AnalChem_Volumetric Analysis
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AnalChem_Volumetric Analysis

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AnalChem_Volumetric Analysis AnalChem_Volumetric Analysis Presentation Transcript

  • 02-412-105 Analytical Chemistry Volumetric Analysis  Titrimetric analysis  Standard solution  Calculation
  •  Volumetric analysis is a quantitative method.  Volumetric analysis is a method in which the amount of a substance is determined by measuring the volume that it occupies and the volume of a second substance that combines with the first in known proportions.  This is also called titrimetric analysis or titration because volume is an important factor in titration. 2
  • Volumetric method  A general term for a method in quantitative chemical analysis.  The amount of a substance is determined by the measurement of the volume.  Commonly used to determine the unknown concentration of a known reactant.  Often referred to as titration, a laboratory technique in which one substance of known concentration and volume is used to react with another substance of unknown concentration. 3
  • The basic principles of volumetric analysis • The one solution to be analyzed contains an unknown amount of chemicals. • The reagent of known concentration reacts with chemical of unknown amount in the presence of an indicator to show the endpoint. • The volumes are measured by a titration which completes the reaction between reagent and solution. • The volume and concentration of reagent used in the titration give the amount of reagent in moles. • The amount of unknown chemical in the measured volume of solution is calculated by using the mole ratio of the equation. • The amount of unknown chemical in the original sample is calculated by the amount of unknown chemical in the measured volume. 4
  • Definition of terms  Volumetric Analysis – Involves the preparations, storage, and measurement of volume of chemicals for analysis  Volumetric Titrimetry – Quantitative chemical analysis which determines volume of a solution of accurately known concentration required to react quantitatively with the analyte (whose concentration to be determined). – The volume of titrant required to just completely 5 react with the analyte is the TITRE.
  •  Titration - A process in which a standard reagent is added to a solution of analyte until the reaction between the two is judged complete  Primary Standard - A reagent solution of accurately known concentration is called a standard solution.  Standardization - A process to determine the concentration of a solution of known concentration by titrating with a primary standard 6
  •  End point – The point at which the reaction is observed to be completed is the end point – The end point in volumetric method of analysis is the signal that tells the analyst to stop adding reagent and make the final reading on the burette. – Endpoint is observed with the help of indicator  Equivalent point – The point at which an equivalent or stoichiometric amount of titrant is added to the analyte based on the stoichiometric equation 7
  • Titration equipment  Burette  Pipet  Erlenmeyer flask  Cylinder  Funnel  Volumetric flask 8
  • Titration processes : reagents • Solution –The solution of unknown concentration (sample) –The solution of known concentration – this is also known as the standard solution Standard Primary Secondary Standardization 9
  • Titration processes : Setup • Capped burette to a clamp stand above a conical flask. • Filled solution (titrant) into burette. • Flowed out solution through stopcock, bubble disappeared. • Contained another solution into conical and dropped indicator. • Slowly poured out a titrant from burette into conical flask. • Swirled conical flask until colour of solution change, stopped pouring out immediately. 10
  • Titration procedure • http://www.youtube.com/watch?v=sFpFCPTDv2w 11
  • Titration Acid-base titration Precipitation Complexometric Reduction-oxidation 12
