DETERMINATION OF AN UNKNOWN DIPROTIC ACID THROUGH VOLUMETRIC ANALYSIS Laboratory 10 DETERMINATION OF AN UNKNOWN DIPROTIC ACID THROUGH VOLUMETRIC ANALYSIS

Objectives Understand the techniques and equipment associated with titrations Apply stoichiometric principles for molarity and molar mass determinations Reinforce the importance of significant figures in measurement and calculations

Titration Volumetric Analysis 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 measured

Titration 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.

Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 liter solution

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/L

Dilution Dilution 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 solution

A stock solution of 1. 00M of NaCl is available A stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a 100.0 mL of 0.750M? M1V1= M2V2 1.00 M X V1 = 0.750 M X 100.00 mL V1 = 75 mL of NaCl

1. Preparation of a 0.25 M NaOH solution 2. 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 a standard solution. Potassium hydrogen phthalate (KHP) 3. Determination of the molar mass of the unknown acid

Assigned Reflection Questions Questions 1 and 2