Pharmaceutical Analytical Chemistry I - Determination of Ammonium Chloride PDF

Pharmaceutical Analytical Chemistry I (CPC 101 & PC 101) Determination of ammonium salts (NH4Cl) Determination of NH4Cl sample by back (residual) titration Principle: (NH4Cl is acidic) Boiling NH4Cl + NaOH NaCl + NH3­ + H2O (sample) know...

Pharmaceutical Analytical Chemistry I (CPC 101 & PC 101) Determination of ammonium salts (NH4Cl) Determination of NH4Cl sample by back (residual) titration Principle: (NH4Cl is acidic) Boiling NH4Cl + NaOH NaCl + NH3­ + H2O (sample) known excess of st. NaOH (25 ml) It depends on boiling NH4Cl with a known excess of stand. NaOH producing ammonia which is removed by boiling. Then the remaining unreacted part of NaOH is back titrated ≠ stand. HCl using M.O. or ph.ph. indicator. NaOH + HCl NaCl + H2O Remaining standard titrant unreacted part 2 Notes: Boiling is necessary for the reaction between NH4Cl & NaOH to remove the produced ammonia and push the reaction forward. This method requires Blank determination which is a separate determination in which we repeat the same steps of the experiment but without the sample and the E.P. in this case is called Blank Reading. Blank determination is required in this experiment because when NaOH is heated and then cooled, its strength is decreased due to: 1- Interaction of NaOH with the glass of the flask. 2- Absorption of atm. CO2. 3 More explanation about blank determination 1st experiment Sample + 25 ml 1 N NaOH ¬ 25 ml of 1 N NaOH ® ¯ x ml Boil part ¯ with glass & CO2 loss due to reaction reacted remaining Cool with the unreacted part sample ¯ (x ml) ↓ titrate the remaining unreacted part Titrate (x ml) ≠ 1 N HCl of 1 N NaOH ≠ 1 N HCl ⇓ ß [E.P. = x ml] E.P. 4  2nd experiment (Blank Determination) {The same experiment but without sample} 25 ml 1 N NaOH ¬ 25 ml of 1 N NaOH ® ↓ Boil ↓ remaining unreacted part loss due to reaction with glass & CO2 Cool ↓ ↓ Titr. ≠ 1 N HCl titrate the remaining unreacted part of ⇓ 1 N NaOH ≠ 1 N HCl the volume consumed ⇓ from the titrant (1 N HCl) Blank reading is called (close to 25 mL) [Blank reading] ex: 24.8, 24.9...etc ∴ The part reacted with the sample = (Blank reading – E.P.) 5 1) Transfer 10 ml of NH4Cl sample into a clean conical flask. (by bulb pipette 10 ml) 2) Add 25 ml of 1 N NaOH. (by bulb pipette 25 ml) 3) Boil for 10 min. inserting a funnel on the neck of the flask to avoid evaporation of the solution. After boiling, ensure that no more NH3 is produced indicated by the disappearance of ammonia odor. 4) Cool and then add 10 drops ph.ph. indicator. 5) Titrate ≠ 0.1 N HCl. { Color change at E.P.: from pink to first colorless } 6  Color Change Titration of NaOH ¹ HCl Before starting titration At the End Point (Pink) (Colorless) NH4Cl + NaOH ® NaCl + NH3­ + H20 1st standard NaOH + HCl ® NaCl + H20 2nd standard The equivalence factor (F) can be calculated on NaOH (1st standard) or on HCl (2nd standard) but it is preferred to be calculated on the 1st standard. 8  Calculation of equivalence factor (F): 1 NaOH ≡ 1 NH4Cl standard sample each 1ml of 1N NaOH ≡ 1 x M.W. of NH4Cl x 1 ≡ …g NH4Cl 1 x 1000 Calculation of concentration: (Blank reading – E.P.) x F x 1000 Concn. of NH4Cl = 10 =……. g/L 9