Palestine Polytechnic University General Chemistry Lab Experiment 11 PDF

Palestine Polytechnic University General Chemistry Lab: Experiment 11: Standardization of Sodium hydroxide Solution Objectives To standardize a solution of NaOH Introduction Standard Solution: - It’s a solution with very known accurate concentration. - It used to dete...

Palestine Polytechnic University General Chemistry Lab: Experiment 11: Standardization of Sodium hydroxide Solution Objectives To standardize a solution of NaOH Introduction Standard Solution: - It’s a solution with very known accurate concentration. - It used to determine the concentration of other solution by titration process A + B  C + D Known Unknown  Standard Solution Primary Standard Secondary Standard - Highly pure - Not match primary standard properties: - Stable Example: NaOH, HCl - Not Hygrocopic NaOH: - Has High Molar mass - React with glass - Crystaline compounds - React with CO2 - Cheap, available, not toxic - Absorb Water from surrounding Example : KIO3, Na2CO3, KHP HCl: - Accurate concentration solution - Evaporate prepared - Not stable - Approximate concentration solution prepared - Need to be standaized against primary standard Titration process is needed for standardization Titration: Slowly addition of one reagent to the other until sufficient amount is added Titration requirements: 1. Reaction is stoichiometric 2. Reaction is rapid 3. No side reaction 4. There is end point for reaction Classification of titration (according to type of reaction): 1. Acid – base Titration 2. Oxidation – Reduction Titration (Redox Titration) 3. Gravimetric Titration (Precipitation Titration) Acid – Base Titration Burrete (titrant) Acid + Base  Salt + water Equivalent point (Stoichiometric point): Point at which clamp sufficient amount of reactant is added to complete the Stopchok reaction. stand End point: Point at which indicator color change Flask (Analyte) - Equivalent point come before end point - For calculation we assume that both are the same white paper Indicator: - Dye that has different color in acidic and basic media - It Change color at or extremely close to end point - It is a weak organic acid or base that color different when it protonated or ionized. (just drops of it added) - It chosen according to pH at end point Indicator examples: Strong acid + Strong base  pH = 7 Bromothymol Blue Strong Acid + Weak base  pH < 7 Methyl Orange Weak Acid + Strong base  pH > 7 Phenolphthalene – change color at pH = 8-8.2 - In acidic media colorless -In basic media pink Notes: - Titration is volumetric analysis use volumetric glass ware: (Burrete, Pipette, volumetric flask) - All reagent should be in solution - Burrete reading graduation is from up to down - Accurate reading is the bottom edge of liquid meniscus when it at the eye level  In this experiment we want to standardize NaOH solution (secondary standard) against KHP (primary standard). KHC8H4O4 + NaOH → KNaC8H4O4 + H2O Weak Acid Strong Base Known ?? Primary standard Secondary standard Pure ,Stable Not Stable Not hygroscopic Hygroscopic High molar mass Low Molar mass (204.24 g/mol) (40 g/mol) Procedure - Before using burrete 1. Wash with distilled water 2. Wash with ≅ 0.1 M NaOH solution 3. Fill with ≅ 0.1 M NaOH solution 4. Be sure that tip is filled with solution  NaOH around 0.1 M Add 0.2-0.3 g KHP (Record acurate mass) 30 ml of water (Heat to dissolve) 1-2 drops ph.ph Titrate with NaOH untill color change to faint pink (perminant color for 30s)  NaOH + KHC8H4O4 → KNaC8H4O4 + H2O From the balanced equation: MNaOH = nNaOH/VNaOH n NaOH = n KHP 𝑚𝑎𝑠𝑠 𝐾𝐻𝑃 MNaOH * V NaOH (L) = 𝑀𝑚 𝐾𝐻𝑃 𝑚 𝐾𝐻𝑃 M NaOH = 𝑀𝑚 𝐾𝐻𝑃 × 𝑉 𝑁𝑎𝑂𝐻 𝐿 - The mass of KHP is known (weighted) - The volume of NaOH will be known from the burette experimentally = vf – vi / 1000 Example: To standardize NaOH solution, a student weight 0.274 g of KHP, dissolve in water and add 2 drops of phenolphthalein. Then he titrate it with NaOH. He start titration with burrete reading of 0.2 ml and color change to faint pink when burrete volume is 13.8 ml. Calculate NaOH molar concentration. m KHP = 0.274 g Vi = 0.2 ml Vf = 13.8 ml VNaOH = (Vf – Vi)/1000 = (13.8-0.2)/1000 = 0.0136 L 𝑚 𝐾𝐻𝑃 0.274 M NaOH = = = 0.0986 mol/L 𝑀𝑚 𝐾𝐻𝑃 × 𝑉 𝑁𝑎𝑂𝐻 204.24 ×0.0136 𝐿 Thank You