3 ACID AND BASE THEORIES: A) Arrhenius Acids produce H+ and bases produce OH not always used because it only IDs X OH as basic species B) Bronsted and Lowry Acid = H + donor > CB = formed after H + dissociates Base = H + acceptor > CA = formed after H + is accepted 3) Lewis Acid/Base Theory: Acid = e pair acceptor Base = electron pair donor H +1 Ex: H + + H N H > H N H H H 1
Examples of Bronsted Lowry Acids/Bases: HF(aq) + H 2 O(l) <=> NH 3 (g) + H 2 O <=> HCO 3 1 + H 2 O <=> HCO 3 1 + H 2 O <=> ID/Define the following important terms: Amphoteric? Hydronium ion? 2
K a and K b of acids and bases: Ex: HA + H 2 O < > H 3 O + + H A) In the above eqn, ID the 2 bases, ID the 2 acids B) If acids and bases are on both sides, which way will equilibrium lie? C) Write the K a for HA dissociation: D) Since we don't include H 2 O in equil. expressions, we shorten the dissociation of acids to HA < > H + + A Ex: B + H 2 O < > BH + + OH A) ID A, B, CA and CB in the above equation B) Determine the K b expression for the above equilibrium: 3
Write Ka and Kb expressions for the following: A) HCl < > Cl + H + B) NH 4 + < > NH 3 + H + C) NH 3 + H 2 O < > NH 4 + + OH 4
Acid Strength: A) Strength depends on where equilibrium lies HA < > H + + A Strong acid equilibrium lies right (most acid in soln will be dissociated) ** A strong acid has a weak conjugate base (CB) Larger the Ka value = stronger acid Weak acid equilibrium lies left (most acid in soln will be HA) *** A weak acid has a strong CB Smaller the Ka value = weaker acid Facts to remember: 1) The stronger the acid, the weaker is its conjugate base The stronger the base, the weaker is its conjugate acid 3) Larger the K a = stronger the acid 4) Water = acid or a base, depending on the strength of who it is with stronger base than the CB of a strong acid weaker base than the CB of a strong acid Ex: Place the following in order of increasing acidity, using the K a values provided H 2 O, F, Cl, NO 2, CN Provided Species Ka HF 7.2 x 10 4 HCl HNO2 4.0 x 10 4 HCN 6.2 x 10 10 5
BRONSTED LOWRY THEORY There is a competition for the H + between the bases H 2 O and A. The stronger one will get the proton and the direction and degree on the equilibrium will be determine by this. ALWAYS STRONGER to WEAKER Predict whether the equilibrium lies predominately to the left or to the right in the following reaction. HCl(aq) + H 2 O(l) <=> HPO 4 2(aq) + CH 3 CO 2 (aq) <=> PO 4 3(aq) + CH 3 CO 2 H(aq) 6
Water amphoteric and "Self ionizing" A) Amphoteric can be an acid or a base B) Self ionizing: H 2 O + H 2 O < > H 3 O + + OH C) Formulas and constants for water, acids and bases: K w = [OH ][H 3 O + ] [OH ] = [H 3 O + ] K w = 1.0 x 10 14 [OH ] < [H 3 O + ] ph = log [H + ] [OH ] > [H 3 O + ] poh = log[oh ] ph + poh = 14.00 pk w = ph + poh ph is? [H+] = 10 ph [OH ] = 10 poh Calculate/Answer the following: A) State if the following is acidic, basic or neutral: 1) [OH ] = 1.0 x 10 5 M 2) 10.0 M H + B) Calculate the ph and poh of the following: 1) 1.0 x 10 3 M [OH ] 2) 1.0 M [H+] C) The ph of human blood is 7.41 at 25 C. Calc the [H+], [OH ], and poh of it. 7
More information on acids A) Acids can have more than 1 acidic H + Monoprotic acid Diprotic acid Triprotic acid B) pka: i) pk a = log (k a ) ii) Strong acids have a low pk a, but a high k a iii) Spectrum of pka values pka: <1 1 3 3 5 5 15 >15 Very strong Strong Weak Very weak Even weaker **H 2 O has a pk a of ~15 8
