Proton Transfer Acids - Base. Dr. Fred Omega Garces Chemistry 201. Miramar College

Transcription

1 16.2 Acids Base Proton Transfer Dr. Fred Omega Garces Chemistry 201 Miramar College Important Notes: K a when H 3 O + is produced, K b when OH is produced 1 Acids Bases; Proton Transfer

2 BrønstedLowry AcidsBases Brønsted Lowry definition Acid Proton H + (H 3 O + ) donor. Base Proton H + (H 3 O + ) acceptor. example: acids: HCl (aq) g H+ (aq) + Cl (aq) Bases: NH 3 (aq) D NH 4 + (aq) HCl (aq) + NH 3 (aq) g NH 4 + (aq) + Cl (aq In an acid base reaction, H + & OH always combine together to form water and an ionic compound (a salt): HCl (aq) + NaOH (aq) g H 2 O (l) + NaCl (aq) 2 Acids Bases; Proton Transfer

3 Consider a Neutralization Reaction Neutralization reaction is a reaction between acids and base NH 3 + HCl g NH 4+ + Cl Base 1 Acid 2 Conj Acid 1 Conj Base 2 Conjugates differ by a proton (H + ) Base NH 3 : BronstedLowery: proton acceptor (Conjugate Base 1 ) Cl : BronstedLowery: proton accept (Conjugate Base 2 ) Acid HCl: BronstedLowery: proton donor (Conjugate Acid 2 ) NH 4+ : BronstedLowery: proton donor (Conjugate Acid 1 ) 3 Acids Bases; Proton Transfer

4 more Conjugate AcidBase pairs Other Examples of Conjugate Pairs: 1. HCO 3 (aq) + H 2 O (l) D H 3 O + (aq) + CO 3 2 (aq) Acid Base Conj.. acid Conj. base of H 2 O of HCO 3 2. HCO 3 (aq) + HS (aq) D H 2 S (aq) + CO 3 2 (aq) Acid Base Conj. acid Conj. base of HS of HCO 3 3. CH 3 CO 2 H (aq) + CN (aq) D HCN (aq) + CH 3 CO 2 (aq) Acid Base Conj. acid Conj. base of CN of CH 3 CO 2 H 4. HSO 3 (aq) + NH 3 (aq) D NH 4 + (aq) + SO 3 2 (aq) Acid Base Conj. acid Conj. base of NH 3 of HSO 3 4 Acids Bases; Proton Transfer

5 Conjugate (Example) Q: Give the conjugates of NH 3, IO, C 2 H 3 O 2, HAsO 4 2 A: NH 4, HIO, HC 2 H 3 O 2, H 2 AsO 4 Q: Identify the acids and bases on both sides of the following equations and show the conjugate acidbase pairs. A: HNO 2 + H 2 O g NO 2 + H 3 O + B: HPO NH 4+ g H 2 PO 4 + NH 3 5 Acids Bases; Proton Transfer

6 Conjugate (Example) Q: Give the conjugates of NH 3, IO, C 2 H 3 O 2, HAsO 4 2 A: NH 4+, HIO, HC 2 H 3 O 2, H 2 AsO 4, AsO 4 3 Q: Identify the acids and bases on both sides of the following equations and show the conjugate acidbase pairs. A: HNO 2 + H 2 O D NO 2 + H 3 O + Acid 1 Base 2 ConjBase 1ß ConjAcid 2 B: HPO NH 4+ D H 2 PO 4 + NH 3 Base 1 Acid 2 ConjAcid 1 ConjBase 2 6 Acids Bases; Proton Transfer

7 Relative Strengths and Net Directions The net direction of an acidbase reaction depends on the relative strengths of the acids and bases involved. A reaction proceeds to a greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base. The strength of the conjugate acidbase pairs are shown as the stronger the acid, the weaker the conjugate base. The strongest acid appears on the top left and the strongest base on the bottom right. When an acid reacts with a base farther down the list, the reaction proceeds in the forward (right) direction. 7 Acids Bases; Proton Transfer