  •  Acid-base titration - Reaction between acid (strong / weak) and base (strong / weak). - Acid react with base to form neutral called neutralization - Equivalent point is the reaction of acid and base equivalent by mole ratio. - End point is the change of solution in conical flask. - Example HCl + NaOH --------> NaCl + H2O 13
  • 14
  •  Redox titration - Based on oxidation-reduction reaction - E-cell  Complexometric - Complex between neutral and ligands  Precipitation - Solubility and Ksp 15
  • Titrimetric methods Type of titration Chemical Acid-based acid or base Redox oxidizing or reducing agent Precipitation ions that to be form insoluble salt Complexometric metal ions that to be form complex Reagents alkali or acid reducing or oxidizing agent ions that to be form insoluble salt comlexing agent (ligand) Indicator pH indicator redox indicator conductivity metal indicator 16
  • Requirement For Successful Volumetric Titration Reaction must be stoichiometric, well defined reaction between titrant and analyte. Reaction should be rapid. Reaction should have no side reaction, no interference from other foreign substances. Must have some indication of end of reaction, such as color change, sudden increase in pH, zero conductivity, etc. Known relationship between endpoint and equivalence point. 17
  • What is the end point? • The end point is the point where an indicator changes color. • This is the visual sign to stop or end the titration. If the indicator is selected correctly, the change is at or very near the equivalence point! • Example - phenolphthalein
  • Acid-base indicator 19
  • Indicator used to observe the end point (at/near the equivalence point) 20
  • Titration Direct Substance is directly titrated with titrant by using simple indicator. Indirect Substance is not directly titrate but product formed has to be titrated. Back Use for the substances which are not water soluble or which are weak acids or bases. 21
  • การไทเทรตแบบโดยตรง (Direct titration) KHC8H4O4 + NaOH -----> KNaC8H4O4 + H2O จากปฏิกิริยา NaOH - Known conc. (M1) 1 โมล KHC8H4O4  1 โมล NaOH - Known volume (V1) จาก M 1V 1 = M 2V 2 Μ1 V1 Μ2  V2 KHC8H4O4 • Unknown conc. ??? (M2) • Known volume (V2) 22
  • Sometime not feasible due to: • Reaction kinetic or the reaction rate is slow. • No suitable indicator in the direct titration. • The color change is slow or delay. • The end point is far from the equivalent point. Indirect titration Back titration 23
  • Indirect titration  สาร A ไม่สามารถทา direct titration ได้ D  แต่ A ทาปฏิกิริยากับ B เป็ น • Known conc. ??? (M2) • Known volume (V2) C  และ C สามารถทาปฏิกิริยา กับ D ที่เป็ นไทแทรนต์ได้ C + D -----> Product  จากปฏิกิริยา สาร D 1 mol A + B(excess) ---> C ทาปฏิกิริยาพอดีกบ A 1 ั • Unknown conc. ??? (M2) mol • Known volume (V2) 24
  •  ต้ องการหาความเข้ มข้ นของ Cu2+ โดยการเทรต แต่ทา direct titration ไม่ได้  จึงให้ Cu2+ ทาปฏิกิริยากับ I- เกิดเป็ น I2 ดังสมการ 2Cu2+ + 4I- → 2CuI (s) + I2 (1)  I2 ที่เกิดขึ ้นไทเทรตด้ วย ไธโอซัลเฟต ดังสมการ I2+ 2S2O32-→ 2I- + S4O62(2)  จากปฏิกิริยา (2) S2O32- 2 mol  I2 1 mol  จากปฏิกิริยา (1) I2 1 mol เกิดขึ ้นจากการใช้ Cu2+ 2 mol 25
  • Back titration Back titration is generally a two-stage analytical technique:  Reactant A of unknown concentration is reacted with excess reactant B of known concentration. A + B(excess) ------> C  Titration is performed to determine the amount of reactant B in excess. B(excess) + D -------> E 26
  • Back titration • http://www.youtube.com/watch?v=YsavyASe7ho 27
  • Standard solution  Primary standard – Known concentration from preparation by weight and dissolve - High purity 99-99.99% Stability (low reactivity) Low hygroscopicity High equivalent weight (can be weighed easily) Non-toxicity Ready and cheap availability Soluble in our solvent Reacts rapidly and stoichometrically with interested analyte  Secondary standard 28
  • Some primary standards 29
