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Chemical calculations II

Chemical calculations II. Vladimíra Kvasnicová. Calculation of pH. pH = - log a(H 3 O + ) a = γ x c a = activity γ = activity coefficient c = concentration (mol /L) in diluted (mM) solutions : γ = 1  a = c pH = - log c( H 3 O + )

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Chemical calculations II

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  1. Chemical calculations II Vladimíra Kvasnicová

  2. Calculation of pH pH = - log a(H3O+) a = γ x c a = activity γ = activity coefficient c = concentration (mol /L) in diluted (mM) solutions: γ = 1  a = c pH = - log c(H3O+) c(H3O+) = [H3O+] = molar concentration

  3. Dissociation of water: H2O ↔ H+ + OH- H2O+H+ + OH-↔H3O++ OH- H2O + H2O ↔ H3O+ + OH- Kdis = [H3O+] x [OH-] [H2O]2 Kdis x [H2O]2 = [H3O+] x [OH-] Kdis x [H2O]2 = constant, because [H2O] is manifold higher than [H3O+] or [OH-] Kw = constant = ionic product of water Kw = [H3O+] x [OH-]

  4. Kw = [H3O+] x [OH-] = 10-14 pKW = pH + pOH = 14 pK = - log K pH = - log [H3O+]pOH = - log [OH-] 10-14= [H3O+] x [OH-] / log log10-14= log ([H3O+] x [OH-] ) log 10-14= log [H3O+] + log [OH-] -14= log [H3O+] + log [OH-] / x (-1) 14= - log [H3O+] - log [OH-] ↓ ↓ ↓ pKW =pH+ pOH 14 = 7 + 7 in pure water

  5. pKW = pH + pOH = 14 => water: [H3O+] = 10–7 (pH = 7) [OH-] = 10–7 (pOH = 7) simplification: [H3O+] = [H+] = c(H+) => pH = – log c(H+) pH = 0 – 14 pH 0 -------------- 7 --------------14 acidic neutral basic If [H+] decreases, [OH-] increasesKW is 10-14 If [OH-] decreases, [H+] increases(= constant !)

  6. strong acids (HA) [HA] = [H+] HA → H+ + A- pH = - log c(H+) = - log cHA strong bases (BOH)[BOH] = [OH-] BOH → B+ + OH- pOH = - log cBOH pH = 14 - pOH

  7. weak acids (HA)[HA] ≠ [H+] Kdis ≤ 10–2 HA ↔ H+ + A- Kdis = [H+] [A-][H+] = [A-] [HA] = cHAKdis = Ka[HA] Ka= [H+]2cHA KaxcHA = [H+]2/log log (KaxcHA ) = 2 x log [H+] log Ka + log cHA = 2 x log [H+] / ½ ½ log Ka + ½ log cHA = log [H+] / x (-1) -½ log Ka - ½ log cHA = - log [H+] - log Ka = pKa ½ pKa - ½ log cHA = pH =>pH = ½ pKa - ½ log cHA

  8. weak acids (HA)[HA] ≠ [H+] Kdis ≤ 10–2 HA ↔ H+ + A- pH = ½ pKa - ½ log cHA weak bases (BOH)[BOH] ≠ [OH-]Kdis = [B+] [OH-] BOH ↔ B+ + OH-[BOH] pOH = ½ pKb - ½ log cBOH => pH of basic solutions: pH + pOH = 14 pH = 14 - pOH

  9. Important equations pH = - log c(H+)pK = - log K pH + pOH = 14 ACIDS: pH = - log cHA pH = ½ pKa - ½ log cHA BASES: pOH = - log cBOH pOH = ½ pKb - ½ log cBOH pH = 14 – pOH

  10. Exercises 1) 0,1M HCl, pH = ?, [H+] = ? [10-1 M, pH =1] 2) 0,01M KOH, pH = ?, [H+] = ? [10-12 M, pH = 12] 3) 0,01M acetic acid, K = 1,8 x 10–5 , pH = ? [pK = 4,74; pH = 3,4] 4) 0,2M NH4OH; pK = 4,74; pH = ? [pOH = 2,72; pH = 11,3] 5) 0,1M lactic acid; pH = 2,4; Ka = ? [pK=3,8; Ka= 1,58 x 10-4]

  11. 6) strong acid: pH = 3 c = ? [10–3 M ] 7) strong base: pH = 11 c = ? [pOH = 3; c = 10–3 M ] 8) dilution of a weak acid: c1 = 0,1 c2 = 0,01 ? ∆ pH [∆ pH = 0,5 ] 9) dilution of a strong acid: c1 = 0,1 c2 = 0,01 ? ∆ pH [∆ pH = 1 ]

  12. BUFFERS = solutions which have the ability to absorb small additions of either a strong acid or strong base with a very little change of pH. • buffers are used to maintain stable pH • composition of buffers: „conjugated pair: acid /base“ * weak acid + it`s salt * weak base + it`s salt * 2 salts of a polyprotic acid * amphoteric compound (e.g. protein)

  13. „bicarbonate buffer“HCO3- NaHCO3 ↔ Na+ + HCO3- H2CO3H2CO3 ↔ H+ + HCO3- NaHCO3 mixed → Na+ + HCO3- H2CO3H+ + HCO3- + H2CO3 + HCl+ NaOH (H+ + Cl-) (Na+ + OH- ) Na+ + HCO3- Na+ + HCO3-H+ + H2CO3H2O + HCO3-Cl- + H2CO3Na+ +H2CO3 HCO3- + H+↔ H2CO3H+ + OH- ↔ H2O

  14. Henderson-Hasselbalch equation pH = pKa + log (cs / ca)(for acidic buffer ) pOH = pKb + log (cs / cb )(for basic buffer) pH = 14 - pOH pK = dissociation constant of the weak acid (pKa) or base (pKb) cs = actual concentration of salt ca = actual concentration of weak acid cb = actual concentration of weak base c = c´ x V c´ = concentration before mixing the components V = volume of a component (acid or base or salt)

  15. Exercises 10) 200ml of 0,5M acetic acid + 100ml of 0,5M sodium acetate => buffer; pKa = 4,76 pH = ? [pH = 4,46 ] 11) 20ml of 0,05M NH4Cl + ? ml 0,2M NH4OH => buffer of pH = 10; Kb = 1,85 x 10–5 pK = ? [pK = 4,73; 27 ml]

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