Titration Analysis

By: Adrienne

What is it??



Titration analysis is an experiment used to find the concentration of a substance in a solution. Although it sounds complex, it is quite simple and relatively easy to perform.

::An Illustration and explanation of a Titration::    First, we fill up the “burette” or “pipetie” at 0 so that we can easily measure how much was used later. As the titration process proceeds, solution from the “pipetie” will slowly drip into the solution in the beaker or “titration dish” Eventually the experiment will reach an endpoint such as the observable colour suddenly changing completely or conductivity. This indicates that a certain amount of the solution in the “pipetie” has reacted with all the solution in the bowl and we have reached the end of the experimental part.

Digging Deeper..



The latter example was a relatively simple experiment that could be performed without much equipment. There are, however, more precise experiments with stopcocks located near the bottom of the burette. These stopcocks control the experiments so carefully that we get much more precise results.

Are we done??

 No! We are nowhere near done! This experiment must be repeated several times until 3 trials have volume results within a range of 0.2 mL of each other (the amount of solution used to react with all of the other solution, this is known as the “equivalence point”). Then, these results are averaged out to get how much of the solution was used. This makes the results much more accurate and precise. But wait! There’s still more!!

What Next?

Once the experiment part is done, we have to do the calculations. I will show an example experiment lab report that I performed and wrote on the next several slides. This will explain the calculation part of the experiment as well.

I. Purpose

 The purpose of this experiment is to show one how to perform titration and to decipher between acids and bases. Hopefully the experiment will show how to perform a titration and decipher between acids and bases.  Hypothesis: The vinegar will be an acid.  The vinegar should be an acid if it turns the solution a pinkish colour. Vinegar seems like it should be an acid because it has a more “sour” taste than “bitter.”  The topic is of interest because it shows how to perform a careful titration as well as a quick one and how to figure out the density and concentration of certain substances.

II. Equipment

           1. Eyedropper 2. Mass scale 3. Distilled water (available at any grocery store) 4. White sheet of paper (no lines) 5. Stirring rod (or a small spoon) 6. A few leaves of red cabbage (it must be red cabbage, not regular cabbage) 7. 2 beakers (If beakers are not available, one can use a short, fat glass that is transparent, and a small pot to boil water in.) 8. Graduated cylinder (Measuring cups and spoons will work, but the experiment will be much harder.) 9. Clear ammonia solution (This is sold with the cleaning supplies in most supermarkets. It must be clear. A colored solution will mess up the endpoint.) 10. Clear vinegar (Once again, colored vinegar will mess up the endpoint.) 11. Safety goggles

III. Procedure

 1. Rinse the beakers, graduated cylinders, and medicine dropper with distilled water thoroughly. If there is any contamination in them, it will mess up the endpoint of the experiment.  2. Take the small beaker (or boiling pot) and rinse it out with distilled water thoroughly.  3. Place the red cabbage leaves in the small beaker and then fill it with about 70 mL (a qualitative measurement) of distilled water. If a boiling pot is being used, three times as much water is probably needed because the water will evaporate faster in a pot than in the small beaker.  4. Boil the cabbage leaves and water in the beaker for approximately three minutes.  5. Let the beaker to cool down, and then carefully take out the leaves from the beaker. The water should be a pinkish-red color. It may be blue-green, but that shows that the water is contaminated a bit. This is the indicator. In the presence of acid (or neutral water) it is a pinkish-red color. In the presence of base, it should be a blue-green color.   6. Take the large beaker and use the graduated cylinder to add 10.0 mL (a quantitative measurement) of the clear ammonia to the beaker. (If measuring spoons are being used, this is 2 teaspoons.) 7. Add 90.0 mL of distilled water (about ½ cup) to the beaker as well. Stir the solution.

III. Procedure Continued…  8. Add a little less than half of the indicator to the beaker. The indicator should turn greenish-blue immediately. This shows the presence of a base, ammonia (NH 3 ). Now a titration will be performed to determine the concentration of the ammonia in the solution.

