Presentation on theme: "14.1 Arrhenius Definition n Acids produce hydrogen ions (H + ) in aqueous solution. n Bases produce hydroxide ions (OH - ) when dissolved in water. n."— Presentation transcript:
14.1 Arrhenius Definition n Acids produce hydrogen ions (H + ) in aqueous solution. n Bases produce hydroxide ions (OH - ) when dissolved in water. n Limits to aqueous solutions. n Only one kind of base. n NH 3 ammonia could not be an Arrhenius base.
Bronsted-Lowry Definitions n And acid is an proton (H + ) donor and a base is a proton acceptor. n Acids and bases always come in pairs. n HCl is an acid. n When it dissolves in water it gives its proton to water. n HCl(g) + H 2 O(l) H 3 O + + Cl - n In this equilibrium, water is a base, makes hydronium ion.
Conjugate Acid/Base Pairs n HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) n Acid + Base Conjugate acid + Conjugate base n Conjugate base: everything that remains of the acid molecule after a proton is lost. n Conjugate acid: formed when the proton is transferred to the base. n This is an equilibrium. n Competition for H + between H 2 O and A - n The stronger base controls direction. n If H 2 O is a stronger base it takes the H + n Equilibrium moves to right.
Practice Identify the acid, base, conjugate acid, conjugate base in the following reactions. #32 n HONH 3 + H 2 OHONH 2 - + H 3 O + n HOCl + C 6 H 5 NH 2 OCl - + C 6 H 5 NH 3 +
Acid Dissociation Constant K a n The equilibrium constant for the general equation. n HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) n K a = [H 3 O + ][A - ] [HA] n (YDVD) H 3 O + is often written H + ignoring the water in equation (it is implied).
Acid dissociation constant K a n HA(aq) H + (aq) + A - (aq) n K a = [H + ][A - ] [HA] n We can write the expression for any acid. n Strong acids dissociate completely. n Equilibrium far to right. n Conjugate base is therefore weak.
Practice Write the dissociation reaction and the corresponding Ka equilibrium expression for the following acids in water. #30 n HC 2 H 3 O 2 n CH 3 NH 3 + n Co(H 2 O) 6 3+
14.2 Acid Strength n Strong acids n K a is large n [H + ] is equal to [HA] n A - is a weaker base than water n Weak acids n K a is small n [H + ] <<< [HA] n A - is a stronger base than water n Strong Acids: n HBr, HCl, HI n HNO 3, HClO 3, HClO 4, H 2 SO 4, HIO 3
Weak Acids n Its equilibrium lies far to the left. (CH 3 COOH) Yields a much stronger (it is relatively strong) conjugate base than water. (CH 3 COO ) Yields a much stronger (it is relatively strong) conjugate base than water. (CH 3 COO ) n (YDVD)
Types of Acids n Monoprotic - One acidic proton. n Polyprotic Acids - more than 1 acidic hydrogen (diprotic, triprotic). n Oxyacids - Proton is attached to the oxygen of an ion. n Organic acids contain the Carboxyl group -COOH with the H attached to O n Generally very weak.
Water as an Acid and a Base n Wateris amphoteric - it behave as both an acid and a base. Water also autoionizes. (BDVD) n 2H 2 O(l) H 3 O + (aq) + OH - (aq) n K W = [H 3 O + ][OH - ] = [H + ][OH - ] n At 25ºC K W = 1.0 x10 -14 n In EVERY aqueous solution. n Neutral solution [H + ] = [OH - ] = 1.0 x10 -7 n Acidic solution [H + ] > [OH - ] n Basic solution [H + ] < [OH - ]
Practice n Order the following from strongest to weakest acid. n H 2 O n HCl n HOC 6 H 5 Ka = 1.6 x 10 -10 n HFKa = 7.2 x 10 -4
Practice #38 Which is the stronger base? n Cl - or H 2 O n H 2 O or NO 2 - n CN - or OC 6 H 5 - HNO 2 Ka = 4.0 x 10 -4 HCNKa = 6.2 x 10 -10 HOC 6 H 5 Ka = 1.6 x 10 -10
Practice n Calculate [H+] or [OH-] for each of the following solutions at 25°C and state whether the solution is neutral, acidic, or basic. Ex.14.3 1. 1.0 x 10 -5 M OH - 2. 1.0 x 10 -7 M OH - 3. 10.0 M H + n At 60°C, K w is 1 x 10 -13 Ex.14.4 1. Using LeChateliers principle, predict whether the reaction 2H 2 O(l) H 3 O + (aq) + OH - (aq) is exothermic or endothermic. 2. Calculate [H+] and [OH-] in a neutral solution at 60°C.
