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Experiment 13 - TItration Lab.docx - Napa Valley College Labs... · Web viewExperiment 13 –...

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Experiment 13 – Titrations Name __________________ Lab Section __________________ Experiment 13 – Titrations Introduction Neutralization occurs when equal number of moles of acid and base combine to form water. In general it follows the equation HA ( aq ) +BOH ( aq ) →AB ( aq ) +HOH ( l) Where, on the reactant side, HA is an acid and BOH is a base; and on the product side, AB is an aqueous salt solution and liquid water is formed. This water is often written as HOH instead of the familiar H 2 O, as it aids in balancing the reaction. This is merely a convention. In this experiment, the molarity (concentration) of a vinegar solution will be determined using titration. Titration is the act of neutralizing and acid with a base, or vice versa. Acid solutions and a base solution are clear and colorless by nature; and the products, water and an aqueous salt, are also clear and colorless. In order to know when the reaction is done (recall the evidences of chemical reaction) it is necessary to use an indicator in the reaction. An indicator is a compound that is not part of the reaction, but signals when the reaction is complete. This experiment will use phenolphthalein as its indicator. Phenolphthalein is colorless in an acidic solution but it turns pink in a basic solution. The endpoint, where all the acid and base have neutralized one another, is a very sharp point. Merely one drop of excess base is enough to cause the indicator to turn pink and signify the end of the titration. This experiment will be done in two parts. The first part of the experiment will be to standardize a base solution against a known acid. Standardization means to determine the concentration (molarity) of the solution. You will make up an unknown base solution and use a known mass of potassium hydrogen phthalate, KHC 8 H 4 O 4 (abbreviated as KHP with a molar mass of 204.23 g mol ), to
Transcript

Experiment 13 – Titrations Name __________________Lab Section __________________

Experiment 13 – Titrations

Introduction

Neutralization occurs when equal number of moles of acid and base combine to form water. In general it follows the equation

H A (aq )+BOH (aq )→AB (aq )+HOH (l)

Where, on the reactant side, HA is an acid and BOH is a base; and on the product side, AB is an aqueous salt solution and liquid water is formed. This water is often written as HOH instead of the familiar H2O, as it aids in balancing the reaction. This is merely a convention.

In this experiment, the molarity (concentration) of a vinegar solution will be determined using titration. Titration is the act of neutralizing and acid with a base, or vice versa.

Acid solutions and a base solution are clear and colorless by nature; and the products, water and an aqueous salt, are also clear and colorless. In order to know when the reaction is done (recall the evidences of chemical reaction) it is necessary to use an indicator in the reaction. An indicator is a compound that is not part of the reaction, but signals when the reaction is complete. This experiment will use phenolphthalein as its indicator. Phenolphthalein is colorless in an acidic solution but it turns pink in a basic solution. The endpoint, where all the acid and base have neutralized one another, is a very sharp point. Merely one drop of excess base is enough to cause the indicator to turn pink and signify the end of the titration.

This experiment will be done in two parts. The first part of the experiment will be to standardize a base solution against a known acid. Standardization means to determine the concentration (molarity) of the solution. You will make up an unknown base solution and use a known mass of potassium hydrogen phthalate, KHC8H4O4 (abbreviated as KHP with a molar mass of 204.23 gmol ), to titrate the base and determine its molarity. This titration will follow the balanced

equation:

KHP (aq )+NaOH (aq )→KNaP (aq )+HOH (l)

The second part of the experiment will use the standardized base to titrate an unknown strength acetic acid, HC2H3O2, solution. This titration will follow the balanced equation:

H C2H 3O2 (aq )+NaOH (aq )→NaC2H3O2 (aq )+HOH (l)

It is important to remember that in any titration, indeed in working with pipets and burets in general, that conditioning of the glassware takes place prior to doing the experiment. Conditioning is essentially rinsing the glassware; except you rinse it with a portion of the liquid you will be using (not always water).

Experiment 13 – Titrations Name __________________Lab Section __________________

Experiment 13 – Titrations Name __________________Lab Section __________________

Figure 13.1: a) titration set up, b) initial buret reading (1.60 mL), c) beginning of titration, d) flashes of pink during titration, e) endpoint of titration, f) final buret reading (31.65 mL)

Experiment 13 – Titrations Name __________________Lab Section __________________

Examples

1) A 1.125 g sample of KHP is dissolved in 50 mL of water and titrated with NaOH. The initial buret reading is 1.10 mL and the final reading is 27.55 mL. What is the molarity of the NaOH solution?

Recall the chemical equation:

KHP (aq )+NaOH (aq )→KNaP (aq )+HOH (l)

1.125g KHP [ 1mol KHP204.23 g KHP ][1mol NaOH

1mol KHP ][ 10.02645L NaOH ]

= 0.20826 M NaOH → 0.208 MNaOH

(Note: the volume is the final volume of base minus the initial volume of base, and that is converted to liters to determine molarity.)

2) A 25.0 mL sample of HC2H3O2 is titrated with 0.310 M NaOH. If it takes 20.50 mL to reach the endpoint, what is the concentration of HC2H3O2?

