Your TA will randomly assign you one of the commercial brands of antacid tablets and you will analyze two samples of it. Before starting, calculate the acid/base ratio of the tablet that you have been assigned that you will neutralize. Start thinking how this may be an error in your final calculation.

1. Weigh one whole tablet using the analytical balance. Record the mass of the intact tablet.

2. Break or cut the tablet to obtain pieces roughly the size indicated in the table below (e.g.. ½, ⅓, etc.) and weigh two of these pieces separately on sheets of weighing paper. (Don't forget to subtract or tare the weight of the paper.) After weighing, fold over the paper and gently crush the sample fragments using the round portion of your padlock, so it dissolves faster.

3. Use a transfer pipet to dispense 25.00 mL of standardized HCl into a clean 125 mL Erlenmeyer flask and add about 20 mL of deionized water. This solution will be used to dissolve the fraction you weighed of the antacid tablet.

4. Transfer one of the weighted crushed samples (without any spilling) into the flask containing the HCl solution. Repeat for the other weighed fraction in another flask and label each flask (1 and 2) to keep track of which sample and which portion is being titrated.

5. Warm gently to dissolve the sample and then boil solution for about a minute. Some components of the samples may remain undissolved, but these will not cause problems.

6. Let the flaks cool down sitting on the bench for couple minutes, and then cool the outside of the flask with tap water.

7. When room temperature is reached add a few drops of methyl purple indicator solution. The flask should now be purple in color. (If it is green instead of purple, you have used too large a fragment and have neutralized all of the HCl. (In this case add 5mL of HCl and observe the color change. If needed add measured amounts of HCl little by little until solution is purple. Remember to record the amount of HCl added and add to the first 25mL of HCl.)

8. Put a clean magnetic stir bar inside flask and stir solution on stir plate. (It is important that you read Titration Tips in Supporting Information below before you titrate, or you may have to repeat titration several times.)

9. Titrate with your standardized NaOH solution until you reach the endpoint, a change in color from purple to green (or from purple to clear). Some samples may not give color changes as sharp as for the HCl standardizations; for these use your best judgment to estimate the endpoint. Endpoints will generally be sharper for quick titrations than for slow ones.

10. Repeat the above procedure with a new sample of the same antacid. Enter your data in an excel spreadsheet in the workrooms, check the web page later on the week and make a final comment on the overall cost to neutralize one mole of HCl for various brands. THIS IS PART OF YOUR REPORT AND IF DATA IS NOT FOUND ON THE WEB YOU SHOULD CONTACT YOUR TA. (No excuses)

 *Active ingredient in all of these antacids is CaCO3CaCO3 size 12{ ital "CaCO" rSub { size 8{3} } } {}.

Molarity is the most commonly used concentration term when one is interested in the amount of materials involved in a chemical reaction in solution. Molarity (M) is defined as the number of moles of solute per liter of solution.

M = moles(solute)Liters(solution)moles(solute)Liters(solution) size 12{ { {# ital "moles" \( ital "solute" \) } over { ital "Liters" \( ital "solution" \) } } } {} (1)

The number of moles is calculated by dividing the mass of the sample in grams by the gram formula weight (GFW or molar mass). One GFW is the same as one mole.

Number moles = grams(solute)GFWgrams(solute)GFW size 12{ { { ital "grams" \( ital "solute" \) } over { ital "GFW"} } } {} (2)

For example in a 0.150 M HNO3HNO3 size 12{ ital "HNO" rSub { size 8{3} } } {} solution, there are 0.150 moles of HNO3HNO3 size 12{ ital "HNO" rSub { size 8{3} } } {} in one liter of this solution. The following factors may then be used in chemical calculations: 

0.150molHNO31L(solution)0.150molHNO31L(solution) size 12{ { {0 "." "150" ital "molHNO" rSub { size 8{3} } } over {1L \( ital "solution" \) } } } {} or 1L(solution)0.150molHNO31L(solution)0.150molHNO3 size 12{ { {1L \( ital "solution" \) } over {0 "." "150" ital "molHNO" rSub { size 8{3} } } } } {}

In a chemical reaction that takes place in solution, the volume and the molarity of one reactant and the molarity of the second reactant can be used together with the stoichiometry of the equation to find the volume of the second reactant needed to react completely with the first reactant.

Titration is a process in which a solution of one reagent, usually the base, is added to an accurately measured volume of another solution, usually the acid, until the reaction is complete. The concentration of one of the reagents is known. From the known concentration and the measured volumes, the concentration of the second solution can be calculated.

In acid-base reactions the end of the reaction or equivalence point is detected by adding a compound that undergoes a color change as it changes from its acid form to its basic form. This compound is called an indicator. An indicator is an organic dye that changes color at a characteristic H+H+ size 12{H rSup { size 8{+{}} } } {} ion concentration. A dye can be an indicator if it has an intense color that changes when it gains or loses H+H+ size 12{H rSup { size 8{+{}} } } {} ions.

HIn+OH−→In−+H2OHIn+OH−→In−+H2O size 12{ ital "HIn"+ ital "OH" rSup { size 8{ - {}} } rightarrow ital "In" rSup { size 8{ - {}} } +H rSub { size 8{2} } O} {} (3)

The change of an indicator dye from Color A to Color B depends upon the concentration of H+H+ size 12{H rSup { size 8{+{}} } } {} (or OH−OH− size 12{ ital "OH" rSup { size 8{ - {}} } } {}) ions. Care must be taken in selecting an indicator to be sure that the color change (endpoint) occurs at the H+H+ size 12{H rSup { size 8{+{}} } } {} ion concentration that corresponds to the equivalence point. Phenolphthalein, methyl red, methyl orange, and litmus are examples of indicators.