        9. Rinse the graduated cylinder thoroughly with distilled water. 10. Now, measure the mass of the graduated cylinder. 11. Fill the graduated cylinder up with 50.0 mL (a quantitative measurement) of vinegar and measure the new mass of the cylinder plus the vinegar. 12. FInd the mass of the vinegar by difference. 13. Use the mass and the volume (50.0 mL) to calculate the density of vinegar. (If measuring cups are being used, use ¼ of a cup. One must also calibrate the medicine dropper the way it was done in Experiment 5.1, using the ¼ cup as a good approximation of 50.0 mL.) 14. Place the beaker full of ammonia solution on the white piece of paper to make it easy to see the color of the solution. 15. Take the medicine dropper and fill it up from the graduated cylinder full of vinegar. 16. Squirt the vinegar into the ammonia solution and stir. Make sure to watch for any color change. (If measuring cups are being used, One cannot do it this way. Read the following instructions, but only do the instructions in step #27). Keep this until the color of the solution turns pinkish. It will not turn as vividly pinkish-red as the original color of the indicator, but it should turn noticeably pink. This shows that the endpoint of the titration is passed.

III. Procedure Continued…

 17. Read the number of mL of vinegar left in the graduated cylinder and, by difference, find how much vinegar that was added in the titration.        18. What was just done is referred to as a “rough titration.” more acid was added than necessary. Start over and do a careful titration. 19. Pour out the titration solution down the drain, and thoroughly rinse the beaker with distilled water. 20. Fill it up again with 10.0 mL of ammonia and 90.0 mL of distilled water. 21. Stir the solution, then add most of what is left of the indicator solution. 22. Put the beaker on the white piece of paper again and then fill the graduated cylinder up again. Make sure the graduated cylinder has precisely 50.0 mL of vinegar in it. 23. Now the careful titration can begin. Use the medicine dropper to quickly add vinegar to the beaker until it gets to within 5 mL of the amount that was added during the rough titration. 24. Now slow down the experiment. Add the vinegar drop by drop and stir thoroughly between each drop. Keep a careful watch on the color. The color should change from green to blue, to a very light blue, and then to a pinkish violet. The pinkish violet color tells you that you have reached the endpoint. The best thing to look for is the very first sign of pink color that exists after the solution has been stirred. As soon as that is spotted, the endpoint has been reached.

III. Procedure Continued…

 25. Squirt the remainder of the medicine dropper

back into the graduated cylinder.

 26.

Read the graduated cylinder and, by difference, calculate the number of mL that was added to reach the true endpoint of the titration.  27.

If measuring cups are not being used, skip this step. If measuring cups are being used, do a careful titration the first time. Add the vinegar drop by drop, still making sure to keep track of the number of drops that were added. Watch for the endpoint. When it is reached, the number of drops times the dropper calibration will show the number of mL that was needed to add to get to the endpoint.  28.

Clean up all the mess.

IV. Observations

                       -When the red cabbage leaves were boiling, the water slowly turned a pinky-purple colour -After about 3 minutes of boiling, it took about 5 minutes to cool.

-The cabbage leaves did not seem any lighter in colour although the water was more pink. -When the ammonia was poured into the beaker and the distilled water was added, no colour changes took place -The solution in the boiling pot was pinky-purple and when it was poured into the clear solution in the other beaker, it immediately turned a bright lime green -The graduated cylinder measured to be about 261 grams -The mass of the vinegar including the cylinder was about 310 grams -The mass of the vinegar was 310 grams – 261 grams = 49 grams -(49 grams/ 50.0 mL) = 0.98 g/mL which is the density of the vinegar -About 23.2 grams of Vinegar was added to the solution, squirt by squirt. -Each squirt turned part of the water pink but when it was stirred in, it just became a lighter green than before.