H C2H 3O2 (aq )+NaOH (aq )→NaC2H3O2 (aq )+HOH (l)

0.310mol NaOHLNaOH [0.02050 LNaOH

1 ][ 1mol HC 2H 3O2

1mol NaOH ][ 10.0250L HC2H3O2 ]

= 0.2542 M H C2H3O2 → 0.254 M H C2H3O2

If the density of HC2H3O2 is 1.01 gmL and the molar mass is 60.06

gmol , what is the %HC2H3O2

of the solution? (remember your definitions of molarity and density)

0.254mol H C2H 3O2 solute1000mLHC 2H3O2 solution [ 60.06gH C2H3O2 solute

1mol HC2H 3O2 solute ][ 1mLHC2H 3O2 solution1.01gH C2H3O2 solute ] x 100%

= 1.51042 % H C2H3O2 → 1.51 % H C2H3O2

Experiment 13 – Titrations Name __________________Lab Section __________________

Procedure

Standardization of the Base

Prepare your base solution by adding approximately 25 mL of 6 M NaOH to a Florence Flask. Use a large beaker to measure between 500 and 600 mL of DI water and add this to the Florence flask to dilute the base. Be sure to keep the base stoppered when not in use.

Place an Erlenmeyer flask on the balance and tare (zero) the balance. Weigh between 0.7 and 1 g of KHP into the Erlenmeyer flask. Dissolve the KHP in approximately 25 mL of DI water. If crystals adhere to the side of the flask, rinse them down into the solution using a squirt bottle.

Add 2-3 drops of phenolphthalein to the KHP solution.

Condition a buret with your base solution. Fill the buret until the meniscus reads between 0 and 3 mL. Record this initial volume of base. Place the KHP solution on top of a white sheet of paper under the buret. This will aid in the detection of the color change of phenolphthalein.

Titrate the KHP solution with your NaOH solution until the solution turns a faint pink. Record the final volume of base.

Calculate the molarity of your base. Note: the molar mass of KHP is 204.23 gmol .

Repeat these steps until three molarities are all within 0.005 M of one another.

Titration of Acetic Acid

Obtain 50 mL of an unknown strength acetic acid solution in a clean, dry, 50 mL beaker from your instructor and record its number.

Condition a 10.0 mL pipet with the unknown solution. Transfer 10.0 mL of the solution into an Erlenmeyer flask. Add 20-25 mL of deionized water to the flask.

Add 2-3 drops of phenolphthalein to the KHP solution.

Condition the buret with your NaOH solution. Fill the buret until the meniscus reads between 0 and 3 mL. Record this initial volume of base. Place the acetic acid solution on top of a white sheet of paper under the buret. This will aid in the detection of the color change of phenolphthalein.

Titrate the acetic acid solution with your NaOH solution until the solution turns a faint pink. Record the final volume of base.

Experiment 13 – Titrations Name __________________Lab Section __________________

Repeat these steps two more times for reproducibility. Prelaboratory Questions

1) Please define the following:

Indicator

Titration

Neutralization

Endpoint

Conditioning

Standardization

2) A 50.0 mL sample of phosphoric acid, H3PO4, is titrated with 42.50 mL of 0.250 M NaOH. What is the molarity of the acid?

3) Assuming the density of the phosphoric acid is 1.07gmL , what is the % H3PO4 of the solution?

Experiment 13 – Titrations Name __________________Lab Section __________________

Data Table

Standardization of NaOH

Mass of KHP

Final buret reading

Initial buret reading

Volume of NaOH

Trial 1

_________

_________

_________

_________

Trial 2

_________

_________

_________

_________

Trial 3

_________

_________

_________

_________

Trial 4

_________

_________

_________

_________

Trial 5

_________

_________

_________

_________

Calculate the molarity of the NaOH for each trial:

Calculated Molarity

Trial 1

________

Trial 2

________

Trial 3

________

Trial 4

________

Trial 5

________

Average Molarity _______M

Experiment 20 – Titrations Name __________________Lab Section __________________

Data Table

Standardization of Acetic Acid

Volume of Acetic Acid

Final buret reading

Initial buret reading

Volume of NaOH

Trial 1

_________

_________

_________

_________

Trial 2

_________

_________

_________

_________

Trial 3

_________

_________

_________

_________

Calculate the molarity of the acetic acid for each trial, the average molarity, and the average %

acetic acid (the density of acetic acid is: 1.01 gmL ):

Calculated Molarity

Trial 1

________

Trial 2

________

Trial 3

________

Experiment 20 – Titrations Name __________________Lab Section __________________

Average Molarity ________ Average % Acetic Acid ________Postlaboratory Questions

1) An extra strength antacid tablet contains 750 mg of active ingredient, CaCO3. If it takes 22.25 mL of HCl to neutralize the tablet, how strong is the acid?

2HCl (aq )+CaCO3 (aq )→CaCl2 (aq )+HOH (l )+CO2(g)

2) A beaker containing 25.0 mL of 0.360 M H2SO4 spills on the counter. How much baking soda, NaHCO3 will be needed to neutralize the acid?

H 2SO4 (aq )+2NaHC O3 (aq )→Na2C O3 (aq )+2HOH ( l)

3) If a 35.75 mL sample of 0.175 M HCl solution is neutralized with 0.125 M Mg(OH)2, how much Mg(OH)2 will be needed

HCl (aq )+Mg ¿


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