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TITRIMETRIC ANALYSIS - Coggle Diagram

- - - - Kw= [H+] [OH-]
      - while 25°C: 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (or M²)
      - Neutral solution and 25°C: [H+] = [OH-]
  - - - pH= -log [H+]
    - - pOH= -log [OH-]
    - - pKw= -log Kw
      - pH + pOH = pKw
        
        https://padlet.com/snakmar/a261-ska3013-h-s024vn1s2otv5l628ut3
  - - - when C > 100Ka
      - HA = c ; H+ = x
    - - Same as weak acid
- - - - An ultrapure compound used as a reference material to prepare or establish a standard solution.(oxalic acid)
    - - A solution whose concentration is determined by titration against a primary standard solution.(NaOH, Its concentration is calibrated using reference materials such as oxalic acid dihydrate.)
  - - - Example: HCl + NaOH, At 25°C, equivalence-point pH = 7 under the usual assumptions.
    - - Example: HCN + NaOH or CH₃COOH + NaOH, The equivalence-point solution is generally basic, so pH > 7.
    - - General pattern in the slides: equivalence-point pH < 7.
  - - - One reactant remains in excess. Determine pH from the remaining excess reactant, or account for a buffer in a weak acid/strong base titration.
    - - Reactants have reacted in stoichiometric amounts. The pH depends on the type of acid and base.
    - - The titrant is in excess. For a strong acid/strong base example, calculate pH using the remaining excess H⁺ or OH⁻.
- - - - step 1: H3PO4 + H2O ---> H2PO4- + H3O+ Ka1 = 7.11×10^-3
      - step 2: H2PO4- + H2O---> HPO4 2- + H3O+ Ka2= 6.34×10^-8
      - step 3: HPO4 2- + H2O---> PO4 3- + H3O+ Ka3= 4.5×10^-13
  - - - Kai · Kb(n-i+1)= Kw
  - - - First buffer pair
        
        H₃A / H₂A⁻
      - Second buffer pair
        
        H₂A⁻ / HA²⁻
      - Third buffer pair
        
        HA²⁻ / A³⁻
  - - - as a base, accept proton HA- + H2O ---> H2A + OH-
      - as a acid, donate proton HA- + H2O ---> A2- + H3O+
    - - • Example: H₂A titrated with NaOH.
        • Two stepwise ionizations.
        • Two buffer regions.
        • Two equivalence points may be observed.
        • pH changes as NaOH is added.
- - - - 2MnO4- + 5C2O4 2- + 16H+ --->2Mn2+ + 10CO2 + 8H2O
        
        MnO4- + 5e- + 8H+ ---> Mn2+ H2O
        
        C2O4 2- ---> 2CO2 + 2e-
  - - - Standard electrode potential is the characteristic potential for a half-reaction under standard conditions, expressed relative to a reference electrode.
      - • Characteristic potential of a half-reaction.
        • Standard potentials are tabulated as reduction potentials.
        • E° does not change when a half-reaction is multiplied by a coefficient.
      - Nernst Equation
        
        E= E0 - (0.0592 / n) lgQ
        
        E= E0 - (RT / nF) lnQ
        
        R: Gas Constant
        
        T: thermodynamic temperature
        
        F: Faraday constant
        
        \(E\) \= electrode potential under the given conditions
        
        \(E^\circ\) \= standard electrode potential
        
        \(n\) \= number of electrons transferred in the balanced half-reaction
        
        \(Q\) \= reaction quotient, written for the half-reaction as written
        
        Ce4+ + e- ---> Ce3+
        
        E= E0 (Ce4+ / Ce3+) - (0.0592 / 1) lg(Ce3+ / Ce4+)
        