8 Strong Acids more Acids and Bases conjugate pairs Conjugate Acid Conjugate Base Name Formula Formula Name Perchloric acid HClO 4 ClO 4 Perchlorate ion Sulfuric acid H 2 SO 4 HSO 4 Hydrogen sulfate ion Hydrochloric acid HCl Cl Chloride ion Nitric acid HNO 3 NO 3 Nitrate ion Hydronium ion H 3 O + H 2 O Water Hydrogen sulfate ion HSO 4 SO 2 4 Sulfate ion Phosphoric acid H 3 PO 4 H 2 PO 4 Dihydrogen phosphate ion Acetic acid CH 3 CO 2 H CH 3 CO 2 Acetate ion Carbonic acid H 2 CO 3 HCO 3 Hydrogen carbonate ion Hydrogen sulfide H 2 S HS Hydrogen sulfide ion Dihydrogen phosphate ion H 2 PO 4 HPO 2 4 Hydrogen phosphate ion Ammonium ion NH + 4 NH 3 Ammonia Hydrogen cyanide HCN CN Cyanide ion Hydrogen carbonate ion HCO 3 CO 2 3 Carbonate ion Water H 2 O OH Hydroxide ion Ethanol C 2 H 5 OH C 2 H 5 O Ethoxide ion Ammonia NH 3 NH 2 Amide ion Hydrogen H 2 H Hydride ion Methane CH 4 CH 3 Methide ion Strong Base 8 Acids Bases; Proton Transfer

9 Concentration Calculations for Strong Acids/Bases Since strong acids or base undergo 100% ionization, the concentration of H + or OH is the analytical concentration of these acid or base solution. That is [H 2 SO 4 ] = [H 3 O + ] yields ph of solution [HNO 3 ] = [H 3 O + ] yields ph of solution and [KOH] = [OH ] yields poh then ph [Ca(OH) 2 ] = 2 [OH ] yields poh then ph Consider 1. Acid: HClO 4 + H 2 O g H 3 O + + ClO M Since this proceeds 100%, then [H 3 O + ] also equals M The ph of the solution = log[0.0010] = Base: Ca(OH) 2 g 2OH + Ca M Since this proceeds 100%, then [OH ] equals M The poh of the solution is log[0.0020] = 2.70 and ph = Acids Bases; Proton Transfer

10 Strong Base (Example) Calc the ph of a soln' when 2.00 g of Li 2 O (29.2 g/mol) is added to L water. Li 2 O + H 2 O g 2OH + Li + [OH ] Conc. is solve via stoichiometry: 2.00 g 1 mol 2 mol OH 1 = M OH 29.9 g Li 2 O 1 mol Li 2 O 0.600L poh = log [0.223 M] = ph = = Acids Bases; Proton Transfer

11 Weak Acids / Bases Weak acids and bases ionizes less than 100%? Dissociation properties Electrolyte Substances which dissociate in water. Strong electrolyte completely dissociates to ions Weak electrolyte undergoes partial dissociation. Acid Examples : H 2 S + H 2 O D H 3 O + + HS Less 100% Dissociation (Weak Acid) Base Examples : NH 3 + H 2 O D OH + NH 4 + Less 100% Dissociation (Weak Base) Concentration of Acid / Base depends on the percent dissociation. Both of these are an equilibrium Problem!!! 13 Acids Bases; Proton Transfer

12 Weak Acids and Bases, Ka values (revisited) The % ionization is dependent of the K eq constant. For acids, this is called Kacid or K a. 14 Acids Bases; Proton Transfer

13 Solving WeakAcid Equilibria As in equilibrium type problems, there are also two type of acid base equilibrium problems. Given the equilibrium concentrations, find K a or K b Given the K a (or K b ) and concentrations information, find the other equilibrium concentrations. Problem Solving Strategy: Balance eqn & Mass action Expression Construct ice table define x as the unknown Make simplifying assumptions Solve Mass action expression for x Check assumption using the 5% rule. do second iteration if error is greater than 5%. Assumptions [H ] from water, ~ is negligible compare to x. [H 3 O + ] eq = M + x = x A weak acid has a very small K a, therefore x is very small compared to the original acid concentration. [HA] eq = [HA] int x = [HA] int 15 Acids Bases; Proton Transfer

14 Weak Acids : ph Calculation Consider HF: HF (aq) + H 2 O (l) D H 3 O + (aq) + F (aq) 100 molc K a = molc HF g 97molc HF & 3F & 3H + (Analytical Conc) (Solution at Equilb) HF (aq) + H 2 O (l) D H 3 O + (aq) + F (aq) i 0.10M lots C e 16 Acids Bases; Proton Transfer

