Strong acids and bases

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1 h.8 & 9 Systematic Treatment of Equilibrium & Monoprotic Acid-base Equilibrium Strong acids and bases.0 onc. (M) ph ? We have to consider autoprotolysis of water: H O Kw OH - H harge Balance The sum of the positive charges in solution equals the sum of the negative charges in solution. ation n ii i j n j anion j n > charge, >concentration Mass Balance Material balance. The quantity of all species in a solution containing a particular atom (or group of atoms) must equal the amount of that atom (or group) delivered to the solution. Systematic Treatment of Equilibrium - Strong acids and bases harge balance. [H [OH - [l - Mass balance. [l M [OH - [H - (.0-8 M) Kw [H [OH - [H ([H (.0-8 M)).0 - [H (.0 8 )[H Kw0 [H.05-7 M, M ph 6.98 H O Kw OH - H When the (strong) acid or base concentration is around ~ M, we have to be concerned with autoprotolysis. For etremely low strong acid or base concentration (< -8 ), ph 7.00 Problem: What is the poh, and ph of a.8-7 M solution of NaOH? ANS: 6.55; 7.5 Systematic Treatment of Equilibrium - Weak acids and bases Weak Acid: acid dissociation constant () H O A - H O [ A O [ A A - H A A Weak Base, hydrolysis reaction: base hydrolysis constant (Kb) B H O Kb BH OH - [ BH [ OH Kb [ B Relationship between and Kb for conjugate acid/base pairs: Kw Kb O [ OH [ OH

2 Eample: alculate the ph of a L solution prepared by dissolving 0. moles of acetic acid? HOAc H O A typical weak-acid problem Kw H OAc OH - H [ OAc 5 OAc.75 Kw [H [OH A typical weak-acid problem Eample: alculate the ph of a L solution prepared by dissolving 0. moles of acetic acid? Simplifying assumptions () A H H H [ [ [H [A [ [ A [ H 0. M and.75-5,. - M harge Balance: [H [OH - [OAc - Mass Balance: [HOAc [OAc - 0. () [ A ~ A M Fraction of dissociation () For a monoprotic weak acid: [ A typical weak-acid problem Eample: alculate the ph of a L solution prepared by dissolving 0. moles of acetic acid? [ A [ A [ A A [ A ( [ A ) ( ) A A typical weak-base problem Eample: Find the ph of 0. M ammonia. NH Kb H O NH OH H O Kb [ OH [ NH 5.75 [ NH Kw OH - Kw [H [OH H harge Balance: [OH - [H [NH Mass Balance: [NH [NH 0. () () A typical weak-base problem Eample: Find the ph of 0. M ammonia. Simplifying assumptions [ OH [ NH Kb [ NH [ OH [ NH ~ Kb [ NH Kb [OH [NH [ OH OH [ [ OH 0. M and Kb.76-5,. - M M A typical weak-base problem Eample: Find the ph of 0. M ammonia. Fraction of dissociation () For a monoprotic weak base: [ NH [ NH [ NH [ NH [ NH ( ) ( )

3 Buffers A buffer solution resists changes in ph when acids or bases are added or when dilution occurs. A buffer solution generally consists of a miture of an acid and its conjugate base. Henderson-Hasselbalch Equation [ A [ OH B K a K b A [ B For a buffer solution containing a weak acid and its conjugate base: ph a [ A log A For a buffer solution containing a weak base and its conjugate acid, ph a [ B log B or poh b B log [ B Henderson-Hasselbalch Equation ph p [A - [ For every power of change in the ratio [A - /[, the ph changes by one unit. As the acid increases, the ph goes down. ph a [ A log A Effect of adding acid to a buffer Eample. alculate the ph of a buffer prepared by adding.00 ml of 0. M acetic acid and 0.00 ml of 0. M sodium acetate. ANS :5.06 Eample. What is the ph if.00 ml of M Hl is added to the above buffer solution? ANS :5.0 Buffer apacity β db dph da dph Diprotic Buffers For H A, we can write two Henderson-Hasselbalch equations: A [ A ph log ph log A A Buffer capacity is a maimum when ph p Best to chose a buffer system whose p is within ± ph unit of your desired ph E: What is the ph of a solution prepared by dissolving.0 g KHP and.0 g Na P in 50.0 ml water? ANS : 5.7

4 h. Polyprotic Acid-base Equilibrium Diprotic acids/dibasic bases Polyprotic acids/polybasic bases H A H - A A - H A - [ A A K, H PO H H PO - K 7. - H PO - H HPO HPO - H PO Amino Acid H L (Acidic form) What s the ph of a M solution of H L? A zwitterion: a molecule with both positive and negative charges Leucine Assume that H L acts as a monoprotic acid L L. - M, ph.88 heck our assumption by calculating [L - H [ L L [ L L.80 [ M L - (Basic form) What s the ph of a M solution of L -? The relations between acid and base equilibrium constants for a diprotic acid/base: Kb Kb Kw Kw H L HL H K HL L - H L H O HL OH KbKw/ HL H O H L OH KbKw/K L - (Basic form) What s the ph of a M solution of L -? L H O HL OH KbKw/ HL H O H L OH KbKw/K. - assume that L - acts as a monobasic species with KbKb [ OH L Kb [ L.6 - M, [H Kw/ M, ph. heck our assumption by calculating [H L L [ OH Kb HL [ Kb L Kb. L [ OH M

5 HL (Intermediate Form) HL is amphiprotic, meaning it can both donate and accept a proton H L HL H K HL L - H K.80 - L H O HL OH KbKw/ HL H O H L OH KbKw/K. - We have to use the systematic treatment to determine ph. HL (Intermediate Form) HL L - H K.80 - HL H O H L OH KbKw/K. - harge balance: [H [H L [L - [OH - Equilibrium onstants: HL (Intermediate Form) HL L - H K.80 - HL H O H L OH KbKw/K. - HL (Intermediate Form) HL L - H K.80 - HL H O H L OH KbKw/K. - [HL If >>Kw, than the nd term in the numerator can be dropped and if K<<, the st term in the denominator can also be neglected: heck our assumptions: ANS: 6.0 For diprotic weak acid For H A H - and - H A - H A A A A [ A A K A A [ A [ H K K K K a K a Isoionic Point: The ph of pure, neutral polyprotic acid (the natural zwitterion) Isoelectric Point: ph when average charge is zero (amino acids) A [ A A A [ A K a K K K K a 5

6 Type Keq Formula Eample Strong acids Keq [H O a Hl, HNO, HlO Strong Bases Weak Acids Weak Bases Buffer Amphiprotic salt(na) Polyprotic acids Keq O [ A [ OH [ B Kb [ BOH O [ A O A K H A O [ A [OH - b O [ OH K b H O [ NaA O K Treated as monoprotic weak acids NaOH, KOH c, HOOH, HlO NHOH c-naac, NH - NH H O NaHO, NaHPO, NaH PO, amino acid 6