        • E depends on the concentrations of species in the redox couple.
        • Use the balanced half-reaction to identify n.
        • E° is written as a reduction potential.
        • At 25°C, use the 0.0592/n form with lg.
  - - - The redox equilibrium constant describes the relative amounts of products and reactants at equilibrium.
      - • Describes the equilibrium position of a redox reaction.
        • Related to the standard cell potential.
        • A larger Keq indicates a stronger preference for products at equilibrium.
      - lgKeq= nE0cell / 0.0592
        
        E0cell= E0 cathode - E0 anode
    - - • Both analyte and titrant react in stoichiometric amounts.
        • Determine Eeq using the relevant redox couples.
        • The calculation depends on the balanced reaction and electron numbers.
        • Eeq is used to construct the redox titration curve.
      - Eeq= (n1E01 + n2E02) / (n1 + n2)
  - - - Before equivalence
        
        Analyte is in excess
        
        How Esys is determined: From the analyte redox couple
      - At equivalence
        
        Analyte and titrant react in stoichiometric amounts
        
        From Eeq
      - After equivalence
        
        Titrant is in excess
        
        From the titrant redox couple
    - - • Measure system potential during titration.
        • Use an appropriate indicating electrode.
        • Plot Esys against titrant volume.
        • Identify the equivalence region from the curve.
- - - - Donates one electron pair to a metal ion through one donor site.(NH3)
    - - Coordinates through two donor sites.(glycine, ethylenediamine)
    - - Coordinates through multiple donor sites.(EDTA)
    - - A chelate is a complex formed when a multidentate ligand binds to a metal ion through multiple donor sites, usually creating one or more rings.
  - - - Mn+ + Y4- ---> MY(n-4)
    - - • Binds metal ions through multiple donor sites.
        • Forms stable metal complexes.
        • Often reacts in a single overall stoichiometric step.
        • Useful for quantitative determination of metal ions.
  - - - The formation constant,Kf, describes the equilibrium formation of a complex from a metal ion and a ligand. A larger Kf generally indicates a more stable complex under the specified conditions.
      - eg: Ca2+ + Y4- ---> CaY2-
        
        Kf= [CaY2-] / [Ca2+] [Y4-]
      - Effect of pH on EDTA
        
        • EDTA exists in different protonation forms.
        • Low pH favours protonated EDTA forms.
        • Higher pH increases the fraction of Y⁴⁻.
        • EDTA titrations commonly use buffered solutions near or above pH 10.
  - - - Pre-equivalence
        
        Metal ion remains in excess
        
        Calculate free metal concentration from the excess metal
      - At equivalence
        
        Metal and EDTA have reacted in stoichiometric amounts
        
        Consider complex dissociation and formation equilibrium
      - Post-equivalence
        
        EDTA is in excess
        
        Consider excess EDTA and complex equilibrium
- - - - H⁺ \= Lewis acid, accepts an electron pair.
        
        NH₃ \= Lewis base, donates a lone pair of electrons.
        
        A coordinate covalent bond forms between the electron-pair donor and acceptor.
- - - - CH3COONa == Na+ + CH3COO- CH3COO- + H2O ---> CH3COOH + OH-
        
        Hydrolysis Constant (Kh)
        
        Kh = Kb
    - - NH4Cl == NH4+ + Cl- NH4+ + H2O ---> NH3 + H3O+
        
        Hydrolysis Constant (Kh)
        
        Kh = Ka
  - - - Types of Buffer Solutions
        
        Acidic Buffer: Weak acid + its conjugate base (salt) CH3COOH / CH3COO-
        
        Basic Buffer: Weak base + its conjugate acid (salt) NH3 / NH4+
  - - - Henderson-Hasselbalch equation
        
        pH= pKa + log(conjugate base/weak acid)
- - - - HCN + NaOH = NaCN + H2O
        
        pH= pKa + lg(CN- / HCN)
      - • Calculate initial moles and apply the reaction stoichiometry.
        • Identify limiting and excess reagents.
        • Strong acid/base excess may determine the resulting pH.
        • Weak acid/base systems may require equilibrium calculations.
        • Equivalence point does not always mean pH = 7.