14.3 The pH Scale n pH= -log[H + ] n Used because [H + ] is usually very small n As pH decreases, [H + ] increases exponentially n Sig. figs - only the digits after the decimal place of a pH are significant n [H + ] = 1.0 x 10 -8 pH= 8.00 2 sig figs n pOH= -log[OH - ] n pKa = -log K
Relationships n K W = [H + ][OH - ] n -log K W = -log([H + ][OH - ]) n -log K W = -log[H + ]+ -log[OH - ] n pK W = pH + pOH = 14.00 n K W = 1.0 x10 -14 n 14.00 = pH + pOH n [H + ],[OH - ],pH and pOH Given any one of these we can find the other three.
Practice Ex. 14.5 n n Calculate the pH and pOH for each solution at 25°C. n n A) 1.0 x 10 -3 M OH - n n B) 1.0 M H + n n C) 2.5 x 10 -2 M H +
Practice n n Calculate the [H + ] and [OH - ] for each solution at 25°C. Identify each solution as acidic, basic or neutral. n n pH= 3.20 n n pH=11.2 n n pOH= 9.6
Practice The pH of ammonia was measured to be 11.89 at 25°C. Calculate the pOH, [H + ], [OH - ] for the sample. The pH of ammonia was measured to be 11.89 at 25°C. Calculate the pOH, [H + ], [OH - ] for the sample. The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate the pOH, [H + ], [OH - ] for the sample. Ex. 14.6 The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate the pOH, [H + ], [OH - ] for the sample. Ex. 14.6
14.4 Calculating pH of Solutions n Always write down the major ions in solution. n Remember these are equilibria. n Remember the chemistry. n Dont try to memorize there is no one way to do this. Read p. 665. Very important!
Strong Acids n ALWAYS WRITE THE MAJOR SPECIES. n Completely dissociated. n [H + ] = [HA] n [OH - ] is going to be small because of equilibrium. n 10 -14 = [H + ][OH - ] n If [HA]< 10 -7 water contributes H +
Practice #49 n A solution is prepared by mixing 50.0 mL of 0.050 M HCl and 1500.0 mL of 0.10 M HNO 3. Water is added until the final volume is 1.00 L. Calculate [H + ], [OH - ] and the pH for this solution.
Practice #50 (WA) n A solution is prepared by mixing 90.0 mL of 5.00 M HCl and 30.0 mL of 8.00 M HNO 3. Water is added until the final volume is 1.00 L. Calculate [H + ], [OH - ] and the pH for this solution.
14.5 Weak Acids n Ka will be small. n List major species in solution. n Choose species that can produce H + and write reactions. n Based on K values, decide on dominant equilibrium. n Write equilibrium expression for dominant equilibrium. n List initial concentrations in dominant equilibrium.
Weak Acids n Define change at equilibrium (as x ). n Write equilibrium concentrations in terms of x. n Substitute equilibrium concentrations into equilibrium expression. n Solve for x the easy way. n Verify assumptions using 5% rule. n Calculate [H + ] and pH.
Practice n Calculate the pH of 0.100 M hypochlorous acid, HOCl, with a Ka = 3.5 x10 -8 Ex.14.8 n Calculate the pH of 2.0 M acetic acid, HC 2 H 3 O 2, with a Ka = 1.8 x10 -5. What is the pOH, [OH - ], and [H + ]?
How about a quadratic? #57 n Calculate the pH of 0.020 M HF solution, with a Ka = 7.2 x10 -4. What is the pOH, [OH - ], and [H + ]?
