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 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

REVIEW 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-] = 1,008·10-14 at 25°C pOH = -log[OH-] pH + pOH = 14 = pKW! KA KA KB KA = [CH3COOH]

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

pH of strong acids and bases 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! c(HA) = [H+] = [A-] pH = -log[H+]

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

pH of weak acids and bases Dissociation of weak acids (Ka < 10-4) HA + H2O A- + H3O+ Ka = = = c-x x x [A-][H3O+] x2 x2 [HA] 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 pKa = -logKa c >> x for diluted weak acids [H3O+] = x = (Ka c)1/2 / log pH = -log[H3O+] pH = -log [H3O+] = ½ [pKa – log(c)]

pH of weak acids and bases Dissociation of weak bases c-x x x [BH+][OH-] x2 x2 B + H2O BH+ + OH- Kb = = = [B] c-x c 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)]  

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

Salts of weak acids and strong bases [CH3COO-] [H+] CH3COONa  CH3COO- + Na+ KA = [CH3COOH] [CH3COOH] [ OH- ] CH3COO- + H2O CH3COOH + OH- KH = [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

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

Salts of weak bases and strong acids NH4Cl  NH4+ + Cl- KB = NH4+ + H2O NH3 + H3O+ KH = [H+][OH-] = Kv [NH4+] [OH-] [NH3] [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-x = c c KB c-x = concentration of a cation of a weak base at equilibrium x = concentration of products at equilibrium

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

Salts of weak acids and weak bases Anions and cations 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-x c-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 = KA2 KH = KW · KA/KB [H3O+]2 = KW · KA KB pH = 7 + ½[pKA - pKB]

BUFFERS BUFFERS = conjugated pair of 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 or mixtures of weak bases and their salts with strong acids Buffer systems in organism are of a great importance (blood, intercellular space, cells)

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

pH calculations Calculate the pH of 1 mM KOH Calculate the pH of 0.01 M formic acid (HCOOH) at 25°C, pKa = 3.8! Calculate the pH of 0.001 M NH3 at 25°C, pKb = 4.8! Calculate the pH of 0.1 M NaCN at 25°C, pKa = 9.21! Calculate the pH of 0.7 M NH4Cl at 25°C, pKb = 4.8! Calculate the pH of 5 mM ammonium lactate CH3CH(OH)COONH4 at 25°C, pKa = 3.86, pKb = 4.8 Calculate the pH of a buffer solution that contains 0.1 M CH3COONa and 0.1 M CH3COOH, pKa = 4.8! Calculate the pH of a buffer solution that contains 0.1 M NH4Cl and 1 M NH3, pKb = 4.8!

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

2. c(HCOOH) = 0.01 M, pKa = 3.8 HCOOH ↔ HCOO- + H+ 0.01-x=c x x x = conc. of products at 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

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

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

5. c(NH4Cl) = 0.7 M, pKb = 4.8 NH4Cl  NH4+ + Cl- NH3 NH4+ + 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 H2O

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 = KA2 KH = KW · KA/KB pH = 7 + ½[pKA - pKB]= 7 + ½ [3.86 – 4.8] = 6.53 (1/KH)1/2

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

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