15 Weak Acids : ph Calculation Consider HF: HF (aq) + H 2 O (l) D H 3 O + (aq) + F (aq) 100 molc K a = molc HF g 97molc HF & 3F & 3H + (Analytical Conc) (Solution at Equilb) HF (aq) + H 2 O (l) D H 3 O + (aq) + F (aq) i 0.10M lots C x x +x +x e 0.10 x lots x x Amount Dissociation: Determine conc. K a = [H 3 O + ] [F ] << 1 by K a [HF] Since K a << 1, The major component in solution is the HF 17 Acids Bases; Proton Transfer

16 Weak Acids : ph comparison with strong acid The ph value of weak acids is higher than those of strong acids for solutions with the same concentration. Example: Determine the ph of 0.10 M HF and of 0.10 M HCl HF (aq)! H + (aq) + F (aq) k a = i 0.10 M less than 100% dissociation C x +x +x e 0.10x x x HCl (aq)! H + (aq) + Cl (aq) 100% dissociation i 0.10 M C 0.10M M M e M 0.10M Assumption method: k a = ( ) = x x 0.10 x = x x 0.10 x 2 = assume x x 0.10 x.010 after applying simplifying assumption ph = log[ 0.10]= 1.00 solving for x x =[H + ] = M, 2nd iteration: ( ) = x x 0.10 x = = x x.0918 ( ) x x ( ) Verifyinging assumption = 8.24% > 5% rule.10 assume x x The Quadratic Formula can also be used to solve this problem Quadratic Formula method: k a = = x x 0.10 x solving for x, x =[H + ] = M, > 5% of 0.05 Using quadratic equation : x x = 0, x =[H + ] = M, ph = 2.10 solving for x Verifyinging assumption x =[H + ] = M, ph = 2.10 ( ) =.38% < 5% rule Two iteration of the simplifying assumption method leads to the correct solution 18 Acids Bases; Proton Transfer

17 Find Ka, Given Concentrations (Try on your own) Type problem: HA + H2O Mass Action Expression: 19 (l) D H3O+ (aq) + A (aq) Ka = [H3O+] [A ] [HA] Acids Bases; Proton Transfer

18 Find Concentrations, Given K a Type problem: Concentrations provided for reaction HPr + H 2 O (l) D H 3 O + (aq) + Pr (aq) What is the [H 3 O+] of 0.10M, Propanoic acid, K a = Use ice table to express equilibrium concentrations in terms of x Solve Mass action expression and solve for concentrations. 20 Acids Bases; Proton Transfer

19 Find Concentrations, ph and K a, Given percent dissociation or ionization (a) A M solution of a weak acid HX is 9.40 percent ionized (a). Using this info., calculate [H + ], [X ], [HX] and K a for HX. Percent dissociation (a) = [HA] Dissoc = X. [HA] Ini [HA] Init * 9.40% of M is M (x = ) HX (aq) + H 2 O (l) D H 3 O + (aq) + X (aq) i lots *C e lots [e] 0.181M lots M M Note that the K a for this unknown acid can be determined: K a Determination K a = [H 3 O + ] [X ] = (0.0188) 2 [HX] K a = Acids Bases; Proton Transfer

20 Weak Acids: percent dissociation (a) Lower concentration of weak acids will dissociate to a greater extent because according to LeChatelier, the system will go in the direction to offset the lower number of particles (concentrations) by breaking up to more particles.* For a weak acid the percent ionization varies as the inverse squareroot of the acid concentration. HF (aq)! H + (aq) + F (aq) less than 100% dissociation i [HF] o C x +x +x e [HF] o x x x k a = x x [HF] o x = x2 [HF] o assume x << [HF] o A graph illustrating that the percent ionization, a, is inversely proportional to the original acid concentration. x = [ k a [HF] o ] 1/2 % α = x [HF] o 100 % α = [ k a [HF] o ] 1/2 100 = [HF] o [ k a ] 1/2 [HF] 100 1/2 o Note: % Ionization is inversely related to [HF] o concentration. 22 Acids Bases; Proton Transfer