A mixture of Weak Acids n The process is the same. n Determine the major species. n The stronger will predominate. n Bigger Ka if concentrations are comparable n Calculate the pH of a mixture 1.00 M HCN (Ka = 6.2 x 10 -10 ) and 5.00 M HNO 2 (Ka = 4.0 x 10 -4 ). What is the concentration of the CN - ion in this solution at equilibrium. EX. 14.9
Percent dissociation n = amount dissociated x 100 initial concentration n For a weak acid percent dissociation increases as acid becomes more dilute. n Calculate the % dissociation of 1.00 M and.00100 M Acetic acid (Ka = 1.8 x 10 -5 n As [HA] 0 decreases [H + ] decreases but % dissociation increases. n Le Chatelier
Calculating Ka n What is the K a of lactic acid (HC 3 H 5 O 3 ) that is 3.7 % dissociated as 0.100 M solution? Ex. 14.11 n A 0.15 M solution of a weak acid, is 3.0% dissociated. Calculate K a #65 n The pH of a 1.00 x 10 -2 M solution of cyanic acid (HOCN) is 2.77 at 25°C. Calculate K a for HOCN from this result. #67
Calculating Ka WA n In a 0.100 M solution of HF, the percent dissociation is 8.1%. Calculate the K a #66 n A 0.050M solution of trichloroacetic acid (CCl 3 CO 2 H) is the same pH as 0.040M HClO 4 solution. Caculate the K a for trichloroacetic acid. #68
14.6 Bases n The OH - is a strong base. n Hydroxides of the alkali metals are strong bases because they dissociate completely when dissolved. n The hydroxides of alkaline earths Ca(OH) 2 etc. are strong dibasic bases, but they dont dissolve well in water. n Used as antacids because [OH - ] cant build up.
Bases without OH - n Bases are proton acceptors. n NH 3 + H 2 O NH 4 + + OH - n It is the lone pair on nitrogen that accepts the proton. n Many weak bases contain N n B (aq) + H 2 O(l) BH + (aq) + OH - (aq) n K b = [ BH + ][ OH - ] [ B ]
Strength of Bases n Hydroxides are strong. n Calculate the pH of a 5.0 x 10 -2 M NaOH solution Ex. 14.12 n Calculate the pH, [OH] and pOH of a 4.0 x 10 -4 M Ca(OH) 2 solution. #78a n Calculate the concentration of of an aqueous Sr(OH) 2 solution that has a pH of 10.5 #82 n Others are weak.Smaller K b - weaker base. n Which is the stronger base: n NH 3 or Cl - n NH 3 or C 5 H 5 N
Practice: weak base n Calculate the pH of 15.0 M ammonia NH 3 with a K b = 1.8 x10 -5 Ex. 14.13 n Calculate the pH of 1.0 M methylamine with a K b = 4.38 x10 -4 Ex. 14.14 n Calculate the [OH - ], [H + ], and pH of 0.20 M pyridine with a K b = 1.7x10 -9 #86b
14.7 Polyprotic acids n Always dissociate stepwise. n The first H + comes off much easier than the second. n Ka for the first step is much bigger than Ka for the second. n Denoted Ka 1, Ka 2, Ka 3
Polyprotic acid n H 2 CO 3 H + + HCO 3 - Ka 1 = 4.3 x 10 -7 n HCO 3 - H + + CO 3 -2 Ka 2 = 5.6 x 10 -11 n Base in first step is acid in second. n In calculations we can normally ignore the second dissociation.
Calculate the Concentration n Calculate the pH of a 5.00 M Phosphoric acid, H 3 PO 4, solution and the equilibrium concentrations of each species. Ex. 14.15 n Ka 1 = 7.5 x 10 -3 n Ka 2 = 6.2 x 10 -8 n Ka 3 = 4.8 x 10 -10
Sulfuric acid is special n In the first step, it is a strong acid. n In the second step, Ka 2 = 1.2 x 10 -2 n For solutions more dilute than 1.0 M, the dissociation of HSO 4 - is important. Solving requires the quadratic equation. n Calculate the pH in a 1.0 M solution of H 2 SO 4 Ex. 14.16 n Calculate the pH in a 1.0 x 10 -2 M solution of H 2 SO 4 Ex. 14.17
14.8 Salts as acids and bases n Salts are ionic compounds. n Salts of the cation of strong bases and the anion of strong acids are neutral. n For example NaCl, KNO 3 n There is no equilibrium for strong acids and bases. n We ignore the reverse reaction.
