Unit 7: Acids & Bases

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1 Definitions of Acids & Bases Unit 7: Acids & Bases Chapter 16 Arrhenius Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions. Definitions of Acids & Bases Brønsted Lowry Acid: Proton donor must have a removable proton (H + ) Base: Proton acceptor must have a nonbonding pair of electrons E.g. NH 3 If it can be either...it is amphiprotic (amphoteric). HCO 3 HSO 4 H 2 O What Happens When an Acid Dissolves in Water? Water acts as a Brønsted Lowry base and abstracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed. Conjugate Acids and Bases: Acids create conjugate bases Bases create conjugate acids 1

2 Acid and Base Strength Identify the conjugate Strong acids completely dissociate in water. Their conjugate bases are quite weak, with VERY little tendency to gain a proton. Weak acids only partially dissociate in water. Their conjugate bases are weak bases. HClO 4 ClO - 4 H 2 S HS - CN - HCN H 2 O H 3 O + or OH - Acid and Base Strength Reminder Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong. E.g. CH 4 strong acids HCl, HBr, HI, HClO 3, HClO 4, HNO 3, H 2 SO 4 strong bases Group 1 and heavy group 2 (Ca, Sr, Ba) hydroxides Acid and Base Strength Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. C 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) H 2 O is a much stronger base than Cl, so the equilibrium lies so far to the right K is not measured (K>>1). Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K<1). 2

3 Autoionization of Water Ion-Product Constant As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. The equilibrium expression for this process is K w = [H 3 O + ] [OH ] H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) At 25 C, K w = ph = log [H 3 O + ] In pure water, [H 3 O + ] = [OH ] K w = [H 3 O + ] [OH ] = ph of Acids & Bases An acid has a higher [H 3 O + ] than pure water, so its ph is <7 A base has a lower [H 3 O + ] than pure water, so its ph is >7. [H 3 O + ] = ( ) 1/2 = ph = log ( ) = 7.00 px = -log X p Scales Some similar examples are poh = log [OH ] pk w = log K w Special Note! Because [H 3 O + ] [OH ] = K w = , we know that log [H 3 O + ] + log [OH ] = log K w = or, in other words, ph + poh = pk w =

4 How Do We Measure ph? For less accurate measurements, one can use Litmus paper Red paper turns blue above ~ph = 8 Blue paper turns red below ~ph = 5 An indicator More accurate measurements require a ph meter. Strong Acids Seven strong acids HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. Strong electrolytes because they exist totally as ions in aqueous solution. Strong Bases Strong bases soluble hydroxides alkali metal heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Calculate ph, poh, [H 3 O + ], [OH - ] 1.50 M HCl 1.50 M H 2 SO M NaOH 1.50 M Ca(OH) 2 Strong because they dissociate completely Dissociation Constants Weak acids do not completely dissociate, they move to equilibrium HA(aq) + H 2 O(l) A (aq) + H 3 O + (aq) Dissociation Constants The greater the value of K a, the stronger the acid. the equilibrium expression would be [H 3 O + ] [A ] K c = [HA] This equilibrium constant is called the acid-dissociation constant, K a. 4

5 Calculating K a from the ph Calculating Percent Ionization The ph of a 0.10 M solution of formic acid, HCOOH, at 25 C is Calculate K a for formic acid at this temperature. Ka = 1.8 x 10-4 [H 3 O + ] eq Percent Ionization = 100 [HA] initial In this example [H 3 O + ] eq = M [HCOOH] initial = 0.10 M Calculating Percent Ionization Calculating ph from K a Percent Ionization = = 4.2% Calculate the ph of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25 C. HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) K a for acetic acid at 25 C is Calculating ph from K a The equilibrium constant expression is K a = [H 3 O + ] [C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] Calculating ph from K a RICE table! Reaction [C 2 H 3 O 2 ], M [H 3 O + ], M [C 2 H 3 O 2 ], M Initial Change x +x +x Equilibrium 0.30 x 0.30 x x x will be very small, so 0.30 x can be approximated as

6 Calculating ph from K a [H 3 O + ] [C 2 H 3 O 2 ] K a = [HC 2 H 3 O 2 ] (x) = 2 (0.30) Calculating ph from K a ph = log [H 3 O + ] ph = log ( ) ph = 2.64 ( ) (0.30) = x = x = x Polyprotic Acids Weak Bases Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the ph generally depends only on the first dissociation. Bases react with water to produce hydroxide ion. Weak Bases Weak Bases The equilibrium constant expression for this reaction is K b can be used to find [OH ] and, through it, ph. K b = [HB] [OH ] [B ] where K b is the base-dissociation constant. 6

7 ph of Basic Solutions ph of Basic Solutions What is the ph of a 0.15 M solution of NH 3? NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) [NH 4+ ] [OH ] K b = = [NH 3 ] 5 RICE Table Reaction [NH 3 ], M [NH 4+ ], M [OH ], M Initial Change -x +x +x Equilibrium x 0.15 x x ph of Basic Solutions ph of Basic Solutions K b = = [NH 4+ ] [OH ] [NH 3 ] = (x) 2 (0.15) ( ) (0.15) = x = x = x 2 Therefore, [OH ] = M poh = log ( ) poh = 2.80 ph = ph = K a K b = K w Acid-Base Properties of Salt Solutions pk a + pk b = pk w = Many ions react with water to form an acidic or basic solution Hydrolysis Reactions A - (aq) + H 2 O (l) HA (aq) + OH - (aq) B + (aq) + H 2 O (l) BOH (aq) + H + 7

