Titration Tutorial Lab Chem 113 PDF

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Description

Lab #5: Titration Tutorial The purpose of this lab was to understand and perform a titration. This refers to an experiment in which one uses the concentration of a known substance to determine the concentration of an unknown substance. When a titration occurs between an acid and base, H + ions from the acid combine with OH - ions from the base to form neutral water (H 2O), and neutralization occurs. A redox reaction is a chemical reaction in which a transfer of electrons takes place. The unknown concentration of the solution can be determined using the following equation: M1V1 = M2V2. The titrant is the solution in which the concentration is known while the solution of the unknown concentration is called the analyte. The titrant in this experiment was the strong base, sodium hydroxide (NaOH), and the analyte was the strong acid, hydrochloric acid (HCl). A colored indicator was used to determine the end point of titration. Given amounts of 0.100M NaOH and HCl (unknown concentration) were added to a burette. A procedure was individually performed for a coarse titration and fine titration. An indicator can be used to determine the end point of a redox reaction when it is not obvious. A coarse titration is not as accurate in determining the end point of a redox reaction while a fine titration is helpful in identify the volume of the titrant required to reach end point. We found that the molarity of hydrochloric acid differed between the results obtained from the coarse titration and those from the fine titration experiment.

Fine Results

Data 1.

Coarse Titration Results Volume of NaOH in burette at the start (mL) Volume of NaOH in burette at the end (mL) Volume of NaOH solution disepensed (mL) Volume of HCl solution in flask (mL) Volume of NaOH in burette at the start (mL) Volume of NaOH in burette at the end (mL) Volume of NaOH solution disepensed (mL) Volume of HCl solution in flask (mL)

50.0 40.0 12.0 10.0 50.0 38.8 11.8 10.0

molarity of the HCL solution using your coarse titration results. M1 = 0.125 M NaOH V1 = (50mL – 40mL) = 12mL V2 = (12mL – 10mL) = 2mL M1V1=M2V2 (0.125)(10) = M2(2) M2 = 0.625 M Molarity of HCL = 0.625 M 2. Calculate the molarity of the HCL solution using your fine titration results. M1 = 0.125 M NaOH V1 = (50mL – 38.8mL) = 11.2mL V2 = (11.8mL – 2.0mL) = 1.8mL M1V1=M2V2 (0.125)(11.2) = M2(1.8) M2 = 0.778 M Molarity of HCl = 0.778 M

Titration

Analysis Calculate the

3. Suppose you titrated a sample of acetic acid with a 0.125 M solution of NaOH. Given the data below, what is the concentration of the acetic acid solution? Volume of 0.125 M NaOH dispensed – 22.40mL Volume of acetic acid solution used – 15.00mL Volume of water added to the HCl solution – 15.00mL Volume of dispensed 0.125M NaOH = 22.40mL Volume of acetic acid = 15.00mL Moles of base = M x V = 0.125 x 22.40 = 2.8mmol NaOH mmol of acid = mmol of base, 2.8mmol NaOH = 2.8mmol acetic acid [Acid] = mmol/V = 2.8mmol/15.0mL = 0.1867M...