REVIEW Q: 1) Which acid is stronger? A) HBr or HBrO B) HC 2 H 3 O 2 or HClO 2) Which base is stronger? HCO 3 1 or NH 3 3) Which is the stronger acid: one with a pk a = 3 or one with a pk a = 4? 4) What is the conjugate base of HS 1? What is the conjugate acid of HSO 3 1? 5) Which direction does this equilibrium system lie? A) HSO 4 1 + H 2 PO 4 1 <=> H 2 SO 4 + HPO 4 2? B) C 6 H 5 COO + HPO 4 2 <=> C 6 H 5 COOH + PO 4 3 9
Calculating ph of STRONG acid and base solutions: A) Always write equation, use ICE box set up B) Focus on major species in math stepss Practice determine the ph of the following: 1) 3.4 x 10 2 M HNO 3 2) 1.0 x 10 10 M HCl 3) 0.085 M NaOH 4) 1.43 x 10 3 M Ca(OH) 2 10
1) Write ionization equation(s) for each 2) Find the value(s) of K a (s) acids or K b bases from handout 3) Label the A/CB or B/CA in each (A) acetic acid (B) sulfuric acid (C) carbonic acid (D) ammonia 11
1) Write equations that represent the: A) The K a for chlorous acid B) The K b for aqueous ammonia. C) The K b for the bicarbonate ion. D) The K a for acetic acid. E) The K b for dimethylamine. F) The K a s for phosphoric acid 12
Calculating the ph of WEAK acid solutions: A) Always write/id the major species in soln B) Treat system as equilibrium problem (math) Practice: 1) Calculate the ph of a 0.0450M HNO 2 solution if the K a is 4.5 x 10 4. 2) Calculate the ph of a 0.100 M solution of hypochlorous acid, with a K a = 3.5 x 10 8 13
Calculations involving weak acids and weak bases: 1) Calculate the ph of a 0.420M benzoic acid solution if its K a = 6.3 x 10 5. 14
2) The ph of a 0.025M solution of butanoic acid is 3.21. Calculate the value of the ionization constant, K a, for butanoic acid. What is its percent ionization? 15
4) The ph of a 0.0100M aqueous solution of methylamine is 11.30. Calculate the K b for this weak base. 16
5) (A) Calculate the percent ionization of 0.45M acetic acid solution if the K a = 1.8 x 10 5. (B) Calculate the percent ionization of a 0.045M acetic acid solution. 17
6) A 0.0730M solution of a monoprotic acid is known to be 1.07% ionized. Calculate the value of the ionization constant. What is the ph of this solution? 18
Calculating Ka from % dissociation: In a 0.100 M aqueous Lactic acid solution (HC 3 H 5 O 3 ), it has 3.7% dissociation. Calculate its K a from this information. 19
Calculating ph of a solution with 2 weak acids: A) Follow same approach as for a single weak acid B) If one acid is a STRONGER weak acid than other, focus on it for your math calc of ph, as it will dominate the solution Ex: You have a mixture of 1.00 M HCN (K a = 6.2 x 10 10 ) and 5.00 M HNO 2 (K a = 4.0 x 10 4 ) **Calculate the ph of the solution and the [CN ] concentration 20
STATEMENTS: 1) Write equations with water as one of the reactants 2) The bigger the K a, the stronger the acid 3) The bigger the K b, the stronger the base 4) The smaller the pk a, the stronger the acid 5) The smaller the pk b, the stronger the base 6) The stronger the acid, the weaker is its conjugate base 7) The weaker the acid, the stronger is its conjugate base 8) Water is a stronger base than the conjugate base of a strong acid but weaker than the conjugate base of a weak acid 9) If H 3 O + is produced in the equilibrium equation, then a K a will be used 10) If OH is produced in the equilibrium equation, then a K b will be used 11) Question to be answered: a) Is it an acid, base, or salt? b) If an acid or a base, it is strong or weak? c) If a salt, is there a part that hydrolyzes (with water)? 12) If the ion from the salt comes from a weak acid or base or a polyprotic acid it will hydrolyze. 13) The more dilute the acid, the more it ionizes (the larger is the % ionization) 14) WRITE EQUATIONS!!!!!!!!! 21