8 Anions of polyprotic acids (H 2 PO 4 2- ) can act as either an acid or a base Metal cations act as weak acids in solution exceptions: strong hydroxide bases Qualitatively predicting ph Remember: acid + base salt + water Strong acid + strong base neutral, ph = 7 weak acid + strong base basic, ph > 7 weak base + strong acid acidic, ph < 7 weak acid + weak base If K a > K b, acidic, ph < 7 If K b > K a, basic, ph > 7 Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound (more willing to donate H + ). Acidity increases from left to right across a row and from top to bottom down a group. Factors Affecting Acid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. Factors Affecting Acid Strength Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic. For a series of oxyacids, acidity increases with the number of oxygens. 8

9 Chapter 17 Additional Aspects of Aqueous Equilibria Remember Le Chatlier A system in equilibrium that undergoes a stress will react to relieve that stress A + B C + D Add C or D? Equilibrium moves left Add A or B? Equilibrium moves right The Common-Ion Effect HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 What happens if some NaC 2 H 3 O 2 is added? NaC 2 H 3 O 3 Na + + C 2 H 3 O 2 [C 2 H 3 O 2 ] increases Reaction will move left to move the [C 2 H 3 O 2 ] back to equilibrium [H 3 O + ] decreases ph decreases The Common-Ion Effect The amount of dissociated acid decreases % ionization decreases Calculate the ph of a 0.16 M HNO 2 (K a = 4.5 x 10-4 ) with 0.10 M KNO 2. What is the [CHO 2- ] and ph of M HCHO 2 (K a = 1.8 x 10-4 ) and M HNO 3? ph = 3.14 What is the ph of the original solution? (No KNO 2 added? [CHO 2- ] = 9.0 x 10-5 M ph = 1.00 ph =

10 Buffers: How does a buffer work? HF + H 2 O H 3 O + + F - Solutions of a weak conjugate acid-base pair. Resist changes in ph because it contains both acidic species to neutralize OH - and basic species to neutralize H 3 O +. Addition of a small amount of base (OH - ) OH - reacts with the H 3 O + Reaction will shift right to reestablish equilibrium [F - ] increases, [HF] decreases If [HF]/[F - ] is large, the change in ph will be very small Buffer Capacity If acid is added, the F reacts to form HF and water. Equilibrium moves left [HF] increases, [F - ] decreases [HF]/[F - ] is still large ph is still not affected very much The amount of acid or base that can be added to a buffer before the ph changes significantly. Buffer Calculations Remember that for HA + H 2 O H 3 O + + A - Buffer Calculations Henderson Hasselbalch equation ph = pk a + log [base] [acid] K a = [H 3 O + ] [A ] [HA] ph = pk a + log [A- ] [HA] 10

11 Henderson Hasselbalch Equation ph Range What is the ph of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a for lactic acid is ph = 3.77 The ph range is the range of ph values over which a buffer system works effectively. Choose a buffer with an acid that has a pk a close to the desired ph. You wish to prepare 1.0 L of a buffer with a ph of Which combination of acid/conjugate base should be selected from the list below? Acid Conjugate Base Ka pka HSO 4 - SO x CH 3 CO 2 H CH 3 CO x HCO 3 - CO x What is the change in ph when 1.00 ml of 1.00 M HCl is added to 1.00 L of an acetic acid/sodium acetate buffer with [HC 2 H 3 O 2 ] = M and [C 2 H 3 O 2- ] = M? new ph = 4.68 What is the change in ph if 1.00 ml of HCl is added to 1.00 L of pure water? new ph = 3.00 Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base). A ph meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. Titration 11

12 Choosing an indicator depends on the ph of the equivalence point. Titration of a Weak Acid with a Strong Base With weaker acids, the initial ph is higher and ph changes near the equivalence point are more subtle. Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation. Solubility Products In a saturated solution, the undissolved solute is in contact with the solution. Equilibrium occurs. NaSO 4 (s) 2 Na + (aq) + SO 4 2- (aq) K sp = [Na + ] 2 [SO 4 2 ] K sp = Solubility product constant *not the same as solubility Calculate the water solubility of CaCO 3 (Ksp = 4.9 x 10-9 ) in molarity. 7.0 x 10-5 M Factors Affecting Solubility: Common Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) 12

13 Factors Affecting Solubility: ph If a substance has a basic anion, it will be more soluble in an acidic solution. Mg(OH) 2 (s) Mg 2+ (aq) + OH - (aq) Mg(OH) 2 (s)+ 2H 3 O + (aq) Mg 2+ (aq) + 4 H 2 O (l) adding acid reacts with the hydroxide ion equilibrium moves right - more salt dissociates Substances with acidic cations are more soluble in basic solutions. Factors Affecting Solubility: Complex Ions Metal ions can act as Lewis acids (accept e - pair) and form complex ions with Lewis bases in the solvent AgCl (s) Ag + (aq) + Cl - (aq) Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2+ (aq) by adding the base, a complex ion forms more AgCl will dissolve to replace the Ag + ions taken by the complex ion Factors Affecting Solubility: Amphoterism Reaction Quotient, Q Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Q = [B+][A-] just like K sp, but use ACTUAL concentrations, not equilibrium concentrations Will a Precipitate Form? If Q = K sp system is at equilibrium solution is saturated. Q < K sp unsaturated solution more solute will dissolve until Q = K sp. If Q > K sp the salt will precipitate until Q = K sp. The concentration of barium ion in solution is M. K sp of BaSO 4 is 1.1 x What concentration of sulfate ion is required to just begin precipitation? 1.1 x 10-8 M When the concentration of sulfate ion reaches M, what concentration of barium ion will remain in solution? 7.3 x 10-9 M 13

14 100.0 ml of M BaCl 2 is mixed with 50.0 ml of M Na 2 SO 4. Will BaSO 4 (K sp = 1.1 x ) precipitate? Selective Precipitation of Ions Yes 14

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