Prof. Zvi C. Koren

Acids & Bases 1 Prof. Zvi C. Koren 20.07.2010

Definitions Arrhenius Acid releases H + in water: HCl(aq) H + + Cl - Base releases OH - in water: NaOH(aq) Na + + OH - Brønsted-Lowry (don t need water) Acid donates proton (H + ) Base accepts proton HCl(g) + NH 3 (g) NH 4 + Cl - (s) Lewis (don t need protons) Acid accepts an e-pair for sharing Base donates an e-pair for sharing BF 3 + :NH 3 F 3 B NH 3 coordinate covalent bond 2 Prof. Zvi C. Koren 20.07.2010

Brønsted-Lowry Conjugate Acid-Base Pairs HF(aq) + NH 3 (aq) NH 4 + + F - acid base acid base amphoteric amphiprotic HCO 3 - + H2 O H 3 O + + CO 3 2- (hydronium ion) acid base acid base HCO 3 - + H2 O OH - + H 2 CO 3 base acid base acid Conjugate acid-base pairs: HA/A -, B/BH + difference within a pair is H + All Bronsted-Lowry acid-base reactions consist of 2 conjugate pairs 3 Prof. Zvi C. Koren 20.07.2010

The Acid-Base Properties of Water: Auto-Ionization H 2 O + H 2 O H 3 O + + OH - or H 2 O H + + OH - (Note: H 3 O + and H + are the same.) Water ionization constant K w = [H + ][OH - ] = 1.0 10-14 @ 25 o C In neutral solutions and pure water: [H + ] = [OH - ] = 1.0 10-7 M In acidic solutions: [H + ] > [OH - ] In basic solutions: [H + ] < [OH - but [H ] + ][OH - ] = 1.0 10-14 4 Prof. Zvi C. Koren 20.07.2010

Sorensen: The potential (or power) of H ph log[h + ] [H + ] = 10 ph Also: poh log[oh ] and also: pk logk The ph Scale For a neutral solution at 25 o C: ph = poh = 7.00 For acidic solutions at 25 o C: ph < 7.00 For basic solutions, at 25 o C: ph > 7.00 From before: [H + ][OH ] = K w log[h + ] + log[oh ] = logk w log[h + ] + log[oh ] = logk w Note about ph and Significant Figures: If [H + ] = 1.0 10 9 M ph = 9.00 2 sig. figs. 2 sig. figs. Quick calculation regarding ph: If [H + ] = 5.0 10 9 M ph = 9 log5.0 (Note: small p big H ) ph + poh = pk w = 14.00 @ 25 o C Sig. Figs. begin after the decimal point! 5 Prof. Zvi C. Koren 20.07.2010

Strong Acids HClO 4 HClO 3 HCl H 2 SO 4 HNO 3 H 3 O + perchloric acid chloric acid hydrochloric acid sulfuric acid (1 st proton) nitric acid hydronium ion HClO 4 (aq) H + + ClO 4 - or HClO 4 (aq) + H 2 O(l) H 3 O + + ClO 4 - Dissociates completely to release H + K >> 1 Strong Acids & Bases Strong Bases Metal hydroides: MOH (from Group I) very strong M(OH) 2 (from Group II) strong Others (ions): CH - 3 methide ion H - hydride ion NH - 2 amide ion C 2 H 5 O - ethoide ion OH - hydroide ion NaOH(aq) Na + + OH - Ca(OH) 2 (aq) Ca 2+ + 2OH - Dissociates completely to release OH - : K >> 1 (In oyacids, the acidic H is bonded to O) 6 Prof. Zvi C. Koren 20.07.2010

Weak Acids & Bases Weak Acids HA(aq) + H 2 O H 3 O + + A - HA(aq) H + + A - acid ionization HAc(aq) Weak Bases constant K a [ H ][ A [ HA] ] 1 H + + Ac - (HAc or HOAc is HC 2 H 3 O 2, acetic acid ) [ H ][ Ac ] -5 K a ( HAc) 1.810 [ HAc] B(aq) + H 2 O BH + + OH - [ BH ][ OH ] base hydrolysisconstant 1 K b NH 3 (aq) + H 2 O NH 4 + + OH - [ B] [ NH 4 ][ OH ] -5 K b ( NH3) 1.810 [ NH3] 7 Prof. Zvi C. Koren 20.07.2010

Compare the acid strengths of HNO 3 vs. HNO 2 and H 2 SO 4 vs. H 2 SO 3 8 Prof. Zvi C. Koren 20.07.2010