21 Behavior of Polyprotic Acids Many acids have more than one ionizable H atom. Polyprotic acids are acids with multiple ionizable protons Consider the following Acids: 1. HNO 2 + H 2 O g H 3 O + + NO 2 Ka1 = Monoprotic 2. H 2 SO 3 + H 2 O g H 3 O + + HSO 3 HSO 3 + H 2 O D H 3 O + + SO 3 2 Ka1 = Ka2 = Diprotic 3. H 3 PO 4 + H 2 O D H 3 O + + H 2 PO 4 H 2 PO 4 + H 2 O D H 3 O + + HPO 4 2 HPO H 2 O D H 3 O + + PO 4 3 Ka1 = Ka2 = Ka3 = Triprotic Note that the proceeding K a s are always smaller than the previous. This makes sense in that the removal of the next H + is from a negatively charge specie. The proceeding K a s (for successive H + ionization) are generally smaller by factors of Acids Bases; Proton Transfer

22 K a s of Polyprotic Acids The ph estimate of a polyprotic acid can be estimated by only the pk a1 for polyprotic acids if pk a1 differs by a factor of 10 3 or greater to pk a2 24 Acids Bases; Proton Transfer

23 Series Equation for Polyprotic Consider the ionization of Carbonic acid: The reaction is never: H 2 CO 3 + H 2 O g 2 H 3 O + + CO 3 2 rather it is a two step process: step i ) H 2 CO 3 + H 2 O g H 3 O + + HCO 3 K a1 = K a1 = [H 3 O + ] [HCO 3 ] / [H 2 CO 3 ] g [H 3 O + ] (1) g ph (1) step ii) HCO 3 + H 2 O g H 3 O + + CO 3 2 Ka2 = K a1 = [H 3 O + ] [CO 3 2 ] / [HCO 3 ] g [H 3 O + ] (2) g ph (2) Note that when K a1 >> K a2, then [H 3 O + ] (1) >> [H 3 O + ] (2) and therefore [H 3 O + ] (T) = [H 3 O + ] (1) + [H 3 O + ] (2) = [H 3 O + ] (1) ph soln = ph (1) negligible When solution can be treated as a mixture of strong acid and weak acid, the contribution from the weaker specie is negligible. 25 Acids Bases; Proton Transfer

24 Example: ph Estimate for Polyprotic Acids Phosphoric acids is a triprotic acid. Calculate the ph of a M solution of phosphoric acid. Name Formula K a1 K a2 K a3 Phosphoric H 3 PO Acid H 3 PO 4 + H 2 O g H 3 O + + H 2 PO 4 H 2 PO 4 + H 2 O g H 3 O + + HPO 4 2 K a1 = K a2 = HPO H 2 O g H 3 O + + PO 4 3 K a3 = K a1 = [H 3 O + ] [H 2 PO 4 ] = x 2 g (0.100*) = x 2 a x = [H 3 O + ] = M [H 3 PO 4 ] * check assumption K a2 = [ x] [HPO 2 4 ] = x = a x = [H 3 O + ] = M [ x ] K a3 = [ x] [PO 4 3 ] = x = a x = [H 3 O + ] = M [ x ] [H 3 O + ] = M M M ~ M ph = Acids Bases; Proton Transfer

25 Lecture notes 18.2 p 30 Sulfuric acid is a diprotic acid. Its two stages of ionization is shown below. H2SO4 (aq) " H+ + HSO4(aq) HSO4 (aq)! H+ pka =??? pka = 1.991, Ka = 1.02e2 Use your text to find relevant information to determine the following: p766 Silberberg Calculate the concentration of HSO4 & SO42 ion in a 0.5 M H2SO4 solution Determine the ph of the solution. b) H2SO4 is a strong acid so 0.5 M H2SO4 will produce 0.5M [H3O+] [H3O+] = 0.50 M, ph =.301 a) H2SO4! H3O+ + HSO4 I.50M.50M.50M HSO4! H3O+ + SO42 ii.50m.5om 0 D x +x +x [e] x x = ka of [HSO4] x = M HSO4 = 0.490M SO42 = 1.02 e2 M H3O+ = 0.51 M ph = SO42 TakeHome (InClass) Assignment: ph Estimate for Polyprotic Acids Sulfuric acid is a diprotic acid. Its two stages of ionization is shown below. H 2 SO 4 (aq) g H + + HSO 4 (aq) pka =??? HSO 4 ( (aq) D H + + SO 4 2 pka = p766 Silberberg Use your text to find relevant information to determine the following: a) Calculate the concentration of HSO 4 & SO 4 2 ion in a 0.5 M H 2 SO 4 solution b) Determine the ph of the solution. 27 Acids Bases; Proton Transfer

