Techniques of Volumetric Analysis SMKN 13 Bandung
Objectives After this session, the students are expected to be able : To explain analysis of volumetric To explain reactant for analysis of volumetric To explain steps of techniques of volumetric analysis To explain calculation in volumetric analysis Teknologi dan Rekayasa
Introduction Volumetric analysis is the process for determining the amount in the solution Can be achieved by accurately measuring volumes and concentration Teknologi dan Rekayasa
Standard Solution Standard solution is one in which the concentration of all solute species is accurately known Standard solution can be prepared in two ways: 1. Directly from a primary standard 2. By standardization Teknologi dan Rekayasa
Primary standard substance Criteria for primary standard : 1. It must be obtainable in a high state of purity and have a known formula 2. It should be easily stored and without reacting with the atmosphere Teknologi dan Rekayasa
Primary standard substance 3. It should be readily available and relatively inexpensive 4. It should have a reasonably high formula weight to minimize errors in weighing Teknologi dan Rekayasa
Primary standard substance 5. For example : sodium oxalate (Na2C2O4) sodium carbonate (Na2CO3) sodium tetra borate (Na2B4O7.10 H2O) Teknologi dan Rekayasa
Secondary standard substance Secondary standard substances are of a lower degree of purity than the primary ones The concentration of the active agent of a secondary standard substance is found by comparison against a primary standard or by the use of a suitable analytical method of known accuracy Teknologi dan Rekayasa
Secondary standard substance Secondary substances are frequently used in practice; they must be kept under condition such that their composition does not change For example : KMnO4 ; HCl ; H2SO4 ; HNO3; NaOH Teknologi dan Rekayasa
Titration A titration is the process in which a solution of one reactant (A) is added from a burette to a fixed volume, measured by pipette, of a solution of the other reactant (B) Teknologi dan Rekayasa
End Point and Equivalence Point Definition End point : the point at which the colour of indicator is change Equivalence point : the point where chemically equivalent amount of each reactant, according to the equation for the reaction are present Teknologi dan Rekayasa
Steps of techniques volumetric analysis Weighing sample Diluting sample into volumetric flask Transfer by volumetric pipette from volumetric flask into Erlenmeyer flask Add a few drops of indicator and then titrated from burette until the colour change Do the titration until the volume required is constant (the difference between two titration not more than 0.02 mL) Teknologi dan Rekayasa
Steps of techniques volumetric analysis 5. Do the titration until the volume required is constant (the difference between two titration not more than 0.02 mL) 6. Calculate the data (as concentration) and report it Teknologi dan Rekayasa
Standarization of HCl Using * primary standard : sodium carbonate (Na2 CO3) * Indicator : methyl red Teknologi dan Rekayasa
Standarization of HCl A solution of approximately 0.1 M HCl was standardized using sodium carbonate 1.325 g of Na2 CO3 was dissolved in a little water in a 250 mL of volumetric flask and the solution made up to the mark with water After thorough mixing, 25 mL of the Na2 CO3 solution was pipetted into a 250 mL of Erlenmeyer flask, add several drops of methyl red Teknologi dan Rekayasa
Standarization of HCl The HCl solution was slowly added from a burette until the colour change from yellow to reddish-orange The volume was 23.45 mL. What was the concentration of the HCl solution? Teknologi dan Rekayasa
The equation for the reaction is 2 HCl (aq) + Na2CO3 (aq) 2 NaCl (aq) +CO2 (g) +H2O(l) The relative formula mass of Na2 CO3 = 106 Volume of Na2CO3 solution = 250 mL n(Na2CO3) = mol = 0.0125 mol In 25 mL of Na2CO3 solution, n(Na2CO3) = x 0.0125 mol = 0.00125 mol Teknologi dan Rekayasa
Standarization of HCl From equation representing the reaction of HCl and Na2CO3, n(HCl) = 2 x n(Na2CO3) = 2 x 0.00125 mol = 0.0025 mol Using , it can be seen that c(HCl) = = 0.1066 M The concentration of the HCl solution is 0.1066 M Teknologi dan Rekayasa