Chem 1721 Brief Notes: Chapters 15 and 16

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1 Chem 1721 Brief Notes: Chapters 15 and 16 Chapter 15: Acids and Bases; Chapter 16: Acid-Base Equilibria Bronstsed-Lowry definitions of acids and bases are based on proton transfer acids are proton donors bases are proton acceptors An acid-base reaction (neutralization reaction) is a proton transfer reaction: Acid + Base Salt (+ water) ex. HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O HNO 3 (aq) + NH 3 (aq) NH 4 Cl (aq) Dissociaton (ionization) equations: acid dissociation equation: HA + H 2 O H 3 O + + A note: acidic solutions are characterized by [H 3 O + ] base dissociation equation: B + H 2 O BH + + OH note: basic solutions are characterized by [OH ] Strong acids and bases are completely ionized in aqueous solution; all in the form of H 3 O + + A, or B n+ + OH Strong Acids Strong Bases HCl HNO 3 LiOH Ca(OH) 2 HBr HClO 4 NaOH Sr(OH) 2 HI H 2 SO 4 KOH Ba(OH) 2 RbOH Mg(OH) 2 CsOH Weak acids and bases are only partially ionized in aqueous solution (frequently 1% or less); the majority of a weak acid or weak base is in its unionized form (HA or B) examples of weak acids examples of weak bases HF NH 3 HC 2 H 3 O 2 N 2 H 4 H 3 PO 4 C 5 H 5 N HNO 2 NEt 3 (where Et = CH 2 CH 3 ) HClO 3 HNMe 2 (where Me = CH 3 ) Conjugate acid/base pairs the conjugate base of an acid is the species that remains after the proton is donated HF + H 2 O F + H 3 O + ; F is the conjugate base of HF; HF and F is a conjugate acid/base pair the conjugate acid of a base is the species that is formed when a base accepts a proton NH 3 + H 2 O NH OH ; NH 4 + is the conjugate acid of NH 3 ; NH 3 and NH 4 + is a conjugate acid/base pair Weak acid dissociation (ionization) equilibria: for weak acids in aqueous solution an equilibrium is established: HA (aq) + H 2 O (l) A (aq) + H 3 O + (aq) this (heterogeneous) equilibrium has an equilibrium constant, K a the acid dissociation (ionization) constant K a = [H 3O + ][A - ] [HA] Comparison of strong and weak acids: Strong Acids Weak Acids completely ionized partially ionized K a is very large K a is small to very small equilibrium position very far to right equilibrium position very far to left at equilibrium [H 3 O + ] = [HA] 0 ; [HA] 0 at equilibrium [H 3 O + ] << [HA] 0 ; [HA] > 0 H 2 O is stronger base than A A is stronger base than H 2 O HA is stronger acid than H 3 O + H 3 O + is stronger acid than HA

2 Auto-dissociation (ionization) of water: H 2 O + H 2 O H 3 O + + OH equilibrium constant, K w the water auto-dissociation constant; K w = [H 3 O + ][OH ] at 25 C K w = 1.0 x For any aqueous solution at 25 C, [H 3 O + ][OH ] = 1.0 x neutral solution: [H 3 O + ] = [OH ] acidic solution: [H 3 O + ] > [OH ] basic solution: [OH ] > [H 3 O + ] a few examples ex. [H 3 O + ] = 2.5 x 10 3 M in lemon juice at 25 C. Calculate [OH ]. Is lemon juice acidic, basic, or neutral? answer: [OH ] = 4.0 x M; acidic because [H 3 O + ] > [OH ] ex. At 50 C K w = 5.5 x What is [H 3 O + ] in a neutral solution at 50 C? answer: [H 3 O + ] = 2.3 x 10 7 M [H 3 O + ] and ph ph is a special measure of [H 3 O + ] in a solution ph = log[h 3 O + ]; OR [H 3 O + ] = 10 ph logarithmic scale so ph decreases by 1 unit as [H 3 O + ] increases by factor of 10 related values: poh = log[oh ] pk a = logk a pk b = logk b ex. Consider a sample of lakewater contaminated by the chemicals associated with acid rain. The ph = 4.5; calculate [H 3 O + ] and [OH ]. ex. What is the ph of a sample of seawater with [OH ] = 1.58 x 10 6 M? answer: [H 3 O + ] = 3.2 x 10 5 M [OH ] = 3.1 x M answer: ph = 8.20 note: relationship between ph and poh K w = [H 3 O + ][OH ] take log of both sides: logk w = log[h 3 O + ] log[oh ] at 25 C: = ph + poh ph calcualtions for solutions of strong and weak acids strong acid solutions weak acid solutions [H 3 O + ] = [HA] 0 [H 3 O + ] << [HA] solve for ph directly equilibrium calculation; use K a and equilibrium table ex. Calculate the ph of M nitric acid. Will the ph of M nitrous acid be higher, lower, or equal to the ph of M nitric acid? answer: 1.60