26 Weak Bases Weak bases produces OH at less than 100%? Base Examples : NH 3 + H 2 O D OH + NH 4 + HS + H 2 O D OH + H 2 S Less 100% Dissociation (Weak Base) In a base equilibrium process, hydroxides are formed. When the hydroxide formation is not stoichiometric, the strategy of equilibrium analysis must be used. 28 Acids Bases; Proton Transfer

27 Types of Weak Bases Two type of behavior for hydroxide formation. 1) Molc with atom with nonbonding pair of electrons serving as proton acceptor site. i.e., amines such as NH 3, CH 3 NH 2, pyridine. (Lone pair) 2) Anions of weak acids (conjugate base of a weak acid). i.e., acetate ion (C 2 H 3 O 2 ) which is the conjugate of acetic acid HC 2 H 3 O 2. (Lone pair via anion) 29 Acids Bases; Proton Transfer

28 Example: ph for weak Base soln Calculate [OH ] and ph for a M solution of ethylamine, C 2 H 5 NH 2. (K b = ) C 2 H 5 NH 2 + H 2 O D C 2 H 5 NH OH i M lots C [e] 30 Acids Bases; Proton Transfer

29 Assumption Check [React] initial / k eq In general, simplifying assumptions works when the initial concentration of the reactant is high but not when it is low. To summarize, we assume that x (D [React]) can be neglected if K eq is small relative to [React] initial. or [React] initial / k eq =?. The benchmark in justifying the assumption is What is the threshold for [React] initial / k eq so that the assumption is justified. If [React] initial / k eq > 400, the assumption is justified; neglecting x introduces an error of < 5% If [React] initial / k eq < 400, the assumption is not justified; neglecting x introduces an error of > 5% 31 Acids Bases; Proton Transfer

30 Example: ph for weak Base soln Calculate [OH ] and ph for a M solution of ethylamine, C 2 H 5 NH 2. (K b = ) C 2 H 5 NH 2 + H 2 O D C 2 H 5 NH OH i M lots C x x + x + x [e] x lots x x K b = = [OH ] [C 2 H 5 NH 3 + ] = x 2 [C 2 H 5 NH 2 ] (0.075 ) = x 2 g x = [OH ] = [C 2 H 5 NH 3 ] = M, poh = 2.16, ph = Verify assumption method, % error = 9.2% > 5% necessitate 2nd iteration Second iteration K b = = x 2 = x yields x = [OH ] = [C 2 H 5 NH 3 ] = M, poh = 2.18, ph = Using the Quadratic Equation, ph = 11.82, confirms second iteration answer. 32 Acids Bases; Proton Transfer

31 Relationship between K a and K b We note that weak acid have conjugate strong base. This suggest there is a relationship between the k a of the weak acid and the K b of the strong base. That is the strength of an acid and the strength of its conjugate base is expressed by: K a K b = K w Consider (1) HF + H 2 O D H 3 O + + F K a = (2) F + H 2 O D OH + HF K b =? kb = [ ] [ F ] [ HF] OH [ ] [ OH ] = HF kb F ka = [ ] Plug [ H3O + ] F [ HF] [ ] [ k a = H 3 O + ] [ F ] OH [ ] [ ] k b F ka kb = [ H3O + ] [ OH ] = K w ka kb = Kw 33 Acids Bases; Proton Transfer

