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Acid-base theory pH calculations

Acid-base theory pH calculations. Joško Ivica. REVIEW QUESTIONS. Write the formulas for hydrochloric acid, potassium hydroxide and their dissociation reactions in water Write the formulas for acetic acid, ammonia and their dissociation reactions in water

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Acid-base theory pH calculations

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  1. Acid-base theorypH calculations Joško Ivica

  2. REVIEW QUESTIONS • Write the formulas for hydrochloric acid, potassium hydroxide and their dissociation reactions in water • Write the formulas for acetic acid, ammonia and their dissociation reactions in water • Write the equation for equilibrium dissociation constant of acetic acid • Write the formula for sodium acetate and its dissociation reaction in water • What is pH? • Ionic product of water

  3. REVIEW H2O KA • HCl H+ + Cl-or HCl + H2O H3O+ + Cl- KOH K+ + OH- • CH3COOH + H2O CH3COO- + H3O+ CH3COOH CH3COO- + H+ NH3 + H2O NH4+ + OH- 3) [CH3COO-] [H+] • CH3COONa CH3COO-+ Na+ • pH = -log[H+] • KW = [H+][OH-] = 1,008·10-14at 25°C pOH = -log[OH-] pH + pOH = 14 = pKW! KA KB KA = [CH3COOH]

  4. ACIDS AND BASES Arrhenius theory: Acids are compounds able to dissociate in water producing a hydrogen ion (H+) and a corresponding anion (only in aqueous solutions) HNO3 H+ + NO3- Bases are compounds which dissociate in water producing a hydroxide ion and a cation NaOH  Na+ + OH- Brønsted - Lowry theory:Acids are compounds that release H+, whereas bases are compounds that are able to bind H+ (applicable also in non-aqueous solutions) acidH+ +base conjugated pair

  5. pH of strong acids and bases c(HA) = [H+] = [A-] HA H+ + A- complete dissociation of an acid pH = -log a(H+) a – activity a(H+) = γ±·c(HA) γ± - mean activity coefficient In very diluted solutions: γ± = 1! pH = -log[H+]

  6. pH of strong acids and bases complete dissociation of a base pOH = -log[OH-] c(BOH) = [OH-] = [B+] BOH B+ + OH- pH = 14 - pOH = 14 + [log a(OH-)] a(OH-) = γ±·c(BOH) In very diluted solutions: γ± = 1! pH = 14 - pOH= 14 + log [OH-]

  7. pH of weak acids and bases c-x x x [A-][H3O+] x2 x2 [HA] Dissociation of weak acids(Ka< 10-4) HA + H2O A- + H3O+Ka= = = c-x c c-x = concentration of an acid at equilibrium x = concentration of products at equilibrium c = concentration of an acid at the beginning c >> x for diluted weak acids pKa = -logKa [H3O+] = x = (Ka c)1/2/ log pH = -log[H3O+] pH = -log [H3O+] = ½ [pKa – log(c)]

  8. pH of weak acids and bases c-x x x [BH+][OH-] x2 x2 B + H2O BH+ + OH- Kb = = = [B] c-x c Dissociation of weak bases c-x = concentration of a base at equilibrium x = concentration of products at equilibrium c >> x for diluted weak bases c = concentration of a base at the beginning pKb = -logKb [OH-] = x = (Kb c)1/2/ log pOH = -log[OH-] pH = 14 - pOH pH = 14 – pOH = 14 – ½ [pKb – log(c)]

  9. Salt hydrolysis • When salts composed of ions of a strong electrolyte (acid or base) and ions of a weak electrolyte are dissolved, complete salt dissociation occurs because ions of a strong electrolyte can exist only in ionized form • Ions originating from a weak electrolyte react with water producing their conjugated particle • Examples: CH3COONa, KCN, NH4Cl, NH4NO3

  10. Salts of weak acids and strong bases [CH3COO-] [H+] KA = [CH3COOH] [CH3COOH] [ OH- ] CH3COO- + H2O CH3COOH + OH- KH = CH3COONa CH3COO- + Na+ [CH3COO-] [H+][OH-] = Kv KH·KA = KW  KH= KW/KA c-x x x CH3COO- + H2O CH3COOH + OH- [CH3COOH] = [OH-] c = concentration of salt at the beginning c-x = concentration of anion of a weak acid at equilibrium x = concentration of products at equilibrium [OH-]2 KW = 10-14 = c-x = c c KA

  11. [OH-]2 = KW · c (salt) pOH = 7 – 1/2[pKA – log(c)] KA pH = 14 - pOH [pKA + log(c)] pH = 7 + ½

  12. Salts of weak bases and strong acids [NH4+] [OH-] [NH3] NH4Cl NH4+ + Cl-KB = NH4+ + H2O NH3 + H3O+ KH = [H+][OH-] = Kv [NH3] [H3O+] [NH4+] KH·KB = KW  KH= KW/KB NH4+ + H2O NH3 + H3O+ [H3O+] = [NH3] c-x x x c = concentration of salt at the beginning KW [H3O+]2 = c c-x = c KB c-x = concentration of a cation of a weak base at equilibrium x = concentration of products at equilibrium

  13. [H3O+]2 = KW· c(salt) KB pH = 7 - ½[pKB + log(c)]

  14. Salts of weak acids and weak bases Anions andcations of weak acids and bases, that produce salt – having concentration c, react with water e.g. NH4CN CN- + H2O = HCN + OH- NH4+ + H2O = NH3 + H3O+ NH4++ CN-HCN + NH3 c-xc-x x x c-x = c KH= [HCN][NH3]/[CN-][NH4+] = [HCN]2/[CN-]2 KH ·KA·KB = KW KH = KW/KA KB KA = [H3O+][CN-]/[HCN]  (1/KH)1/2 [H3O+]2 = KA2KH = KW · KA/KB [H3O+]2 = KW· KA KB pH = 7 + ½[pKA - pKB]