  • Secondary standard  Compound that does not have a high purity.  Suitable primary standards are not always available for a given titration  Determine concentration by standardization. Titrate standard using another standard. • secondary material as titrant • always be standardized by using a primary standard • uncertainly more than primary standard solutions • Example- NaOH , KMnO4, I2 , etc. 30
  • Standardization  Required when a non-primary titrant is used – – – – Prepare titrant with approximately the desired concentration Use it to titrate a primary standard Determine the concentration of the titrant Reverse of the normal titration process!!! Titration Standardization titrant (known titrant (unknown concentration) concentration) analyte (unknown concentration) analyte (known concentration) 31
  • การคานวณปริมาตรวิเคราะห์ การคานวณโดยใช้ หน่ วยโมลาร์ (Molar) aA + bB ------> products A = titrant B = titrated a = no.mole of titrant จานวนโมลที่ทาปฏิกิริยากันพอดี b = no.mole of titrated ค่าอัตราส่วนที่ทาปฏิกิริยากัน (R) จานวนโมลของ B ที่ทาปฏิกิริยาพอดีกบไทแทรนต์ A เท่ากับ 1 โมล ั R = b a 32
  • ขันตอนการคานวณ ้ aA + bB ------> cC + dD 1. จานวนมิลลิโมลของไทแทรนต์ , A mmolA = VAMA 2. จานวนมิลลิโมลของไทเทรต, B mmolB = mmol A x R = VAMA x R = VAMA x (b/a) หรือ (VBMB)/b = (VAMA)/a 33
  • 3. คานวณนาหนักของ B ้ mg B = mmolB x MW.B = VAMA x R x MW.B 4. คานวณ % ของสาร B mg B % B = mg Sample x 100 mg B = VAMA x R x MW.B B= VAMA x R x MW.B x 100 mg sample 34
  • A 25.07 mL sample of vinegar is titrated with 37.31 mL of 0.5119 M NaOH. What is the molarity of the acetic acid in vinegar? Write reaction NaOH + CH3COOH -----> CH3COONa + H2O Solution mmol of NaOH = 0.5119 mol/L x 37.31 mL = 19.10 so mmol of CH3COOH = mmol of NaOH = 19.10 mmol mmol 19.10 Molar CH3COOH = ------------------ = --------- = 0.7618 M Volume (mL) 25.07 Mass of CH3COOH = moles x MW = 0.01910 mol x 60.00 g/mol = 1.146 g acetic acid % CH3COOH = 1.146 g x 100/25.07 g sample = 4.47% 35
  • ตัวอย่ าง KHP 410.4 mg (MW. = 204.2 g/mol) เมื่อนามาละลายน ้าแล้ วไทเทรตกับ NaOH ปรากฏว่าใช้ NaOH ไปปริมาตร 36.70 mL จงคานวณหาความเข้ มข้ น ของ NaOH NaOH + KHP -------> NaHP + H2O Ratio = b/a = 1 (VA)(MA)(R)(MW.B) = (36.70)(MA)(1)(204.2) = MA = MA = mgB 410.4 410.4 (36.70)(1)(204.2) 0.0548 M 36
  • กิจกรรม สาร NaHCO3 หนัก 0.4671 กรัม เมื่อละลายน ้าแล้ วนามาไทเทรตกับกรดเกลือ ปรากฏว่าใช้ กรดเกลือไปเท่ากับ 40.72 mL HCO3- + H+ -----> H2O + CO2 กรดเกลือ (HCl)ที่ใช้ นี ้ถูกนามาหาความเข้ มข้ นที่แน่นอนกับ 37.86 mL Na2CO3 โดยเกิดปฏิกิริยา CO32- + 2H+ -----> H2O + CO2 การไทเทรตจะใช้ HCl เท่ากับ 37.86 mL จงคานวณหาเปอร์ เซนต์ของ NaHCO3 ในสารตัวอย่าง 37
  • วิธีทา mgB ความเข้ มข้ นของ HCl = (VA)(R)(MW.B) R = 1/2 = 187.6 x 2 = 0.09350 M (37.86)(105.99) % NaHCO3 = VAMA x R x MW.B x 100 mg sample (0.09350)(40.72)(1)(84.01) x100 = 0.4671 x 1000 = 68.48 38
  • Back titration - calculation Question 1 A 150.0 mL of 0.2105 M HNO3 was added in excess to 1.3415 g calcium carbonate (CaCO3). The excess acid was back titrated with 0.1055 M NaOH. It required 75.5 mL of the base to reach the end point. Calculate the percentage (w/w) of CaCO3 in the sample. 39
  •  write reaction and balance 2HNO3 + CaCO3  Ca(NO3)2 + CO2 + H2O ….(1) HNO3 + NaOH  NaNO3 + H2O ….(2)  from equations: 1 mole HNO3 ½ mole CaCO3 1 mole HNO3  1 mole NaOH  mmol HNO3 = 0.215 M x 150 mL = 31.575  mmol NaOH = 0.1055 M x 75.5 mL = 7.965  แสดงว่า HNO3 ที่ใช้ ทาปฏิกิริยากับ NaOH = 7.965 mmol  ดังนัน HNO3 ที่ใช้ ทาปฏิกิริยากับ CaCO3 = 31.575-7.965 = 23.61mmol ้ 40
  •  แสดงว่ามี CaCO3 = ½ x 23.61 = 11.805 mmol  calculate - g CaCO3 = mol x MW. g CaCO3 = (11.805 x 10-3) x 100 = 1.1805 g Weight of CaCO3  calculate - % CaCO3 = -------------------------- x 100 Weight of sample 1.1805 % CaCO3 = ----------- x 100 = 87.99 1.3415 41
  • กิจกรรม  Take assignment 3.2 (20 min).  Take exercise in homework (download in website) and keep back on tomorrow before 1.00 p.m.  ค้ นคว้ าการวิเคราะห์โดยอาศัยการไทเทรตแบบย้ อนกลับ พร้ อมแสดงวิธีการทดลองและวิธีการคานวณ กลุมละ 1 ่ เรื่ อง (อธิบายให้ เพื่อนฟั งสัปดาหน้ า) 42