Relative Strengths of Conjugate Acid-Base Pairs HCl + H 2 O H 3 O + + Cl - ACID BASE acid base Acid Base HCl Cl - HAc + H 2 O H 3 O + + Ac - acid base ACID BASE H 3 O + H 2 O HAc + CH 3 NH 2 CH 3 NH 3 + + Ac - ACID BASE acid base CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH - base acid ACID BASE HAc Ac - CH 3 NH 3 + CH 3 NH 2 H 2 O OH - 9 Prof. Zvi C. Koren 20.07.2010

Determining K a & K b values from ph measurements - 1 K a of Weak Acid 0.0300 M HCOOH (formic acid) K a ph=2.66 HCOOH(aq) H + + COOH - [H + ]=2.210-3 M M I C E K a [ H ][ COOH [ HCOOH] Use ICE Bo Method: HCOOH H + COOH - 0.0300 - - - + + 0.0300- =2.210-3 2 3 2 [ H ][ COOH ] (2.210 ) 4 1.710 3 [ HCOOH] 0.0300 0.0300 2.210 10 Prof. Zvi C. Koren 20.07.2010 ]

Determining K a & K b values from ph measurements - 2 K b of Weak Base The ph of a 0.15-M solution of aniline (C 6 H 5 NH 2 ) is 8.89. Calculate the K b of aniline: C 6 H 5 NH 2 (aq) + H 2 O C 6 H 5 NH 3 + + OH - K b 0.15 0.15- [ C 6 H 5 [ C NH 6 H 5 3 ][ OH NH 2 ] ] 2 0.15 Need to calculate = [OH - ] from ph (two methods): ph [H + ] [OH - ] or ph poh [OH - ] [OH - ] = 7.810-6 M K b = 4.110-10 11 Prof. Zvi C. Koren 20.07.2010

Calculation of ph of Strong Acids & Bases Strong Acid: 0.10-M HNO 3 : 0.10 HNO 3 (aq) H + + NO - 3 0.10 0.10 Strong Base: 0.00020-M Ca(OH) 2 : ph = -log[h + ] = 1.00 2.010-4 Ca(OH) 2 (aq) Ca 2+ + 2OH - [OH - ] = 4.010-4 M 2.010-4 4.010-4 [OH - ] poh ph or [OH - ] [H + ] ph ph = 10.60 12 Prof. Zvi C. Koren 20.07.2010

Calculation of ph of Weak Acids & Bases 1.810 1.810 Weak Acid Calculate the ph of a 0.10-M HAc solution: 0.10 HAc(aq) H + + Ac -, 0.10-5 K a [ H ][ Ac [ HAc] ] 2 0.10 2 0.10 [ H Weak Base Calculate the ph of a 0.10-M NH 3 solution: 0.10 NH 3 (aq) + H 2 O NH + 4 + OH -, 0.10-5 [OH - ] K b [ NH 4 ][ OH [ NH3] [H + ] ph poh ph ] 2 0.10 = 11.11 2 0.10 [ OH ] 1.310 3 M ] 1.310 3 ph=2.89 13 Prof. Zvi C. Koren 20.07.2010 M

The Birth of a Salt: A Family Portrait Salt + Water Acid Base Koren s Acid-Base Family Rule: The Strength of the Baby-Salt will be according to the Stronger Parent

Acid-Base Properties of Salts Qualitative Hydrolysis of Salts Reminder: The rn between an acid and a base produces a SALT. For eample, HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) (Daddy) (Mommy) (Baby) Koren s Family Rule: The nature of the baby is influenced by the stronger parent! Types of Salts: (Which are acidic, basic, or neutral?) 1. Salt from a strong acid & a strong base: NaCl 2. Salt from a weak acid & a strong base: NaAc 3. Salt from a strong acid & a weak base: NH 4 Cl 4. Salt from a weak acid & a weak base: NH 4 Ac 15 Prof. Zvi C. Koren 20.07.2010

Hydrolysis of Salts 1. From st A + st B: NaCl Dissociation of Dissolved Salt: NaCl(aq) Na + + Cl - Hydrolysis of Ions: Na + + H 2 O? NaOH(aq) + H + Cl - + H 2 O? HCl(aq) + OH - X X NaCl is a NEUTRAL salt. 2. From w A + st B: NaAc Dissociation of Dissolved Salt: NaAc(aq) Na + + Ac - Hydrolysis of Ions: Na + + H 2 O X Ac - + H 2 O HAc(aq) + OH - NaAc is a BASIC salt. 16 Prof. Zvi C. Koren 20.07.2010