32 Amphoteric Specie which can act as either a base or an acid by donating or accepting protons are considered amphoteric substances. Under certain conditions, chemicals can either give up a proton or accept a proton. i.e., H 2 O, HCO 3, HPO 2 4, H 2 PO 4 Consider Hydrogen carbonate (bicarbonate) : Acid process: H 2 CO 3 + H 2 O g H 3 O + + HCO 3 1st H+, K a1 = HCO 3 + H 2 O g H 3 O + + CO 2 3 2nd H+, K a2 = Base process: CO H 2 O g OH + HCO 3 1st OH, K b1 = HCO 3 + H 2 O g OH + H 2 CO 3 2nd OH, K b2 = Note from the above reaction that K a1 competes against K b2 these two equilibrium constant are related by K w : K b2 = k w / K a1 To determine acidity or basicitiy of an amphoteric solution, the K a s and K b s must be weighted against each other. 34 Acids Bases; Proton Transfer

33 Amphoteric Behavior of Carbonic Acid Ka 1 = H 2 CO 7 Ka 3 HCO 2 = CO 2 3 Kb 2 = Kb 1 = Is the solution Acidic or Basic? Other salts that pose the same problem NaHSO 3 NaH 2 PO 4 K 2 HPO 4 KHC 2 O 4 sodium bisulfite sodium dihydrogen phosphate Potassium biphosphate Potassium oxalate KH 2 C 6 H 5 O 7 NH 4 HCO 3 Potassium citrate Ammonium bicarbonate From polyprotic acids 35 Acids Bases; Proton Transfer

34 Amphoteric Behavior of Carbonic Acid Ka 1 = H 2 CO 7 Ka 3 HCO 2 = CO 2 3 Kb 2 = Kb 1 = From the magnitude of the equilbrium constant for bicarbonate ion For the HCO 3 : K b2 > K a2 In this process, the HCO 3 is a better base than HCO 3 is an acid. Therefore is a solution basic or acidic, if HCO 3 is introduced in a solution i.e., NaHCO 3? i.e., 42 g NaHCO 3 is added to 500 ml of water, What is the ph? NaHCO 3 OH H+ HCO 3 Kb 2 = Ka 2 = From inspection, K b2 >> K a2, therefore this solution is basic. To determine the ph, solve the equilb rxn for both reaction. K b2 = [OH ] [H 2 CO 3 ] g [OH] & K a2 = [H 3 O + ] [CO 32 ] g [H 3 O + ] [HCO 3 ] [HCO 3 ] Excess of OH and H 3 O + will determine the ph of the solution. 36 Acids Bases; Proton Transfer

35 Factors Affecting Solubility: Acidic Soln Solubility of salts influenced by acid, base properties of the salt s cation & anion. Idea LeChatelier s Principle is obeyed. Consider Solubility of BaF 2 in acidic solution: BaF 2(s) D Ba +2 (aq) + 2F (aq) Eqn(1) but F + H 2 O D OH (aq) + HF F reacts further decreasing the amt. of F in eqn 1 As HF is produced, more BaF 2 dissolves to maintain K sp equilibrium. F can be removed further by addition of H 3 O + Solubility increases because H 3 O + removes F and converts it to HF BaF 2(s) Ba +2 (aq) + 2F (aq) Acidic Solution, Solubility increase BaF 2(s) + H 3 O + (in acidic conditions, soluble) D Ba +2 (aq) + 2F (aq) + H 3 O + D HF 37 Acids Bases; Proton Transfer

36 Factors Affecting Solubility: Basic Soln Solubility of salts influenced by acid, base properties of the salt s cation & anion. Idea LeChatelier s Principle is obeyed. Consider Solubility of BaF 2 in basic c solution: BaF 2(s) D Ba +2 (aq) + 2F (aq) Eqn(1) but F + H 2 O D OH (aq) + HF F reacts further to produce HF from eqn 1. Solubility decreases because F is generated when HF reacts with excess OH. Increase F shift equilibrium back to solid. BaF 2(s) Ba +2 (aq) + 2F (aq) Basic Solution, Solubility decrease BaF 2(s) + OH (in basic conditions, insoluble) D Ba +2 (aq) + 2F (aq) + H 2 O D HF (aq) + OH D F (aq) + H 2 O According to LeChatelier s principle, less BaF 2 will not dissociate. 38 Acids Bases; Proton Transfer

