Slide 1Slide 2Slide 3Slide 4Titration VocabularySlide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Laboratory 10DETERMINATION OF AN UNKNOWN DIPROTIC ACID THROUGH VOLUMETRIC ANALYSISObjectives1. Understand the techniques and equipment associated with titrations2. Apply stoichiometric principles for molarity and molar mass determinations3. Reinforce the importance of significant figures in measurement and calculations•Titration–A procedure in which one substance (titrant) is carefully added to another (analyte) until complete reaction has occurred. •The quantity of titrant required for complete reaction tells how much analyte is present.•Volumetric Analysis –A technique in which the volume of material needed to react with the analyte is measuredTitration Vocabulary•Titrant–The substance added to the analyte in a titration (reagent solution)•Analyte–The substance being analyzed•Equivalence point–The point in a titration at which the quantity of titrant is exactly sufficient for stoichiometric reaction with the analyte.•End point–The point in a titration at which there is a sudden change in a physical property, such as indicator color, pH, conductivity, or absorbance. Used as a measure of the equivalence point.•Indicator–A compound having a physical property (usually color) that changes abruptly near the equivalence point of a chemical reaction.A concentration that expresses the moles of solute in 1 L of solutionMolarity (M) = moles of solute 1 liter solutionMolarity (M)Calculate the molarity of a 5L solution containing 126g of HNO3.Calculate the number of moles: 126g HNO3 x 1 mol = 2 mol 63 g HNO3• M= moles of solute liters of solution• M= 2 mol 5L• M= 0.4 mol/LDilutionDilution is the process of decreasing the concentration of a stock solution by adding more solvent to the solution.The equation for dilution is, M1V1= M2V2•M1= molarity of the stock solution• M2= molarity of the diluted solution• V1= volume of stock solution• V2= volume of diluted solutionA stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a 100.0 mL of 0.750M?M1V1= M2V21.00 M X V1 = 0.750 M X 100.00 mL V1 = 75 mL of NaCl1. Preparation of a 0.25 M NaOH solution2. Standardization of the 0.25 M NaOH solution-Determine the concentration of a sodium hydroxide solution to a high degree of accuracy. -This process is called standardization and the resulting solution is astandard solution. COO-COOHCOO-COO-K++ NaOHK+Na++ H2OPotassium hydrogen phthalate (KHP)3. Determination of the molar mass of the unknown acidAssigned Reflection Questions Questions 1 and