1 / 18

CHAPTER 6 TITRIMETRIC METHODS OF ANALYSIS

CHAPTER 6 TITRIMETRIC METHODS OF ANALYSIS.

leda
Download Presentation

CHAPTER 6 TITRIMETRIC METHODS OF ANALYSIS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 6 TITRIMETRIC METHODS OF ANALYSIS

  2. Titrimetric methods include a large and powerful group of quantitative procedures based on measuring the amount of a reagent of known concentration (standard solution-titrant) that is consumed by the analyte. Titrimetry is a term which includes a group of analytical methods based on determining the quantity of a reagent of known concentration that is required to react completely with the analyte.

  3. Basic principle aA +bB → Products A: Titrant (standard solution) B: Analyte (unknown ) a and b are number of moles of each

  4. Requirments inTitration 1- stoichiometric reaction 2- rapid rate reaction 3- quantitative reaction (99.9%complete at stoichiometry) 4-have a defined end or equivalence point

  5. Equivalence and end Points • Equivalence point: • Related to the amounts of reactants consumed. • End point • Related to the physical sign that associate with the condition of chemical equivalence. • Ideally, the end point and equivalent one coincide. (variation may be to color change of indicator)

  6. Standard solutions • Standard Solution • - A primary standard • - A secondary standard • Standardization • A process in which concentration of a volumetric solution is determined by titrating it with a known mass of a primary standard.

  7. Solution of NaOH Solution of NaOH Solution of HCl 5 mL Titration: This is performed by adding a standard solution from a buret or other liquid- dispensing device to a solution of the analyte until the point at which the reaction is believed to be complete.

  8. PRIMARY STANDARD • HIGH PURITY • ATMOSPHERIC STABILITY • INDEPENDENT OF HUMIDITY • MODEST COST • LARGE MOLAR MASS • NON HGYROSCOPIC

  9. Types of titrimetry 1- Precipitation: Ag+ + Cl-→ AgCl (s) 2- Acid Base titration: OH- + HA → A- + H2O 3- Complex formation EDTA2- + Ca2+ → EDTA Ca.xH2O 4- Oxidation Reduction MnO4- + 5Fe2+ + 8H+ → 5Fe3+ + Mn2++4H2O

  10. Calculations with Molarity The mole: is the formula weight of a substance expressed in grams Number of moles = weight in Grams/formula weight Number of m.moles = weight in milligrams/formula weight Molarity= Moles/Liter or mmoles/ml

  11. Moles= Volume (L) . (M) Weight (g) = Volume (L) . (M). form weight(g) m.Moles= Volume (ml) . (M) Weight (mg) = Volume (ml) . (M). form weight(g)

  12. Calculating the results of Titration Needs: the volume and molarity of the titrant aA + bB → Products (Titrant)(substance titrated) No. of m.moles of A = Volume of A titrated (ml) . Molarity of A (M) Or mmolesA = mlA . (MA) No. of m.moles of B is obtained by multiplying the No. of mmoles of A by the combining ratio (b/a)

  13. mmolesB = mmolesA . (b/a) = mlA . (MA). (b/a)

  14. If the weight of substance is needed: from no. of mmoles Wt (mg) = mmoles . Form wt (g) mgB = (mlA) . (MA). (b/a) . (form WtB) If % of B is needed in a sample = (WtB/Wtsample).100 The same is B% =(mgB/mgS).100 B% = (mlA) . (MA). (b/a) . (form WtB)(100)/mgS

  15. Calculating the molarity of a solution from a standardizing titration This means that from the weight of primary standard substance (B )when dissolved and titrated with other solution, the molarity of the solution (A) can be calculated as follows: aA + bB → Products (SolutionA) (primary standard) (mlA) (MA)/a = (mlB) (MB)/b (mlB) (MB) = mmolesB = mgB /Form WtB (MA)/a = mgB/ (mlA) (b/a) (Form WtB)

  16. Exercises Excercize1 How many ml of 0.25 M NaOH will react with 10.0 ml of 0.10 M H2SO4.

More Related