  15. BUFFERS • BUFFERS = conjugated pairof acid or base, which is able to maintain pH in particular (narrow) interval after adding strong acid or base into solution (system) • Buffers are typically mixtures of weak acids and their salts with strong bases ormixtures of weak bases and their salts with strong acids • Buffer systems in organism are of a great importance (blood, intercellular space, cells)

  16. pH calculations of buffer solutions Buffer consisting of a weak acid and its salt with a strong base HA + H2O A- + H3O+ Ka Henderson – Hasselbalch equation pH = pKa + log[A-]/[HA] HA – weak acid A- – conjugated base Buffer consisting of a weak base and its salt with a strong acid B + H2O BH+ + OH- pOH = pKb + log[BH+]/[B] B – weak base BH+ - conjugated acid

  17. pH calculations • Calculatethe pH of 1 mM KOH • Calculatethe pH of 0.01 M formic acid (HCOOH) at 25°C, pKa = 3.8! • Calculatethe pH of 0.001 M NH3at 25°C, pKb = 4.8! • Calculatethe pH of 0.1 M NaCN at 25°C, pKa = 9.21! • Calculatethe pH of 0.7 M NH4Cl at 25°C, pKb = 4.8! • Calculatethe pH of 5 mM ammonium lactate CH3CH(OH)COONH4at 25°C, pKa = 3.86, pKb = 4.8 • Calculatethe pH of a buffer solution that contains 0.1 M CH3COONa and 0.1 M CH3COOH, pKa = 4.8! • Calculatethe pH of a buffer solution that contains 0.1 M NH4Cl and 1 M NH3, pKb = 4.8!

  18. 1. c(KOH) = 0,001 M = [K+] = [OH-] KOH  K+ + OH- pOH = -log [OH-] = 3 pH = 14 – pOH = 11

  19. c(HCOOH) = 0.01 M, pKa = 3.8 HCOOH ↔ HCOO-+ H+ 0.01-x=cx x x = conc. of productsat equilibrium ↓ conc. of HCOOH at equilibrium Ka =[HCOO-][H+]/[HCOOH] = x2/c = [H+]2/0.01 [H+] = (Ka·0.01)1/2 pH = -log[H+] = ½ [3.8 – log(0.01)]= 2.9 2.

  20. H2O NH3NH4+ + OH- 0.001-x x x x = conc. of productsat equilibrium ↓ conc. ofNH3at equilibrium0.001-x = c Kb=[NH4+][OH-]/[NH3] = x2/c = [OH-]2/0.001 [OH-] = (Kb·0.001)1/2 3. c(NH3) = 0.001 M, pKb = 4.8 pOH = -log[OH-] =½[pKb - log(0.001)] pH = 14 - ½[4.8 - log(0.001)] = 14 – 3.9 = 10.1

  21. 4. c(NaCN) = 0.1 M, pKa = 9.21 NaCN  Na+ + CN- HCN H+ + CN- Ka=[H+][CN-]/[HCN] CN- + H2O HCN + OH- KH = [OH-][HCN]/[CN-] c-x = c x x [HCN] = [OH-] Kv = Ka KH  Kv/ Ka = [OH-]2/c  [OH-] = (Kvc/ Ka)1/2 pOH = ½(pKv – pKa + log c)  pH = 14 - ½(pKW – pKa + log c) = pH = 7 + ½ [pKA + log(c)] = 7 + ½ (9.21 + log 0.1) = 11.1

  22. H2O 5. c(NH4Cl) = 0.7 M, pKb = 4.8 NH4Cl  NH4+ + Cl- NH3NH4+ + OH- Kb = [NH4+][OH-]/[NH3] NH4+ + H2O NH3 + H3O+ KH = [NH3][H3O+]/[NH4+] c-x = c x x [NH3] = [H3O+] Kv = Ka KH  Kv/ Ka = [H3O+]2/c  [H3O+] = (Kvc/Kb)1/2 pH = 7 - ½[pKB + log(c)] = 7 – ½ (4.8 – 0.15) = 4.68

  23. 6. c(CH3CH(OH)COONH4) = 0.005 M, pKa = 3.86, pKb = 4.8 CH3CH(OH)COO- + H2O CH3CH(OH)COOH + OH- NH4+ + H2O NH3 + H3O+ CH3CH(OH)COO- + NH4+ CH3CH(OH)COOH + NH3 c-x c-x x x KH = [CH3CH(OH)COOH][NH3]/[CH3CH(OH)COO-][NH4+] = [CH3CH(OH)COOH]2/[CH3CH(OH)COO-]2 KW = KH KA KB KH = KW/KA KB KA = [H3O+][CH3CH(OH)COO-]/[CH3CH(OH)COOH] [H3O+]2 = KA2KH = KW · KA/KB pH = 7 + ½[pKA - pKB]= 7 + ½ [3.86 – 4.8] = 6.53 (1/KH)1/2

  24. 7. 0.1 M CH3COONa, 0.1 M CH3COOH, pKa = 4.8 CH3COOH + H2O CH3COO- + H3O+ Ka pH = pKa + log [CH3COO-]/[CH3COOH] = 4.8 + 0 = 4.8

  25. 8. 0.1 M NH4Cl a 1 M NH3, pKb = 4.8 NH3+ H2O NH4+ + OH- Kb pOH = pKb + log [NH4Cl]/[NH3] = 4.8 – 1 = 3.8 pH = 14 – pOH = 10.2

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