Titration and Dilution Lab PDF

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Lab done 4/2020. I hope this helps....

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CHM 121 - D003 [18837] 4/13/20 Dilution and Titration Purpose: Today’s lab experiment will be spent reviewing the lab techniques of dilution and titration. Students will first learn how to prepare solutions by means of dilution using volumetric pipets and a volumetric flask. They will proceed to titrate the diluted solution in order to find the concentration of an unknown acid. Fundamental principle or theory behind the experiment: The ultimate goal of the experiment is to be able to find the concentration (molarity) of a concentrated stock HCl solution. This measurement can be found upon the completion of titration, yet it is important to note that dilution must take place prior to titrating. Therefore, this lab can be separated into two parts, with the first part focusing on the dilution process. Dilution will result in the amount of solute in the diluted (final) solution being the same as the amount of solute in the original (initial) concentrated solution. Dilution is essentially reached by preparing a concentrated stock and adding more solvent. Since we now know the molarity of the initial solution and the volume of the initial and final solution, the molarity of the diluted solution can be determined using the formula Mf =(Mi * Vi) / Vf. Once dilution is complete, part 2 of the lab consists of using dilution and titration procedures to determine the concentration of an unknown HCl stock solution. First we will prepare a diluted HCl sample. We also found the molarity of NaOH solution in part 1, which will be used as a standard solution to titrate our diluted HCl solution. Once titration is complete, the concentration (molarity) of this diluted HCl and the concentrated stock HCl solution will be

calculated. We will also be learning how to use two types of pipettes, the Mohr pipette and the volumetric transfer pipette. A Mohr pipet is calibrated at each millimeter and is used to deliver any volume of sample within the capacity of the pipet. A volumetric transfer pipet is used to deliver one volume of sample as stamped on the barrel of the pipet. Method: This lab will include using a variety of materials such as unknown concentrated HCl stock solution, deionized water, a 250 mL volumetric flask, a 10 mL Mohr (measuring) pipette, a 25 mL volumetric pipette, a valve-type bulb, and a 150 mL beaker. Prior to beginning the exercise, it is advised that students familiarize themselves with how to use a valve-type rubber bulb, where the A button stands for aspire, S stands for suck, and E stands for empty. We will begin the lab with the preparation of 250 mL diluted unknown acid solution. Students will begin by measuring 20 mL of concentrated HCl stock solution (located in the fume hood) in a 25 mm x 150 mm test tube. Next add deionized water to a 250 mL volumetric flask, filling it to where it is about ⅓ full. A 10 mL Mohr pipet can be used to suspend 10 mL of concentrated HCl stock solution from the test tube previously prepared into the volumetric flask. Add more deionized water to the volumetric flask until it is half full, making sure to now swirl the mixture to hasten this solution. More deionized water can be added until the liquid level almost meets the mark, then allow for a full minute of drainage. Using a dropper, students will continue to add the deionized water until the liquid level bottom of the meniscus is reached. Firmly stopper the volumetric flask and invert it repeatedly to ensure thorough mixing. Students can now begin titrating the diluted HCl solution with standardized NaOH solution. We can start titration by taking the standardized NaOH solution (from Part 1) and using it to refill the buret, making sure to record the initial volume. Students can now add 50.00 mL of

diluted HCl solution from volumetric flask to a 125 mL Erlenmeyer flask by using a 25 mL volumetric pipet. We will now add 2 drops of phenolphthalein indicator to the Erlenmeyer flask to visually determine if the titrated solution has reached neutralization. NaOH can be used to titrate the diluted acid solution, making sure to record the final volume one the ending point is met. Finally students can both calculate the molarity of the diluted acid from titration results as well the molarity of concentrated HCl stock solution. Data Sheet: 1. Prepare 250 mL diluted acid solution: Volume of concentrated acid (Vi) ______10 mL______ Volume of diluted acid (Vf)

________250mL_________

2. Titration of diluted acid with standardized NaOH from Part 1: Titration 1 Initial volume reading of NaOH __________14.12mL__________ Final volume reading of NaOH

_________26.54mL___________

Volume of NaOH solution used

_________12.42mL___________

Volume of diluted acid solution

50.00 mL

Molarity of diluted acid (Mdl)

____2.39x10-3 M_______ ________2.39x10-3 M_________

Average Molarity of diluted acid (Mf)

3. Calculate molarity of unknown HCl stock solution: Volume of concentrated acid (Vi) (from 1) ________10mL_________ Volume of diluted acid (Vf) (from 1)

________250mL_________

Average Molarity of diluted acid (Mf) (from 2)

________2.39x10-3 M_________

Molarity of concentrated stock acid solution (Macid) _______2.3x10-2 M_______ Discussion: The first thing students had to determine was the concentration of concentration of diluted HCl (Mf). The molarity of the diluted solution can be calculated by knowing the molarity of the initial solution and the volume of the initial and final solution. Molarity is not given, therefore students must use the equation M = n/V, where n equals the number of moles of HCl in our titrated 50 mL solution. The number of moles of NaOH are 2.986x10-3 moles, which is equivalent to the amount of moles of HCl. We can come to this conclusion since a 1:1 stoichiometry of neutralization reaction occurred during titration. An error that students could have encountered was adding too much NaOH to the solution. Although the universal indicator was added to the solution, the color changes very drastically making it hard to determine the exact amount of base needed to neutralize the solution. Another error could have been mindlessly preparing the diluted acid solution by adding the water into the concentrated acid which may give off acid splash. Another important technique to keep in mind is to agitate the mixture once the titrant is added to the solution, since some can get stuck to the sides and result in lower molarity.

Conclusion: By using both dilution and titration techniques, students were able to determine the concentration of unknown solutions. Dilution must take place first in order to have an equal amount of solute in the diluted (final) solution as the amount of solute in the original (initial) concentrated solution. Then titration takes place to find the concentration of an unknown acid. Students are working with harmful chemicals that need to be handled carefully. NaOH is

corrosive, while HCl is corrosive, toxic, and a skin and respiratory irritant.

Pre-Lab 1. Calculate the molarity of a solution that contains 0.237 mol of NaOH in 2.00 L of solution: Molarity = 0.237 mol NaOH / 2.00 L = 0.119 M 2. A 250.0 mL diluted NaCl solution was prepared from 10.00 mL 1.243 M concentrated solution. Calculate the concentration of this diluted solution. Md = Vi*Mc / Vf Md = (10mL * 1.243 M) / 250.0 mL

Md = 0.050 M

3. During dilution, water is added to dilute the solution, so that the volume of the diluted solution and the molarity of the dilution solution are different from the original concentrated solution. What is not changed during this process? The amount of solute remains unchanged.

Post-Lab 1. How many grams of NiCl2∙6H2O will be used to prepare a 0.0350 M, 500.0 mL of NiCl2 solution? Molar mass = 237.69 g/mol 237.69 g * 0.5 L * 0.0350 M = 4.16 g 2. How many milliliters of 12.0 M HCl are required to prepare a 2.00 M, 250.0 mL of HCl solution?

Vf = Vi * Mi / Mf Vf = (250.0 mL * 2.00 M) / 12 M Vf= 41.7 mL 3. 15.00 mL of an unknown concentrated solution was diluted to 100.0 mL. Then 35.00 mL of this diluted acid solution was titrated with 23.24 mL of 0.3724 M standard NaOH solution to reach the ending point. What is the molarity of the original concentrated acid solution? Mi = Vf * Mf / Vi Mi = 23.24 mL * 0.3724 M / 35 mL = 0.247 M Mi = 0.247 M * 100mL / 15 = 1.65 M...