Bases and ph calculations: A) Strong bases = XOH when X = grp 1 metal, or Ca, Sr, Ba B) Determine the ph just like acids, except log [OH ] = poh Then ph + poh = 14 C) General rxn for a base: B + H 2 O < > BH + + OH D) Perform WEAK BASE calc. just as Weak acid, using equilibrium Practice: 1) Calculate the ph of a 15.0 M aqueous ammonia solution if the K b is 1.8 x 10 5. 2) What concentration of methylamine would give a ph of 12.188? The K b for methyl amine is 5.0 x 10 4. 22
POLYPROTIC ACIDS: A) K a1 for 1st H + dissociation K a2 for 2nd H + dissociation K a3 for 3rd H + dissociation B) Each H + dissociation will be treated as a 1 step dissociation, do math in steps C) Each H + is increasingly harder to dissociate, there for = a lower Ka value K a1 > K a2 > K a3 Ex: 1) Calculate the ph and the concentration of all of the species in a 1.0 M H 2 SO 4 solution. 23
2) Calculate the ph and the concentration of all of the species in 1.00 x 10 2 M H 2 SO 4 solution. 24
3) Calculate the concentration of all particles of a 0.10M H 3 PO 4 solution if the K a 's are 7.5 x 10 3, 6.2 x 10 8, and 3.6 x 10 13 25
Acid Base Properties of Salts (hydrolysis of salts): A) Salt = ionic compound B) Salts form by Acid + base Acid + Base > Salt + H 2 O SA + SB > ph = 7 SA + WB > ph <7 WA + SB > ph >7 (K WA )(K CB ) = 1 x 10 14 and (K WB )(K CA ) = 1 x 10 14 Practice: 1) Salt of a strong acid and strong base. ex. HCl + NaOH > 26
2) Salt of a strong base and a weak acid: **For a salt formed from a cation of neutral properties (like Na or K) with a CB of a Weak Acid, the solution will be basic ** You need to find the K b of the anion by using K a x K b = K w BEFORE using it in equilibrium math a) Calculate the ph of 0.10M NaC 2 H 3 O 2 b) Calculate the ph of 0.200M NaCN 27
c) Calculate the ph of 0.10M Na 2 SO 3 d) Calculate the ph of 0.10M NaHSO 3 28
3) Salt of a weak base and a strong acid: ** Salts in which the anion is not a base and the cation is the conjugate acid of a weak base produce acidic solutions ** Find the K a of the Conjugate acid 1st, then do equilibrium math a) Find the ph of 0.20M NH 4 NO 3 if the K b for NH 3 is 1.8 x 10 5 b) Find the ph of 0.010 M AlCl 3, which forms Al(H 2 O) 6 +3 in water. The K a for Al(H 2 O) 6 +3 is is 1.4 x 10 5 Special note in general: The higher the charge on the metal, the more acidic the solution 29
What to do if both the anion and cation of the salt can affect ph: ** Both anion and cation are from weak acids/bases Look at the Ka and Kb values and compare them Ka > Kb = acidic Ka< Kb = basic Ka = Kb = neutral (ph = 7) Let's practice: Will the following produce acidic or basic solutions? 1) NH 4 C 2 H 3 O 2 2) NH 4 CN 3) Al 2 (SO 4 ) 3 30
Effect of structure on acid/base properties: A) Acids = donate H+ B) What affects the ability of an H+ to be lost? i) X H bond must be POLAR AND WEAK ** As electronegativity of anion decreases, bond and polarity decrease H F > H Cl > H Br > H I Strongest bond (weakest acid) ii) Oxyacids Weakest bond (strongest acid) as # of Oxygens in a molecule increases, the e density shifts to these O atoms, THUS REDUCING the e density of X H bond, = easy to break = stronger acid Ex: Predict the strength of the acids in increasing acidity HOI HOBr HOCH 3 HOCl 31