Hydrolysis of Salts (continued) 3. From st A + w B: NH 4 Cl Dissociation of Dissolved Salt: NH 4 Cl(aq) NH 4 + + Cl - Hydrolysis of Ions: Cl - + H 2 O X NH 4 + + H 2 O NH 3 (aq) + H 3 O + NH 4 Cl is an ACIDIC salt. 4. From w A + w B: NH 4 Ac Dissociation of Dissolved Salt: NH 4 Ac(aq) NH 4 + + Ac - Hydrolysis of Ions: NH 4 + + H 2 O NH 3 (aq) + H 3 O + Ac - + H 2 O HAc(aq) + OH - NH 4 Ac will depend on which is stronger (or less weak) 17 Prof. Zvi C. Koren 20.07.2010

K b of Conjugate Base 1. A - + H 2 O HA + OH -, K b (A - ) 2. HA + H 2 O H 3 O + + A -, K a (HA) 3. H 2 O + H 2 O H 3 O + + OH -, K w = K a (HA) K b (A - ) Also, K w = K b (B) K a (BH + ), Must be conjugates For eample, K w = K b (NH 3 ) K a (NH 4 + ) Reminder: K a refers to rn of a + H 2 O and NOT with anything else! K b refers to rn of b + H 2 O and NOT with anything else! 18 Prof. Zvi C. Koren 20.07.2010

Acid-Base Properties of Salts Quantitative - 1 1. From st A + st B: NaCl Calculate the ph of a 0.10-M NaCl solution: Dissociation of Dissolved Salt: NaCl(aq) Na + + Cl - Hydrolysis of Ions: Na + + H 2 O X X Cl - + H 2 O only H 2 O produces significant amounts of H + and OH - : H 2 O H + + OH -, K w = [H + ][OH - ] = 1.010-14 And [H + ] = [OH - ] = 1.010-7 M ph = 7.00 19 Prof. Zvi C. Koren 20.07.2010

Acid-Base Properties of Salts Quantitative - 2 2. From w A + st B: NaAc Calculate the ph of a 0.10-M NaAc solution: Dissociation of Dissolved Salt: NaAc(aq) Na + + Ac - 0.1 Hydrolysis of Ions: Na + + H 2 O X K b of Ac - : K b K b ( Ac ) 0.1 0.1 0.1 Ac - + H 2 O HAc(aq) + OH - 0.1- K [ HAc][ OH [ Ac ] a K w ( HAc) ] 2 0.10 10. 10 18. 10 14 5 5.610 10 = 5.610-10 = [OH - ] = 7.510-6 M ph = 8.88 Is a solution containing HAc and NaAc, acidic, basic, or neutral? 20 Prof. Zvi C. Koren 20.07.2010

Acid-Base Properties of Salts Quantitative - 3 3. From st A + w B: NH 4 Cl Calculate the ph of a 0.10-M NH 4 Cl solution: Dissociation of Dissolved Salt: NH 4 Cl(aq) NH 4 + + Cl - Hydrolysis of Ions: 0.1 0.1 Cl - + H 2 O X 0.1 0.1 NH 4 + + H 2 O NH 3 (aq) + H 3 O + 0.1- K a of NH + 4 : 14 K w 10. 10 K a ( NH 4 ) 5.610 5 K ( NH ) 18. 10 K a [ NH3][ H 3O [ NH ] ph = 5.12 4 b ] 3 2 0.10 10 = 5.610-10 = [H 3 O + ]= 7.510-6 M 21 Prof. Zvi C. Koren 20.07.2010

Polyprotic Acids & Bases Acids: H 2 SO 4 (sulfuric acid) H 2 SO 3 (sulfurous acid) H 2 C 2 O 4 (oalic acid) H 2 CO 3 (carbonic acid) H 3 PO 4 (phosphoric acid) others which is stronger? Bases: SO 3 2- (sulfite ion) C 2 O 4 2- (oalate ion) CO 3 2- (carbonate ion) PO 4 3- (phosphate ion) HPO 4 2- (hydrogen phosphate ion) others H 3 PO 4 (aq) H + + H 2 PO 4 -, Ka1 = 7.510-3 H 2 PO 4 - H + + HPO 4 2-, K a2 = 6.210-8 HPO 4 2- H + + PO 4 3-, K a3 = 3.610-13 For H 2 CO 3 : K 1 = 4.210-7 K 2 = 4.810-11 CO 3 2- + H 2 O HCO 3 - + OH -, K b1 = K b (CO 3 2- ) = K w /K a (HCO 3 - ) = Kw /K 2 (H 2 CO 3 ) = 2.110-4 HCO 3 - + H2 O H 2 CO 3 (aq) + OH -, K b2 = K b (HCO 3 - ) = Kw /K a (H 2 CO 3 ) = K w /K 1 (H 2 CO 3 ) = 2.410-8 22 Prof. Zvi C. Koren 20.07.2010