A primary standard is usually a solid reactant that:

(1) is available in a high purity form

(2) does not change chemically when stored or exposed to air

(3) has a high formula weight to minimize errors in weighing

(4) is soluble in the solvent being used.

Sodium carbonate, Na2CO3Na2CO3 size 12{ ital "Na" rSub { size 8{2} } ital "CO" rSub { size 8{3} } } {}, is commonly used as a primary standard base for standardizing acids, while potassium acid phthalate (KHP), KHC8H4O4KHC8H4O4 size 12{ ital "KHC" rSub { size 8{8} } H rSub { size 8{4} } O rSub { size 8{4} } } {}, and oxalic acid dihydrate, H2C2O4⋅2H2OH2C2O4⋅2H2O size 12{H rSub { size 8{2} } C rSub { size 8{2} } O rSub { size 8{4} } cdot 2H rSub { size 8{2} } O} {}, are primary standard acids used for standardizing bases.

Na2CO3(s)+2HCl→H2O+CO2+2NaClNa2CO3(s)+2HCl→H2O+CO2+2NaCl size 12{ ital "Na" rSub { size 8{2} } ital "CO" rSub { size 8{3} } \( s \) +2 ital "HCl" rightarrow H rSub { size 8{2} } O+ ital "CO" rSub { size 8{2} } +2 ital "NaCl"} {} (4)

KHC8H4O4(s)+NaOH→KNaC8H4O4+H2OKHC8H4O4(s)+NaOH→KNaC8H4O4+H2O size 12{ ital "KHC" rSub { size 8{8} } H rSub { size 8{4} } O rSub { size 8{4} } \( s \) + ital "NaOH" rightarrow ital "KNaC" rSub { size 8{8} } H rSub { size 8{4} } O rSub { size 8{4} } +H rSub { size 8{2} } O} {} (5)

(COOH)2⋅2H2O(s)+2NaOH→Na2(COO)2+4H2O(COOH)2⋅2H2O(s)+2NaOH→Na2(COO)2+4H2O size 12{ \( ital "COOH" \) rSub { size 8{2} } cdot 2H rSub { size 8{2} } O \( s \) +2 ital "NaOH" rightarrow ital "Na" rSub { size 8{2} } \( ital "COO" \) rSub { size 8{2} } +4H rSub { size 8{2} } O} {} (6)

The dry solid is carefully weighed on an analytical balance and then diluted in a volumetric flask to give a known molarity. The molarity of an acid or base is determined by titrating a measured volume of the acid or base with the primary standard. Then these acid or base solutions can in turn be used to determine the molarity of another acid or base (this is the case with the standardized solutions of NaOH and HCl used in this experiment).

In the example below a known amount of KHP will be titrated with a solution of NaOH to determine the NaOH solution concentration.

EXAMPLE: If 0.8168 g of KHP requires 39.35 mL of NaOH to reach the endpoint in a titration, what is the molarity of the NaOH? (1 mol KHP = 204.2 g)

1. Clean the buret until it drains smoothly.

2. Make sure the stopcock doesn't leak. The level should hold for ~3 minutes.

3. Remove air bubbles from buret tip before beginning. Rapid spurts usually work.

4. Rinse the buret two or three times with 5 mL portions of titrant (the solution you will use to titrate). Hold it horizontal and rotate to rinse.

5. Don't forget to record the initial reading. It does not need to be 0.00 mL.

6. Use a reading card to find the bottom of the meniscus.

7. Always read and record buret volumes to 0.01 mL.

8. Remember that the buret scale reads down, not up.

9. Put white paper under the flask to see color change easily.

10. Swirl the flask with one hand while turning the stopcock with the other or use stir bar and stir plate.

11. Add titrant slowly near the endpoint. Color that dissipates can be seen when getting close to endpoint.

12. Drops can be "split" by quickly turning the stopcock through the open position.

13. Near the endpoint use deionized water from wash bottle to rinse the flask walls from any splashed drops (adding water does not affect number of moles you have).

14. Don't drain buret below 25 mL (You won’t be able to determine final volume accurately). If necessary, read the buret level, refill, read the level, and continue. 

Amount neutralized by antacid 25ml of HCLThe amount of HCl that was neutralized by the fragment of the whole tablet, was only neutralized in part, this is where the titration comes in, the solution was still acidic, even after the dissolved antacid fragment, so the volume of NaOH used to finally cause the change in color is equal to the excess of HCl that was in the solution of HCl + water + antacid. So the number of moles of NaOH that was used is equal to the number of moles of HCl that was neutralized by the NaOH. So we DO use the formula in the procedure, so in the report form, where it had asked for the number of moles of HCl neutralized by the antacid:

So then, this would be the 2.5×10−32.5×10−3 size 12{2 "." 5 times "10" rSup { size 8{ - 3} } } {} mol HCl – the excess which is calculated by converting the volume of NaOH used into moles, this number of moles then equals the number of moles oh HCl.

Volume NaOH titratedvolume excess HCl = volume titration of NaOH