Acid Base Properties of oxides: H O X substances Acid if O X bond has a high electronegativity X Base if O X bond has a low electronegativity X Covalent oxides in water: form acids, called acidic oxides Ex: SO 3 + H 2 O > H 2 SO 4 CO 2 + H 2 O > H 2 CO 3 Ionic oxides in water : form basic solutions Ex: CaO + H 2 O > Ca(OH) 2 K 2 O + H 2 O > 2KOH 32
ACID STRENGTH: 1) HOCl vs HClO 2 vs HClO 3 2) HOCl vs HOBr vs HOI 3) H 2 SO 3 vs H 2 SeO 3 33
4) H 3 BO 3 vs Al(OH) 3 (H 3 AlO 3 ) 5) HCl vs HBr vs HI vs HF 6) Fe(H 2 O) 6 +2 vs Fe(H 2 O) 6 +3 34
1) Rank the following 0.10M solutions in order of increasing ph: A) HI, HF, NaF, NaI B) NH 4 Br, HBr, KBr, NH 3 C) C 6 H 5 NH 3 NO 3, NaNO 3, NaOH, HOC 6 H 5, KOC 6 H 5, C 6 H 5 NH 2 35
2) Place the following in order of increasing acid strength: a) HIO 3, HBrO 3 b) HNO 2, HNO 3 c) HOCl, HOI d) H 3 PO 4, H 3 PO 3 3) Place in order of increasing base strength: a) IO 3, BrO 3 b) NO 2, NO 3, H 2 O c) OCl, OI 36
4) Place the following in order of increasing acid strength: a) H 2 O, H 2 Se, H 2 S b) CH 3 CO 2 H, F 3 CCO 2 H, F 2 CHCO 2 H, FCH 2 CO 2 H c) NH 4 +, PH + 4 5) Arrange the following 0.10M solutions in order of increasing ph: a) KOH, KCl, KCN, NH 4 Cl, HCl b) CaBr 2, KNO 2, HClO 4, HNO 2 37
Put in Increasing order of ph (all are 0.10M): NaCN, HCl, KC 2 H 3 O 2, LiHSO 3, NaOH, NH 4 Br, NaI 38
Lewis Acid Base Theory Model: Lewis acid e pair acceptor Lewis base e pair donor Ex: Predict the products, and ID the Lewis acid and Lewis base BF 3 + NH 3 > SO 3 + H 2 O > Al +3 + 6H 2 O > 39
ACID/ BASE RXN CALCULATIONS: 1) Calculate the ph of the resulting solution when 50.0 ml of 0.15M KOH is reacted with 50.0 ml of 0.15M HBr 2) Calculate the ph of the resulting solution when 70.0 ml of 0.15M KOH is reacted with 50.0 ml of 0.15M HBr 3) Calculate the ph of the resulting solution when 50.0 ml of 0.15M KOH is reacted with 70.0 ml of 0.15M HBr 40
4) Calculate the ph of the acid before any base has been added. Then calculate the ph of a solution that results from the reaction 60.0 ml of 1.20M LiOH with 30.0 ml of 1.50M HNO 2. 5) Calculate the ph of the acid before any base has been added. Then calculate the ph of a solution that results from the reaction 40.0 ml of 1.20M LiOH with 40.0 ml of 1.20M HNO 2. 41
6) Calculate the ph of a solution that results from the reaction 40.0 ml of 1.20M CH 3 NH 2 with 60.0 ml of 1.50M HNO 3. 42
7) Calculate the ph of a solution that results from the reaction 100.0 ml of 2.50M CH 3 NH 2(aq) with 200.0 ml of 1.25M HNO 3. 43
EQUATION WRITING: Write the equation that would be used to determine the ph of the resulting solution and indicate whether a K a or K b will be used and how it will be obtained. Show as much of the calculation(give the mathematical equation) as possible without solving it. 1) 0.25M chlorous acid 2) 0.25M potassium chlorite solution 3) 0.25M methylamine solution 4) 0.25M ammonium chloride solution 44
5) 0.25M cobalt(iii) bromide solution 6) 0.25M lithium bisulfite solution 7) 0.25M sulfurous acid 45
8) 50.0 ml of 0.25M hydrochloric acid is added to 50.0 ml of 0.25M sodium hydroxide 9) 50.0 ml of 0.25M chlorous acid is added to 50.0 ml of 0.25M sodium hydroxide 10) 40.0 ml of 0.25M chlorous acid is added to 50.0 ml of 0.25M sodium hydroxide 46
10) 40.0 ml of 0.25M hydrochloric acid is added to 30.0 ml of 0.25M aqueous ammonia solution 11) 40.0 ml of 0.25M hydrochloric acid is added to 40.0 ml of 0.25M aqueous ammonia solution 47
12) 50.0 ml of 0.100M nitric acid reacting with 60.0 ml of 0.100M potassium hydroxide. 13) 60.0 ml of 0.100M nitric acid reacting with 50.0 ml of 0.100M potassium hydroxide. 48
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