Polyprotic Acids & Bases (continued) Problem: Calculate the ph of a 0.10-M solution of H 3 PO 4 : Answer: H 3 PO 4 (aq) H + + H 2 PO - 4, Ka1 = 7.510-3 H 2 PO - 4 H + + HPO 2-4, K a2 = 6.210-8 HPO 2-4 H + + PO 3-4, K a3 = 3.610-13 Because K 1 >> K 2 >> K 3, [H + ] 1 >> [H + ] 2 >> [H + ] 3 We only need to consider the first ionization!!! H 3 PO 4 (aq) H + + H 2 PO 4 -, Ka1 = 7.510-3 0.10 0.10- = 2 /(0.10-) 2 /0.10 = [H + ] = 2.710-2 M ph = 1.56 23 Prof. Zvi C. Koren 20.07.2010

Polyprotic Salts Problem: Calculate the equilibrium concentrations of all the species in a 0.10-M Na 2 CO 3 solution. Identify all the species. Answer: Note that we have a salt: 0.10 Dissociation of dissolved salt: Na 2 CO 3 (aq) 2Na + + CO 2-3 Hydrolysis of non-spectator ions: 0.10 CO 2-3 + H 2 O HCO - 3 + OH - 0.10- HCO 3 - + H 2 O H 2 CO 3 (aq) + OH - 0.10 [Na + ] = 0.20 M Because K 1 >> K 2, [OH - ] 1 >> [OH - ] 2 For the calculation of [OH - ] eq, we only need to consider the 1 st ionization (so go back): 2.110-4 = K b (CO 2-3 ) = 2 /(0.10-) 2 /0.10 = [OH - ] = 4.610-3 M from before [CO 2- [H + ] = 2.1810-12 M 3 ] = 0.10-0.10 M ph = 11.66 (continued) 24 Prof. Zvi C. Koren 20.07.2010

Polyprotic Salts (continued) Answer continued: recall: CO 3 2- + H 2 O HCO 3 - + OH - HCO 3 - + H 2 O H 2 CO 3 (aq) + OH - [HCO 3- ]: produced in 1 st rn, but reacts in 2 nd rn. BUT, K 1 >> K 2. [HCO 3- ] [OH - ] = 4.610-3 M Theoretically, which is greater, [HCO 3- ] or [OH - ]? [H 2 CO 3 ]: only produced in 2 nd rn. 2.410 8 K b2 [ H 2 CO 3 [ HCO ][ OH 3 ] ] [H 2 CO 3 ] = K b2 = 2.410-8 M (continued) 25 Prof. Zvi C. Koren 20.07.2010

Polyprotic Salts (continued) Note the BALANCES in this reaction system: Charge Balance & Material Balance Charge Balance: Total concentrations of (+) ions = Total concentrations of `( ) ions Σq i = 0 Σn + C + = Σn C [Na + ] + [H + ] = 2 [CO 3 2- ] + [OH - ] + [HCO 3- ] 0.20 2.1810-12 2 (0.10-4.610-3 ) 4.610-3 4.610-3 Material Balance (Conservation) for Carbonate : Total amount of carbonate material is conserved, )"מה שהיה, הוא שיהיה )" end from the beginning to the [Na 2 CO 3 ] 0 = [CO 3 2- ] eq + [HCO 3- ] eq + [H 2 CO 3 ] 0.10 0.10-4.610-3 4.610-3 2.410-8 26 Prof. Zvi C. Koren 20.07.2010

Polyprotic Salts (continued) Is a solution of NaHSO 3 basic or acidic? NaHSO 3 (aq) Na + + HSO 3 As an Acid: HSO 3 + H 2 O H 3 O + + SO 3 2 K a (HSO 3 ) = K 2 (H 2 SO 3 ) = 6.210 8 As a Base: HSO 3 + H 2 O OH + H 2 SO 3 (aq) K b (HSO 3 ) = K a 14 Kw 1.010 8.310 2 (H SO ) 1.210 2 3 13 HSO 3 is much stronger as an acid than it is as a base 27 Prof. Zvi C. Koren 20.07.2010

Lewis Acids & Bases Identify the Lewis acids and bases in the following reactions: Note: A Lewis acid accepts a pair of e s; a Lewis base donates the e s H + + OH H 2 O H 3 N + BF 3 H 3 N BF 3 NH 3 + H 2 O NH 4+ + OH CO 2 (g) + OH HCO 3 bicarbonate ion Cu 2+ + 2OH - Cu(OH) 2 (s) 28 Prof. Zvi C. Koren 20.07.2010