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

2. 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

3. 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

5. 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.

6. 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.

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

8. Calculate the molarity of a 5L solution containing 126g of HNO3.
Calculate the number of moles: 126g HNO3 x mol = 2 mol
63 g HNO3
M= moles of solute
liters of solution
M= 2 mol 5L
M= 0.4 mol/L

9. 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

10. 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 mL of 0.750M?
M1V1= M2V2
1.00 M X V1 = M X mL
V1 = 75 mL of NaCl

11. 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

12. Assigned Reflection Questions
Questions 1 and 2