-Eventually the solution turned to a pinky colour. -(Starting over with careful titration) -After the cabbage solution was added to the ammonia and distilled water solution, it instantly turned a bright green. -Almost 23.2 grams of Vinegar was added and then the experiment was slowed down to one drop at a time. -As before, each squirt or drop turned part of the water pink but when it was stirred in, it just became a lighter green than before. -After about the 11 th drop, the solution turned to a pinky colour. -This is the reaction: C2H4O2 + NH3  NH4 (+) + C2H3O2 (-) -To find the concentration of the ammonia, one must start with the number of moles of acid added, (20.7 mL/1) x (0.98g/mL) = 2.0x10^1 g. -Now, mass of acid used = (mass of vinegar)x0.0500 = (2.0x10^1g) x 0.0500 = 1.0 g (because there is only 5.00% acid in vinegar. -There are 60.0g of C2H4O2 = 1 m of C2H4O2 so, (1.0g of C2H4O2/1) x (1m C2H4O2/60.0 g C2H4O2) = 0.017 moles C2H4O2. -Now (0.017 moles C2H4O2/1)x(1 mole NH3/1 mole C2H4O2) = 0.017 moles NH3 (because for every one mole of C2H4O2 there is one mole of NH3) -Finally, the concentration = # of moles/# of Liters = 0.017 moles NH3/0.010 L =

1.7 M

V. Conclusion

 The ammonia and the distilled water is the base because when the cabbage water was added to it, it instantly turned green. In the presents of an acid the cabbage water is a pinkish-red and in the presents of a base it is a blue-green color. Therefore, the ammonia and distilled water solution is a base. The ammonia, distilled water, and cabbage water together is therefore a base with a bright, lime green color. When the vinegar was added, the solution of ammonia, distilled water, and cabbage solution (the base) slowly turned lighter and lighter green until it suddenly turned to a very light pink. This is because vinegar is an acid and in the presents of an acid it is a pinkish-red color. Therefore the hypothesis is correct. So, now the titration is complete which means there was enough acid to react with all of the base (or visa versa).

 In the careful titration, everything was the same except that the amount of vinegar added was more precise. Once about 20.7mL of vinegar (the acid) was added to the base, it suddenly turned pink and therefore reached the end of the titration. Titration (as defined in the text book) is the process of slowly reacting a base of unknown concentration with an acid of known concentration (or visa versa) until just enough acid has been added to react with all of the base. This process determines the concentration of the unknown base (or acid). This is what was done. Now, since the endpoint of the titration was reached, the concentration of the base (or ammonia solution) can be found.)  This is the process to finding the concentration of ammonia. This is the reaction that happened: C2H4O2 + NH3  NH4^+ + C2H3O2^-. To find the concentration of the ammonia, one must start with the number of moles of acid added, (20.7 mL/1) x (0.98g/mL) = 2.0x10^1 g. Now, mass of acid used = (mass of vinegar)x0.0500 = (2.0x10^1g) x 0.0500 = 1.0 g (because there is only 5.00% acid in vinegar. There are 60.0g of C2H4O2 = 1 m of C2H4O2 so, (1.0g of C2H4O2/1) x (1m C2H4O2/60.0 g C2H4O2) = 0.017 moles C2H4O2. Now (0.017 moles C2H4O2/1)x(1 mole NH3/1 mole C2H4O2) = 0.017 moles NH3 (because for every one mole of C2H4O2 there is one mole of NH3) Finally, the concentration = # of moles/# of Liters = 0.017 moles NH3/0.010 L =

1.7 M

 The concentration was found by finding how many moles of acid that was used and then finding the ratio of moles of base to moles of acid and then dividing moles my liters to get the molarity. The answer is therefore 1.7 M.  For further research, one could experiment using different types of substances instead of vinegar and ammonia and figure out whether they are acids or bases.

::BIBLIOGRAPHY::

 Jenkins, Frank Dr., van Kessel, Hans, Tompkins, Dick, Lantz, Oliver Dr., Chemistry, Alberta 20-30, Nelson, © 2007.  Wile, Jay L. Exploring Creation With Chemistry, 2nd ed. Anderson: Apologia Educational Ministries, 2007.

 http://www.glossary.oilfield.slb.com/DisplayImage.cfm

?ID=409 , Diagram of titration process, Schlumberger, © 2008 Schlumberger limited.