3 ex. Calculate [HBr] if the solution has ph = answer: M ex. Calculate the ph of 0.10 M HCN (aq). For HCN K a = 4.9 x a few considerations: 2 potential sources of H 3 O + : dissociation of HCN AND the auto-dissociation of water [H 3 O + ] tot = [H 3 O + ] HCN-diss + [H 3 O + ] H2O-diss dominant source of H 3 O + determined by larger K value; here K a > K w by 4 orders of magnitude set up an equilibrium calculation based on the acid dissociation equilibrium of the weak acid, HCN: HCN (aq) + H 2 O (l) H 3 O + (aq) + CN (aq) initial [ ] 0.10 M Δ [ ] x x + x equil [ ] (0.10 x) M ---- x M x M K a = [H 3 O + ][CN - ] [HCN] 4.9 x = (x)(x) ( x) 4.9 x = x answer: x = [H 3 O + ] = 7.0 x 10 6 M; ph = 5.15 note: here we use the simplifying assumption that the x in the [HCN] eq term is negligibly small; it is usually safe to use this approximation when K a is small (i.e. ~ 10 5 or smaller) to check if this approximation is valid: x [HA] 0 < 5% if the result of this calculation is > 5% the approximation is not valid; solve for x using the quadratic formula percent dissocation (ionization) in weak acid solutions another measure of acid strength extent of reaction in the forward direction for strong acids, effectively 100% for weak acids <<< 100% related to K a and concentration percent dissociaton = (Δ[HA]/[HA] 0 )*100 for weak acids percent dissociation increases as concentration decreases compare: M HC 2 H 3 O 2 (aq) and M HC 2 H 3 O 2 (aq). For acetic acid K a = 1.8 x M HC 2 H 3 O 2 (aq) acid M HC 2 H 3 O 2 (aq) 6.71 x 10 4 M [H 3 O + ] 4.24 x 10 4 M 3.17 ph % % dissociation 4.24%

4 determination of K a from experimental data: from ph or % dissociation use date to determine x, then calculate K a ex. The ph of a M solution of HF (aq) is Calculate K a for HF. if ph = 2.036, then [H 3 O + ] = x = 9.20 x 10 3 use the value of x to calculate K a K a = [H 3 O+ ][F - ] [HF] K a = x 2 ( x) K a = (9.20 x 10-3 ) 2 ( ) answer: K a = 3.51 x 10 4 ex. A M solution of HNO 2 (aq) is 3.65% dissociated. Calculate K a for HNO 2. if % diss = 3.65, then Δ[HNO 2 ] = x = use to calculate K a answer: K a = 4.69 x 10 4 ph calculations for solutions of strong and weak bases; percent dissociation and K b calculations for weak bases strong base solutions weak base solutions calculate [OH ] based on solution concentration equilibrium exists in solution solve for poh directly, then ph B (aq) + H 2 O (l) BH + (aq) + OH (aq) base dissociation constant, K b [OH ] << [B] recall: at 25 C ph + poh = ex. Calculate the ph of 0.25 M NaOH (aq) and 0.25 M Ca(OH) 2 (aq). Before calculation... which do you expect to be more basic? Have the lower ph? ex. Calculate the ph and % dissociation of 0.40 M NH 3 (aq). For NH 3, K b = 1.8 x think through this calculation in the same way as a weak acid problem 2 sources of OH ; NH 3 base ionization and auto-ionization of water so [OH ] tot = [OH ] NH3 + [OH ] H2O [OH ] NH3 because K b NH 3 > K w answer: NaOH ph = Ca(OH) 2 ph = set up equilibrium calculation for the base dissociation of the weak base NH 3 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH initial [ ] 0.40 M Δ [ ] x x + x equil [ ] (0.40 x) M ---- x M x M K b = [NH 4 + ][OH - ] [NH 3 ] x = [OH ] = x 10-5 = (x)(x) ( x) 1.8 x 10-5 = x answer: ph = 11.43; % diss = 0.68%

