & Polyprotic Acids Monoprotic acids capable of donating one H + ion (HCl) Polyprotic acids capable of donating multiple H + ions Diprotic acids capable of donating 2 H + ions (H 2 SO 4 ) Triprotic acids capable of donating 3 H + ions (H 3 PO 4 ) Anhydrides without water Oxides that can become acids or bases by adding water Oxides of nonmetallic elements produce acids Oxides of metallic elements usually form bases 19.2 Strengths of Acids & Bases 1. Describe the difference between strong and weak when discussin acids & bases. 2. Calculate K a & K b. Strengths of Acids Strong acids ionize completely in water forming many H + ions Weak acids ionize incompletely in water and form few H + ions Acid Ionization Constant Because the strength of an acid or base is related to how completely the acid ionizes in water, a concept similar to the solubility product (K sp ) can be used for quantifying the strength of an acid. HA(g) + H 2 O(l) H 3 O + (aq) + A - (aq) K a = [H 3 O + ][A - ] [HA] Where: K a = the acid dissociation constant for an acid [H 3 O + ] = conc. of hydronium at equilibrium [A - ] = conc. of the conjugate base at equilibrium [HA] = conc. of the acid at equilibrium. Like the equilibrium constant, the greater the numerator in this equation, the greater the extent to which the reaction proceeds. The higher K a, the stronger the acid. o Strong acids ionize completely (single arrow) HA(g) + H 2 O(l) H 3 O + (aq) + A - (aq)
The lower K a, the weaker the acid. Weak acids have a K a of less than 1. Strengths of Bases Strong bases dissociate completely in water to form many OH - ions. Weak bases ionize only partially forming the conjugate acid of the base and hydroxide ions. Base Dissociation Constant This is exactly the same as the acid dissociation constant, except substitute bases. B(aq) + H 2 O(l) HB + (aq) + OH - (aq) K b = [HB + ][OH - ] [B] Where: K b = the base dissociation constant for an base [HB + ] = conc. of the conjugate acid at equilibrium [OH - ] = conc. of the hydroxide ion at equilibrium [B] = conc. of the base at equilibrium. Like the equilibrium constant, the greater the numerator in this equation, the greater the extent to which the reaction proceeds. The higher K b, the stronger the base. o Strong bases ionize completely (single arrow) B(aq) + H 2 O(l) HB + (aq) + OH - (aq) The lower K b, the weaker the base. Weak bases have a K b of less than 1. Strong v. Weak, Concentrated v. Dilute Strong = high degree of ionization (or dissociation) Concentrated = high molarity Concentrated weak acid may have more H + ions than dilute strong acid 19.3 ph 1. Calculate the ion concentrations of [H + ] and [OH - ] for a solution. 2. Describe how ph and poh are calculated. 3. Calculate ph from [H + ] and poh from [OH - ] and vice versa. 4. Calculate K a for strong acids. Ion Product for Water Water self-ionizes to form equal numbers of H + ions and OH - ions H 2 O(l) H + (aq) + OH - (aq) Equilibrium expression: K eq = K w = [H + ][OH - ]
K w = the ion product constant for water Experiments show that in pure water at 298K, [H + ] ions and [OH - ] are both equal to 1.0 x 10-7 M. K w = [1.0 x 10-7 ][1.0 x 10-7 ] = 1.0 x 10-14 For any aqueous solution the product of H + and OH - must always equal the product 1.0 x 10-14. If one concentration increases, the other must decrease. ph and poh ph Since [H + ] is very small, ph is used to make the number easier to work with. ph = -log[h + ] The higher [H + ], the lower the ph A change of 1 ph unit represents a 10x change in concentration poh poh = -log[oh - ] The higher [OH - ], the lower the poh A change of 1 poh unit represents a 10x change in concentration ph + poh Due to the mathematical relationship of ph & poh, poh + ph = 14 Calculating [H + ] from ph ph = -log[h + ] Multiply both sides by -1 -ph = log[h + ] Take the antilog of both sides antilog (-ph) = [H + ] Remember the units are mol/liter Calculating [OH - ] from poh poh = -log[oh - ] Multiply both sides by -1 -poh = log[oh - ] Take the antilog of both sides antilog (-poh) = [OH - ] Remember the units are mol/liter Calculating ph of Strong Acids and Bases For strong acids, the concentration of the acid may be used directly to find the ph because the [H + ] will equal the molarity of the solution
Using ph to calculate K a Weak acids do not ionize completely When calculating the K a for a weak acid, the [H 3 O + ] at equilibrium must be subtracted from the initial parent acid concentration to find the parent acid concentration at equilibrium HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) K a = [H 3 O + ][F - ] [HF] = initial [HF] [H 3 O + ] 19.4 Neutralization 1. Describe the neutralization process including the products of neutralization. 2. Describe the titration process. 3. Calculate amount of acid/base necessary for neutralization to occur. 4. Describe buffers. Reactions between Acids & Bases Neutralization Acid + Base Salt + Water Salt = an ionic compound made from the cation of a base and the anion of an acid NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(g) Acid-Base Titration Method of determining the concentration of a solution Uses mole ratios from balanced neutralization reaction Acid of known concentration used to determine unknown base Base of known concentration used to determine unknown acid Steps in Titration Measured volume of acid/base of unknown concentration placed in flask with indicator (or ph meter) Buret is filled with acid/base (opposite of flask) of known concentration (standard solution) Solution from buret added slowly until indicator shows that stoichiometric point is achieved o End point = where indicator changes color o Stoichiometric point = equivalence point o Equivalence point = point at which moles of H + and OH - are equal
Salt Hydrolysis Salts dissociating in water can form weak acids or bases through their interactions with water. Ions from salts may form H 3 O + or OH - in aqueous solution Hydrolysis resulting in base NaF(s) Na + (aq) + F - (aq) F - (aq) + H 2 O(l) HF(aq) + OH - (aq) Hydrolysis resulting in acid NH 4 Cl(s) NH 4 + (aq) + Cl - (aq) NH 4 +- (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) Buffers Solutions that resist changes in ph when acid or base is added Mixture of either: weak base and its conjugate acid or weak acid and its conjugate base Buffer capacity Amount of acid or base that a buffer can absorb without a change in ph Buffers work by employing LeChatelier s principle o Increases of [H + ] and [OH - ] stress the reaction of the buffer to compensate for changes HX(aq) H + (aq) + X - (aq) o If acid is added, the reaction shifts in reverse to consume added [H + ], effectively absorbing the acid o If base is added, the reaction shifts forward as the added [OH - ] reacts with H + to form H 2 O, effectively absorbing the base Carbonic acid/hydrogen carbonate buffer in blood CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) H + (aq) + HCO3 - (aq)