Basic Salts n If the anion of a salt is the conjugate base of a weak acid, a basic solution will be generated. n In an aqueous solution of NaF n The major species are Na +, F -, and H 2 O n F - + H 2 O HF + OH - n K b =[HF][OH - ] [F - ] n but Ka = [H + ][F - ]= 7.2 x 10 -4 [HF]
Basic Salts n K a x K b = [HF][OH - ]x [H + ][F - ] [F - ] [HF]
Basic Salts n K a x K b = [HF][OH - ]x [H + ][F - ] [F - ] [HF]
Basic Salts n K a x K b = [HF][OH - ]x [H + ][F - ] [F - ] [HF]
Basic Salts n K a x K b = [HF][OH - ]x [H + ][F - ] [F - ] [HF] n K a x K b =[OH - ] [H + ]
Basic Salts n K a x K b = [HF][OH - ]x [H + ][F - ] [F - ] [HF] n K a x K b =[OH - ] [H + ] n K a x K b = K W n Calculate the pH of a 0.30 M NaF solution. The K a for HF is 7.2 x 10 -4 Ex. 14.18
K a tells us K b n The anion of a weak acid is a weak base. n Calculate the pH of a solution of 1.00 M NaCN. Ka of HCN is 6.2 x 10 -10 n The CN - ion competes with OH - for the H + n Which is the stronger acid NH 4 + or CH 3 NH 3 + ? (NH 3 : K b = 1.8x10 -5 CH 3 NH 2 :K b =4.4x10 -4 ) #102 n Arrange the following solutions in order of most acidic to most basic. KOH, KCl, KCN, NH 4 Cl, HCl #99
Acidic salts n A salt with the cation of a weak base and the anion of a strong acid will be acidic. n The same development as bases leads to K a x K b = K W n Calculate the pH of a solution of 0.10 M NH 4 Cl (the K b of NH 3 1.8 x 10 -5 ). Ex. 14.19 n Other acidic salts are those of highly charged metal ions.(Al 3+ ) n More on this later.
Salts with Acidic and Basic ions n K a > K b acidic n K a < K b basic n K a = K b neutral n Predict whether an aqueous solution of each of the following salts will be acidic, basic or neutral. EX.14.21 n NH 4 C 2 H 3 O 2 n NH 4 CN n Al 2 (SO 4 ) 3 n #105 in Text.
14.9 Structure and Acid-Base Properties n Any molecule with an H in it is a potential acid. n The stronger the X-H bond the less acidic (compare bond dissociation energies). n The more polar the X-H bond the stronger the acid (use electronegativities). n The more polar H-O-X bond - stronger acid.
Strength of oxyacids n The more oxygen hooked to the central atom, the more acidic the hydrogen. n HClO 4 > HClO 3 > HClO 2 > HClO n Remember that the H is attached to an oxygen atom. n The oxygens are electronegative n Pull electrons away from hydrogen
Strength of oxyacids Electron Density ClOH
Strength of oxyacids Electron Density ClOHO
Strength of oxyacids ClOH O O Electron Density
Strength of oxyacids ClOH O O O Electron Density
Hydrated metals n Highly charged metal ions pull the electrons of surrounding water molecules toward them. n Make it easier for H + to come off. Al +3 O H H
14.10 Acid-Base Properties of Oxides n Non-metal oxides dissolved in water can make acids. n SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) n Ionic oxides dissolve in water to produce bases. n CaO(s) + H 2 O(l) Ca(OH) 2 (aq)
Lewis Acids and Bases n Most general definition: n Acids are electron pair acceptors. n Bases are electron pair donors. BF F F :N:N H H H
Lewis Acids and Bases n Boron triflouride wants more electrons. BF F F :N:N H H H
Lewis Acids and Bases n Boron triflouride wants more electrons. n BF 3 is Lewis acid NH 3 is a Lewis base. B F F F N H H H
Lewis Acids and Bases Al +3 ( ) H H O Al ( ) 6 H H O + 6 +3
Comparing the Acids n Model AcidBase n ArrheniusH + producerOH - Producer n B-LowryH + donor H + Acceptor n Lewis e - pair acceptor e - pair donor