5 ex. Codeine (C 18 H 21 NO 3 ) is a weak base. The ph of a M solution of codeine is Calculate K b. answer: K b = 1.6 x 10 6 Polyprotic Acids acids with more than one acidic proton (H + ) H + dissociation is step-wise; one H + dissociated per step each dissociation step has its own K a value ex oxalic acid, H 2 C 2 O 4 ; K a1 =5.9 x 10 2 ; Ka 2 = 6.4 x st dissociation step: H 2 C 2 O 4 (aq) + H 2 O (l) HC 2 O 4 (aq) + H + (aq) 2 nd dissociation step: HC 2 O 4 (aq) + H 2 O (l) C 2 O 4 2 (aq) + H + (aq) note: K a1 > K a2 this is ALWAYS true for polyprotic acids; H 2 C 2 O 4 stronger acid than HC 2 O 4 loss of H + from an anion is less favorable than loss of H + from a charge-neutral species polyprotic acids are common ion solutions (more in Ch. 15); [H 3 O + ] 0 in 2 nd dissociation step (is = x from 1 st dissociation step) ex. Determine the equilibrium concentrations of [H 2 CO 3 ], [HCO 3 ], [CO 3 2 ] and [H 3 O + ] as well as the ph of 0.45 M H 2 CO 3 (aq). For carbonic acid K a1 = 4.3 x 10 7, K a2 = 5.6 x st dissociation step: H 2 CO 3 + H 2 O HCO 3 + H 3 O + initial [ ] 0.45 M 0 M 0 M equil [ ] (0.45 x) M x M x M using K a1 : x = [HCO 3 ] = [H 3 O + ] = 4.4 x 10 4 M [H 2 CO 3 ] =.45 x = 0.45 M 2 nd dissociation step: 2 HCO 3 + H 2 O CO 3 + H 3 O + initial [ ] 4.4 x 10 4 M 0 M 4.4 x 104 M equil [ ] (4.4 x 10 4 x) M x M (4.4 x x) M using K a2 : x = [CO 3 2 ] = 5.6 x M finally: [H 3 O + ] = 4.4 x 10 4 M; ph = 3.36 note: x is negligibly small in both the (4.4 x 10 4 x) M and (4.4 x x) M terms Acidic and Basic properties of salts Ionic compounds can be acidic, basic or neutral cations tend to be neutral or acidic anions tend to be neutral or basic neutral cations and anions are related to the strong acids and bases neutral cations: Li +, Na +, K +, Rb +, Cs +, Ca 2+, Sr 2+, Ba 2+ neutral anions: Cl, Br, I, NO 3, ClO 4

6 the conjugate acid of a weak base (i.e. BH + ) is acidic: BH + + H 2 O B + H 3 O + ; K a ex. NH 4 +, C 5 H 5 NH +, N 2 H 5 +, C 6 H 5 NH 4 + the conjugate base of a weak acid (i.e. A ) is basic: A + H 2 O HA + OH ; K b ex. F, ClO, PO 4 3, NO 2, C 2 H 3 O 2 a small, highly charged metal cation may be acidic: M(H 2 O) x n+ + H 2 O M(H 2 O) x 1 (OH) (n 1)+ + H 3 O + ; K a ex. Zn 2+, Al 3+, Cr 3+, Fe 3+ an anion may be acidic if it is the conjugate base (by definition) of a polyprotic acid but still has an acidic H+ ex. HSO 4, HSO 3, HCO 3, H 2 PO 4, HPO 4 2 ; H 2 PO 4 + H 2 O HPO H 3 O + a table to summarize the acid/base properties of ions: Salt Examples Ion that participates in H + Cation conjugate acid of a strong base Anion conjugate base of a strong acid Cation conjugate acid of a weak base Anion conjugate base of a strong acid Cation conjugate acid of a strong base Anion conjugate base of a weak acid Cation conjugate acid of a weak base Anion conjugate base or a weak acid transfer reactions with H 2 O Solution is: NaCl, KNO 3, BaI 2 neither neutral NH 4 Cl, (C 5 H 5 NH)Br, (CH 3 ) 2 NH 2 Cl cation acidic Na(C 2 H 3 O 2 ), KClO, LiF anion basic NH 4 F, NH 4 CN both cation and anion K a and K b of a conjugate acid/base pair are related: K a *K b = K w ; pk a + pk b = pk w neutral if K a = K b acidic if K a > K b basic if K a < K b ex. Write the acid ionization equation for NH 4 + ; write the K a expression; calculate K a for NH 4 +. ex. Calculate the ph of 1.20 M NaNO 2 (aq). For HNO 2, K a = 4.6 x NO 2 (aq) + H 2 O (l) HNO 2 (aq) + OH (aq) initial [ ] 1.20 M equil [ ] (1.20 x) M ---- x M x M use K b for NO 2 ; K b = K w /K a for HNO 2 = 2.2 x ex. Calculate the ph of 0.88 M C 5 H 5 NBr (aq). For C 5 H 5 N, K b = 1.7 x C 5 H 5 NH + (aq) + H 2 O (l) C 5 H 5 N (aq) + H 3 O + (aq) initial [ ] 0.88 M equil [ ] (0.88 x) M ---- x M x M use K a for C 5 H 5 NH + ; K a = K w /K b = 5.9 x 10 6 answer: x = [OH ] = 5.1 x 10 6 ; ph = 8.71 answer: x = [H 3 O + ] = 2.3 x 10 3 M; ph = 2.64