37 Acid Base Reaction, which chemical dominates Consider the following reactions below. Determine which chemical in each equation is the dominant specie in solution. Which direction will the equilibrium favor? H 2 S + H 2 O D H 3 O + + HS H 2 PO 4 + NH + 4 D H 3 PO 4 + NH 3 HBrO 4 + BrO 3 D HBrO 3 + BrO K 4 HCH 3 CO 2 + CN D CH 3 CO 2 + HCN H 2 SO 3 HF NaCN HBrO 3 NaCH 3 CO 2 H 2 S NH 4 + H 3 O + HIO H 2 C 2 O 4 H 3 PO 4 KH 2 PO 4 HCN HBrO 4 NaHCO 3 H 2 PO 4 Strong Acids 39 Acids Bases; Proton Transfer

38 Strong Acids more Acids and Bases conjugate pairs Conjugate Acid Conjugate Base Name Formula Formula Name Perchloric acid HClO 4 ClO 4 Perchlorate ion Sulfuric acid H 2 SO 4 HSO 4 Hydrogen sulfate ion Hydrochloric acid HCl Cl Chloride ion Nitric acid HNO 3 NO 3 Nitrate ion Hydronium ion H 3 O + H 2 O Water Hydrogen sulfate ion HSO 4 SO 2 4 Sulfate ion Phosphoric acid H 3 PO 4 H 2 PO 4 Dihydrogen phosphate ion Acetic acid CH 3 CO 2 H CH 3 CO 2 Acetate ion Carbonic acid H 2 CO 3 HCO 3 Hydrogen carbonate ion Hydrogen sulfide H 2 S HS Hydrogen sulfide ion Dihydrogen phosphate ion H 2 PO 4 HPO 2 4 Hydrogen phosphate ion Ammonium ion NH + 4 NH 3 Ammonia Hydrogen cyanide HCN CN Cyanide ion Hydrogen carbonate ion HCO 3 CO 2 3 Carbonate ion Water H 2 O OH Hydroxide ion Ethanol C 2 H 5 OH C 2 H 5 O Ethoxide ion Ammonia NH 3 NH 2 Amide ion Hydrogen H 2 H Hydride ion Methane CH 4 CH 3 Methide ion Strong Base 40 Acids Bases; Proton Transfer

39 Lecture notes 18.2 p 39 ph of Sodium hydroxide, sodium bisulfite and sodium phosphate 1 Calculate the ph if 20.0 grams of sodium hydroxide is dissolved in enough water to form 1.00L solution. [NaOH] = 20 g * 1 mol * 1 = 0.5 M 40 g 1.0 L poh = ph = 14 poh ph = Calculate the ph if 52.0 grams of sodium bisulfite is dissolved in enough water to form 1.00L solution. [NaHSO3] = 52 g * 1 mol * 1 = 0.5 M 104 g 1.0 L Na2SO3! Na + HSO3 Ka2 = 6.16 e8 pka2 = M 0.5.5M HSO3 " H3O+ + SO32 I.5M 0 0 Δ x +x +x [e].5x x x Ka = X^2 = 6.16 e8 0.5 x^2 = 6.16E08 * 0.5 [H3O+] = 1.75E04 ph = Calculate the ph if 35.5 grams of sodium phosphate is dissolved in enough water to form 1.00L solution [Na3PO4] = 35.5 g * 1 mol * 1 = M 165 g 1.0 L Na3PO4! 3Na + PO43 Ka3 4.5 e13 pka3 = kb1 2.22E02 pkb1 = 1.65 PO43 " OH + HPO42 I.215M 0 0 Δ x +x +x [e].215x x x Kb = X^2 = 2.22E x^2 = 2.22E02 * [OH] = 6.91E02 poh = ph = Midterm Practice Assignment: ph of Sodium hydroxide, sodium bisulfite and sodium phosphate NaOH, NaHSO 3, Na 3 PO 4 1 Calculate the ph if 20.0 grams of sodium hydroxide is dissolved in enough water to form 1.00L solution. ph = Calculate the ph if 52.0 grams of sodium bisulfite is dissolved in enough water to form 1.00L solution. ph = Calculate the ph if 35.5 grams of sodium phosphate is dissolved in enough water to form 1.00L solution ph = Acids Bases; Proton Transfer

40 Summary Strength of Acid or Base K eq or K a, k b value. The competition for amphoteric substance to give OH or H 3 O + Solubility of salt is directly influence by LeChatelier Principle and how reactant or product reacts with the acid. 42 Acids Bases; Proton Transfer