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

Laboratory 10
DETERMINATION OF AN UNKNOWN DIPROTIC
ACID THROUGH VOLUMETRIC ANALYSIS
Objectives
1. Understand the techniques and equipment
associated with titrations
2. Apply stoichiometric principles for molarity and
molar mass determinations
3. 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 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
•
M= moles of solute
liters of solution
•
M=
•
M= 0.4 mol/L
2 mol
5L
1 mol
= 2 mol
63 g HNO3
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. 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.
COO- Na+
COOH
+ NaOH
COO- K+
+ H2O
COO- K+
Potassium hydrogen phthalate
(KHP)
3. Determination of the molar mass of the unknown
acid
Assigned Reflection Questions
Questions 1 and 2