7 ex. Calculate the ph of M Al(H 2 O) 3+ 6 (aq). For Al(H 2 O) 3+ 6, K a = 1.4 x Al(H 2 O) 6 + H 2 O Al(H 2 O) 5 (OH) 2+ + H 3 O + initial [ ] M equil [ ] (.097 x) M ---- x M x M set this up like a normal weak acid calculation answer: x = [H 3 O + ] = M; ph = 2.92 Structure/Strength Relationships for Acids and Bases acid strength is all about how vulnerable (i.e. donatable) the acidic H + is; the more willing the acid (HA) is to donate H +, the stronger the acid factors that affect the strength of the HA bond will affect the strength of the acid weaker HA bond H + donated more easily stronger acid stronger HA bond H + donated less easily weaker acid binary acids, HA vs. oxoacids, HAO n for a set of binary acids with A belonging to the same group of the periodic table, HA bond strength decreases as the atomic radius of A increases (top to bottom of periodic table) acid ΔH BDE K a HF 567 kj/mol 10 4 HCl 431 kj/mol very large HBr 366 kj/mol very large HI 299 kj/mol very large note: of the strong hydrohalic acids HCl, HBr, and HI, bond strength data suggests that HI should be the strongest for a set of binary acids with A belonging to the same period of the periodic table, HA bond strength is related to the polarity of the bond as the electronegativity of A increases, the polarity of the HA bond increases, and the acid strength increases the more electronegative A, the more A is attracted to the electron density in the HA bond; as the electron density in the bond is pulled toward A (electrons not shared equally between H and A), the HA bond is weakened substance, HA electronegativity of A acidic? CH NO NH NO, a weak base H 2 O 3.5 weakly amphoteric HF 4.0 weak acid, K a = 10 4 oxoacids, HAO n structurally, the central atom A is always bonded to one or more hydroxyl (OH) group the acidic proton(s) in an oxoacid are the H s of the OH group(s) below, structures of carbonic, nitric and sulfuric acids O O O H-O C O-H O N H-O S O-H O-H O H 2 CO 3 HNO3 H 2 SO 4

8 for oxoacids with the same number of oxygens, but a different central atom, A, acid strength increases with increasing electronegativity of A the more electronegative A pulls electron density away from the H creating a more polarized OH bond for oxoacids with the same A but a different number of oxygens, acid strength increases with an increasing number of oxygen atoms 2 factors to consider: 1. O s are electronegative, so the more O s the more the general flux of electron density is away from H, weakening the OH bond 2. the more O s, the more stable the resulting oxoanion (conjugate base of the acid) acid electronegativity of A K a acid # of O s K a HOI x HClO x 10 8 HOBr x 10 9 HClO x 10 2 HOCl x 10 8 HClO 3 3 ~ 1 HClO 4 4 very large Lewis definitions of acids and bases Lewis acid/base definitions are related to electron pairs donated and accepted a Lewis acid is an electron pair acceptor a Lewis base is an electron pair donor some examples: Al 3+ (aq) + H 2 O (l) Al(H 2 O) 3+ ; Al 3+ Lewis acid, H 2 O Lewis base Cu 2+ (aq) + NH 3 (aq) Cu(NH 3 ) 2+ ; Cu 2+ Lewis acid, NH 3 Lewis base Al 3+ + O H H [Al-OH 2 ] 3+ Cu 2+ + NH 3 [Cu(NH 3 )] 2+ acidic and basic oxides two general guildelines: 1. oxides of nonmetals are acidic 2. oxides of metals are basic acidic and basic behavior of oxides can be viewed in terms of Lewis acid-base definitions ex. SO 3 (g) + H 2 O (l) H 2 SO 4 (aq); SO 3 (S - Lewis acid) accepts an e pair from H 2 O (O - Lewis base) ex. MgO (s) + H 2 O (l) Mg 2+ (aq) + 2 OH (aq); oxide ion (O 2 Lewis base) donates an e pair to H 2 O (H Lewis acid)

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